Scilab Textbook Companion for
Programming In Ansi C
by E. Balagurusamy1
Created by
Devender Thakur
computer science
Computer Engineering
Shoolini University, Solan
College Teacher
Mr. Amit Nayyar
Cross-Checked by
October 7, 2013
1 Funded
by a grant from the National Mission on Education through ICT,
http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilab
codes written in it can be downloaded from the ”Textbook Companion Project”
section at the website http://scilab.in
Book Description
Title: Programming In Ansi C
Author: E. Balagurusamy
Publisher: Tata McGraw - Hill Education, New Delhi
Edition: 4
Year: 2008
ISBN: 978-0-07-064822-7
1
Scilab numbering policy used in this document and the relation to the
above book.
Exa Example (Solved example)
Eqn Equation (Particular equation of the above book)
AP Appendix to Example(Scilab Code that is an Appednix to a particular
Example of the above book)
For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means
a scilab code whose theory is explained in Section 2.3 of the book.
2
Contents
List of Scilab Codes
4
1 Overview of C
7
2 Constants Variables and Data Types
11
3 Operators and Expressions
16
4 Managing Input and Output Operations
22
5 Decision Making and Branching
31
6 Decision Making and Looping
40
7 Arrays
51
8 Character Arrays and Strings
65
9 User Defined Functions
76
10 Structures and Unions
87
12 File Management in C
95
13 Dynamic Memory Allocation and linked Lists
3
104
List of Scilab Codes
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
1.3
1.4
1.5
1.6
1.7
2.1
2.1cs
2.2
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
2.2cs
2.3
2.4
3.1
3.1cs
3.2
3.2cs
3.3
3.4
3.5
3.6
4.1
4.1cs
4.2
Exa
Exa
Exa
Exa
4.2cs
4.3
4.4
4.5
Printing a mesage . . . . . . . . . . . . . . . . . . . .
Adding two numbers . . . . . . . . . . . . . . . . . . .
Interest Calculation . . . . . . . . . . . . . . . . . . .
Use of subroutines . . . . . . . . . . . . . . . . . . . .
Use of math functions . . . . . . . . . . . . . . . . . .
Representation of integer constants on a 16 bit computer
Case study 1 avg of numbers . . . . . . . . . . . . . .
typical declarations assignments and values stored in
various types of variables . . . . . . . . . . . . . . . .
Case study temprature in Farenheit and Celsius . . . .
use of scanf function . . . . . . . . . . . . . . . . . . .
Interest calculation using scanf . . . . . . . . . . . . .
Use of integer arithmetic . . . . . . . . . . . . . . . .
case study 1 salesmans salary . . . . . . . . . . . . . .
Sequence of squares of numbers . . . . . . . . . . . . .
case study 2 solution of the quadratic equation . . . .
Different kind of operators . . . . . . . . . . . . . . . .
Use of variables in expressions . . . . . . . . . . . . .
Round off errors . . . . . . . . . . . . . . . . . . . . .
Cast to evaluate the equation . . . . . . . . . . . . . .
Use of getchar function . . . . . . . . . . . . . . . . .
Case study 1 Inventory report . . . . . . . . . . . . . .
Whether character is alphabet or digit or special character . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Case study 2 Reliability graph . . . . . . . . . . . . .
Print character in reverse case . . . . . . . . . . . . .
Input formatting options . . . . . . . . . . . . . . . .
Reading of real numbers . . . . . . . . . . . . . . . . .
4
7
7
8
9
9
11
12
12
13
14
14
16
16
17
18
18
19
20
21
22
23
24
25
25
26
27
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
4.6
4.8
4.9
4.10
4.11
5.1
5.1cs
5.2
5.2cs
5.3
5.4
5.5
5.6
5.7
6.1
6.1cs
6.2
6.2cs
6.3
6.3cs
6.4
6.4cs
6.5
6.6
6.7
1.cs
2.cs
3.cs
4.cs
7.1
7.2
7.3
7.4
7.5
1.cs
2.cs
8.1
8.2
Reading of strings . . . . . . . . . . . . . . . . .
Testing for correctness . . . . . . . . . . . . . . .
Output of integer numbers . . . . . . . . . . . .
Printing a real number . . . . . . . . . . . . . . .
Printing of character and strings . . . . . . . . .
Ratio . . . . . . . . . . . . . . . . . . . . . . . .
Case study 1 range of numbers . . . . . . . . . .
counts the number of boys . . . . . . . . . . . . .
Case study 2 Pay Bill Calculations . . . . . . . .
Evaluate the power series . . . . . . . . . . . . .
Largest of the three numbers . . . . . . . . . . .
Reads the customer number and power consumed
Loan applications and to sanction loans . . . . .
square root for five numbers . . . . . . . . . . . .
evaluate the equation . . . . . . . . . . . . . . .
Case study 1 Table of Binomial Coefficients . . .
Multiplication table . . . . . . . . . . . . . . . .
Case study 2 Histogram . . . . . . . . . . . . . .
Uses a for loop . . . . . . . . . . . . . . . . . . .
Case study 3 Minimum Cost . . . . . . . . . . .
Read the marks and print total marks . . . . . .
Case study 4 Plotting of two Functions . . . . .
Use of the break statement . . . . . . . . . . . .
Evaluate the series . . . . . . . . . . . . . . . . .
Use of continue statement . . . . . . . . . . . . .
Case study 1 Median of list of numbers . . . . .
Case study 2 Calculation of standard deviation .
Case study 3 Evaluating a Test . . . . . . . . . .
Case study 4 Production and sales analysis . . .
Sum of squares of 10 numbers . . . . . . . . . . .
Count the number of students . . . . . . . . . . .
Compute and print . . . . . . . . . . . . . . . . .
Multiplication table . . . . . . . . . . . . . . . .
Popularity of various cars . . . . . . . . . . . . .
Case study 1 Counting words in a text . . . . . .
Case study 2 Processing of a customer list . . . .
Read a series of words . . . . . . . . . . . . . . .
Read a line of text . . . . . . . . . . . . . . . . .
5
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
27
28
29
29
30
31
32
33
33
35
36
37
37
38
40
40
42
42
43
44
45
46
48
48
49
51
52
53
54
58
59
60
62
62
65
66
68
68
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
Exa
8.3
8.4
8.5
8.6
8.7
8.8
8.9
1.cs
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.16
1.cs
10.1
10.2
10.3
10.4
10.5
12.1
12.2
12.3
12.4
12.5
12.6
1.cs
2.cs
13.3
13.4
13.5
Copy one string into another . . . . . . . . . . . . . .
Display the string under various format specifications .
Program using for loop . . . . . . . . . . . . . . . . .
Print the alphabet set a to z and A to Z . . . . . . . .
Concatinate the three parts into one string . . . . . .
Compare whether strings are equal . . . . . . . . . . .
Sort a list of names in alphabetical order . . . . . . .
Case study 1 Calculation of Area under a Curve . . .
Multiple functions . . . . . . . . . . . . . . . . . . . .
Include arguments in function calls . . . . . . . . . . .
Return result . . . . . . . . . . . . . . . . . . . . . . .
Computes x raised to the power y . . . . . . . . . . .
Calculate standard deviation of an array values . . . .
Sort an array . . . . . . . . . . . . . . . . . . . . . . .
Autometic variables . . . . . . . . . . . . . . . . . . .
Global variables . . . . . . . . . . . . . . . . . . . . .
Factorial of a number using recursion . . . . . . . . . .
Case study 1 Book Shop Inventory . . . . . . . . . . .
Define a structure type . . . . . . . . . . . . . . . . .
Comparison of structure variables . . . . . . . . . . .
Calculate the subject wise and student wise totals . .
Array member to represent the three subjects . . . . .
structure as a parameter to a function . . . . . . . . .
Read data from keyboard and write it to a file . . . .
Read and write odd and even numbers . . . . . . . . .
Read and write data to and from the file INVENTORY
Error handling in file operations . . . . . . . . . . . .
use of function ftell or mtell and fseek or mseek . . . .
Append additional items to the file INVENTORY . .
Case study 1 Insertion in a sorted list . . . . . . . . .
Case study 2 Building a Sorted List . . . . . . . . . .
Create a linear linked list . . . . . . . . . . . . . . . .
Insert the item before the specified key node . . . . . .
Delete a specified node in the list . . . . . . . . . . . .
6
69
69
70
71
72
73
74
76
77
78
79
80
81
82
83
84
85
87
89
90
91
92
93
95
96
98
99
100
102
104
107
109
111
113
Chapter 1
Overview of C
Scilab code Exa 1.3 Printing a mesage
1
2
3
4
5
//
Example 1 . 3
// SAMPLE PROGRAM 1 : PRINTING A MESSAGE
// P r i n t i n g B e g i n s
printf ( ” I s e e , I remember ” ) // P r i n t i n g u s i n g p r i n t f ( )
function
6 // P r i n t i n g e n d s
7
8
9
//We can a l s o p r i n t a m e s s a g e u s i n g d i s p ( ) f u n c t i o n
in s c i l a b
10 disp ( ” I s e e , I remember ” ) // P r i n t i n g u s i n g d i s p ( )
function
Scilab code Exa 1.4 Adding two numbers
7
1
2
3
4
5
6
//
Example 1 . 4
// SAMPLE PROGRAM 2 : ADDING TWO NUMBERS
number =100;
amount =30.75+75.35 ;
// A d d i t i o n o f two numbers
disp ( number ) ;
// D i s p l a y
v a l u e o f number
variable
7 // D i s p l a y v a l u e o f amount i n f l o a t i n g p o i n t w i t h
f i v e p l a c e s i n a l l and two p l a c e s t o t h e r i g h t o f
decimalpoint
8 printf ( ”%5 . 2 f ” , amount ) ;
Scilab code Exa 1.5 Interest Calculation
1 //
Example 1 . 5
2 // SAMPLE PROGRAM 3 : INTEREST CALCULATION
3
4 PRINCIPAL =5000.00;
5 PERIOD =10;
6 // A s s i g n m e n t S t a t e m e n t s
7 amount = PRINCIPAL ;
8 inrate =0.11;
9 year =0;
10 // Computation u s i n g w h i l e l o o p
11 while ( year <= PERIOD )
12
printf ( ”%2d %8 . 2 f \n ” , year , amount ) ;
13
value = amount + inrate * amount ;
14
year = year +1;
15
amount = value ;
16 end // End o f w h i l e l o o p
8
Scilab code Exa 1.6 Use of subroutines
1
2
3
4
5
6
7
8
9
10
11
12
//
Example 1 . 6
// SAMPLE PROGRAM 4 : USE OF SUBROUTINES
// Program u s i n g f u n c t i o n
function []= mul (a , b )
// mul ( ) f u n c t i o n s t a r t s ( i . e .
definition starts )
y=a*b;
printf ( ” M u l t i p l i c a t i o n o f %d and %d i s %d” ,a ,b , y )
;
endfunction
// mul ( ) f u n c t i o n e n d s
a =5; b =10;
// C a l l i n g mul ( ) f u n c t i o n
mul (a , b )
Scilab code Exa 1.7 Use of math functions
1
2
3
4
5
6
7
8
9
//
Example 1 . 7
// SAMPLE PROGRAM 5 : USE OF MATH FUNCTIONS
// Program u s i n g c o s i n e f u n c t i o n
angle =0; MAX =180;
printf ( ”
Angle
Cos ( a n g l e ) \n ” ) ;
while ( angle <= MAX )
x =( %pi / MAX ) * angle ;
y = cos ( x ) ;
// Use o f c o s i n e f u n c t i o n
9
10
11
12
printf ( ”%15d %13 . 4 f \n ” , angle , y ) ;
angle = angle +10;
end
10
Chapter 2
Constants Variables and Data
Types
Scilab code Exa 2.1 Representation of integer constants on a 16 bit computer
1
2
//
Example 2 . 1
// R e p r e s e n t a t i o n o f i n t e g e r c o n s t a n t s on a 16− b i t
computer .
3
4
5
disp ( ” I n t e g e r v a l u e s ” ) ;
// I n t e g e r v a l u e s l a r g e r t h a n 3 2 7 6 7 a r e n o t s t o r e d
p r o p e r l y on 16− b i t machine
6 printf ( ”%d %d %d \n ” , int16 (32767) , int16 (32767+1) ,
int16 (32767+10) ) ;
7
8
9
disp ( ” Long i n t e g e r v a l u e s ” ) ;
//To s t o r e l o n g i n t e g e r s p r o p e r l y , u s e i n t 3 2 i n t e g e r
type
10 printf ( ” %ld %ld %ld \n ” , int32 (32767) , int32 (32767+1) ,
int32 (32767+10) ) ;
11 // The same r e s u l t a s from a b o v e s t a t e m e n t can be
a c h i e v e d d i r e c t l y from b e l o w commented s t a t e m e n t
12 // p r i n t f ( ” %ld %ld %ld \n ” , 3 2 7 6 7 , 3 2 7 6 7 + 1 , 3 2 7 6 7 + 1 0 ) ;
11
Scilab code Exa 2.1cs Case study 1 avg of numbers
1 //
Case Study :− C h a p t e r 2 Page No. −47
2 //
1 . C a l c u l a t i o n o f A v e r a g e o f numbers
3
4 N =10; sum1 =0; count =0;
// I n i t i a l i z a t i o n
5
6
7
8
9
10
11
12
13
of
variables
printf ( ” E n t e r t e n numbers ” ) ;
while ( count < N )
number = scanf ( ” %f ” ) ; // R e a d i n g number ( u s i n g
scanf () function )
sum1 = sum1 + number ;
count = count +1;
end
average = sum1 / N ;
// Avarage i s c a l c u l a t e d
printf ( ” N = %d Sum1 = %f ” ,N , sum1 ) ;
printf ( ” A v e r a g e = %f ” , average ) ;
Scilab code Exa 2.2 typical declarations assignments and values stored in
various types of variables
1
2
//
Example 2 . 2
// Program shows t y p i c a l d e c l a r a t i o n s , a s s i g n m e n t s
and v a l u e s s t o r e d i n v a r i o u s t y p e s o f v a r i a b l e s .
3
4 // D e c l a r a t i o n s and A s s i g n m e n t s
5 m = int16 (54321) ;
12
6 n = int32 (1234567890) ;
7 k = uint16 (54321) ;
8 // A s s i g n m e n t s
9 x =1.234567890000;
10
11
12
13
14
15
16
17
18
19
// B y d e f a u l t t y p e i s d o u b l e
in s c i l a b
y =9.87654321;
//
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
p =1.0; q =1.0;
// P r i n t i n g
printf ( ” m=%d\n ” ,m )
printf ( ” n=%ld \n ” ,n )
printf ( ” x=%. 1 2 f \n ” ,x )
printf ( ” x=%f\n ” ,x )
printf ( ” y=%. 1 2 f \n ” ,y )
printf ( ” y=%f\n ” ,y )
printf ( ” k=%u p=%f q=%. 1 2 f \n ” ,k ,p , q )
Scilab code Exa 2.2cs Case study temprature in Farenheit and Celsius
1
2
3
4
5
6
7
8
9
10
11
Case Study :− C h a p t e r 2
Page no . −48
2 . S o l u t i o n o f t e m p r a t u r e i n F a r e n h e i t and
Celsius
F_LOW =0;
F_MAX =250;
STEP =25;
fahrenheit = F_LOW ;
// I n i t i a l i z a t i o n
printf ( ” F a h r e n h e i t
C e l s i u s \n\n ” ) ;
while ( fahrenheit < = F_MAX )
celsius =( fahrenheit -32.0) /1.8;
// c o n v e r s i o n
from F a r e n h e i t t o C e l s i u s
printf ( ”%6 . 2 f
%7 . 2 f \n ” , fahrenheit ,
celsius ) ;
fahrenheit = fahrenheit + STEP ;
//
//
13
12
end
Scilab code Exa 2.3 use of scanf function
1
2
3
4
5
6
7
8
9
//
Example 2 . 3
// The program i l l u s t r a t e t h e u s e o f s c a n f ( )
function
disp ( ” E n t e r an i n t e r g e r number : ” ) ;
number = scanf ( ”%d” ) ;
// Read from k e y b o a r d
if ( number <100) then
disp ( ” Your number i s s m a l l e r t h a n 100 ” ) ;
else
disp ( ” Your number c o n t a i n more t h a n two d i g i t s ” )
;
end
Scilab code Exa 2.4 Interest calculation using scanf
1
2
3
4
5
6
7
8
//
Example 2 . 4
// Sample program 3 ( exm1 . 5 ) d i s c u s s e d i n c h a p t e r 1
can be c o n v e r e d i n t o a more f l e x i b l e i n t r a c t i v e
program u s i n g s c a n f ( ) f u n c t i o n
disp ( ” E n t e r i n s i n g l e l i n e s e p a r t e d by s p a c e ” ) ;
disp ( ” I n p u t amount , i n t e r e s t r a t e , and p e r i o d ” ) ;
[ amount , inrate , period ]= scanf ( ” %f %f %d” ) ;
// u s e
of scanf ()
year =1;
// Computation u s i n g w h i l e l o o p
while ( year <= period )
14
9
10
11
12
13 end
value = amount + inrate * amount ;
printf ( ”%2d Rs %8 . 2 f \n ” , year , value )
year = year +1;
amount = value ;
15
Chapter 3
Operators and Expressions
Scilab code Exa 3.1 Use of integer arithmetic
//
Example 3 . 1
// The program shows t h e u s e o f i n t e g e r a r i t h m e t i c t o
c o n v e r t a g i v e n number−
3 // o f d a y s i n t o months and d a y s
1
2
4
5
6
7
8
days = input ( ’ E n t e r d a y s : ’ ) ;
months = int16 ( days /30) ;
// Compute f o r months
days = int16 ( pmodulo ( days ,30) ) ; // compute f o r d a y s
disp ( days , ” Days =” , months , ” Months =” ) ;
Scilab code Exa 3.1cs case study 1 salesmans salary
1
2
3
4
5
//
//
Case Study :− C h a p t e r 3 , Page No : 7 6
1 . Salesman ’ s S a l a r y
BASE_SALARY =1500.00;
// Minimum b a s e s a l a r y
16
6
7
8
9
10
11
12
13
14
15
16
BONUS_RATE =200.00;
// Bonus f o r e v e r y c o m p u t e r
sold
COMMISSION =0.02;
// Commission on t o t a l monthly
sales
printf ( ” I n p u t number s o l d and p r i c e \n [ E n t e r i n
s i n g l e l i n e s e p a r a t e d by s p a c e ] ” ) ;
[ quantity , price ] = scanf ( ”%d %f ” ) ;
// I n p u t
q u a n t i t y and p r i c e
// Computation f o r bonus , c o m m i s s i o n and g r o s s s a l a r y
of a salesman
bonus = BONUS_RATE * quantity ;
commission = COMMISSION * quantity * price ;
gross_salary = BASE_SALARY + bonus + commission ;
printf ( ” Bonus
= %6 . 2 f \n ” ,
bonus ) ;
printf ( ” Commission
= %6 . 2 f \n ” ,
commission )
;
printf ( ” G r o s s
salary
= %6 . 2 f \n ” ,
gross_salary ) ;
Scilab code Exa 3.2 Sequence of squares of numbers
1 //
Example 3 . 2
2 // Program t o p r i n t a s e q u e n c e o f s q u a r e s o f numbers .
3
4 N =100; A =2;
5 a=A;
6 while (a < N )
7
disp ( a ) ;
// P r i n t s s q u a r e o f number
8
a = a ^2;
// compute s q u a r e o f number
9 end
17
Scilab code Exa 3.2cs case study 2 solution of the quadratic equation
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
//
//
Case Study :− C h a p t e r 3 , Page No : 7 7
2. Solution of the Quadratic equation
printf ( ” I n p u t v a l u e s o f a , b , and c \n ” ) ;
a = input ( ” a =” ) ;
b = input ( ” b =” ) ;
c = input ( ” c =” ) ;
discriminant = b ^2 -4* a * c ;
if ( discriminant <0)
printf ( ” \n\nROOTS ARE IMAGINARY\n ” ) ;
else
// Computes r o o t 1 and r o o t 2
root1 = ( - b + sqrt ( discriminant ) ) /(2.0* a ) ;
root2 = ( - b - sqrt ( discriminant ) ) /(2.0* a ) ;
printf ( ” \n\ nRoot1 = %5 . 2 f \n\ nRoot2 = %5 . 2 f \n ” ,
root1 , root2 ) ;
17 end
Scilab code Exa 3.3 Different kind of operators
1
2
//
Example 3 . 3
// The program e m p l o y s d i f f r e n t k i n d o f o p e r a t o r s .
The r e s u l t s o f t h e i r e v a l u a t i o n a r e a l s o shown
f o r comparison
3
18
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
// I n c r e m e n t ( i . e . ++)/ Decrement (−−) o p e r a t o r s a r e
unavailable in Scilab
a = int16 (15) ;
b = int16 (10) ;
a = a +1; // R e p l a c e m e n t f o r ++a
c =a - b ;
printf ( ” a = %d b = %d c = %d\n ” ,a ,b , c ) ;
b = b +1; // R e p l a c e m e n t f o r b++
d=b+a;
printf ( ” a = %d b = %d d = %d\n ” ,a ,b , d ) ;
printf ( ” a /b = %d\n ” ,a / b ) ;
//
Division operator
printf ( ” pmodulo ( a , b ) = %d\n ” , pmodulo (a , b ) ) ; //
Modulus o p e r a t i o r
printf ( ” a ∗b = %d\n ” ,a * b ) ;
//
Multiplication
// I n s c i l a b t h e r i s no c o n d i t i o n a l o p e r a t o r ( ? : ) ,
h e n c e ’ i f ’ can be u s e d i n p l a c e o f ? :
if (c > d ) then
disp (1) ;
end
if (c < d ) then
disp (0) ;
end
Scilab code Exa 3.4 Use of variables in expressions
1
2
//
Example 3 . 4
// The program i l l u s t r a t e s t h e u s e o f v a r i a b l e s i n
e x p r e s s i o n s and t h e i r e v a l u a t i o n .
3
4 a =9; b =12; c =3;
5 // E x p r e s s i o n s and t h e i r
evaluations
19
6 x =a - b /3+ c *2 -1;
7 y =a - b /(3+ c *(2 -1) ) ;
8 z =a -( b /(3+ c ) *2) -1;
9
10 printf ( ” x=%f\n ” ,x )
11 printf ( ” y=%f\n ” ,y )
12 printf ( ” z=%f\n ” ,z )
13 // d i s p can be u s e d . .
14 // d i s p ( x , ” x =”)
15 // d i s p ( y , ” y =”)
16 // d i s p ( z , ” z =”)
Scilab code Exa 3.5 Round off errors
1
2
//
Example 3 . 5
// Output o f program shows round− o f f e r r o r s t h a t can
o c c u r i n c o m p u t a t i o n o f f l o a t i n g p o i n t numbers
3
4 //Sum o f n t e r m s o f 1/ n
5 count =1;
6 sum1 =0;
7 n = input ( ” E n t e r v a l u e o f n : ” ) ;
8 term =1.0/ n ;
9 while ( count <= n )
10
sum1 = sum1 + term ;
11
count = count +1;
12 end
13 printf ( ”Sum= %f ” , sum1 ) ;
20
Scilab code Exa 3.6 Cast to evaluate the equation
1 //
Example 3 . 6
2 // Program u s i n g a c a s t t o e v a l u a t e t h e e q u a t i o n .
3
4 sum1 =0;
5 for n = int8 (1:10)
6
sum1 = sum1 +1/ double ( n ) ;
// c o n v e r s i o n from ’ i n t ’
7
8 end
to ’ double ’ or ’ f l o a t ’
printf ( ”%2d %6 . 4 f \n ” ,n , sum1 ) ;
21
Chapter 4
Managing Input and Output
Operations
Scilab code Exa 4.1 Use of getchar function
//
Exaymple 4 . 1
// The program shows t h e u s e o f g e t c h a r f u n c t i o n i n
an i n t r a c t i v e e n v i r o n m e n t .
3 // I n S c i l a b i n p l a c e o f g e t c h a r f u n c t i o n s c a n f
f u n c t i o n can be u s e d t o g e t
4 // c h a r a c t e r a s t h e r e i s no g e t c h a r f u n c t i o n i n
Scilab .
1
2
5
6
7
8
9
10
11
12
13
disp ( ” Would you l i k e t o know my name ? ” ) ;
disp ( ” Type Y f o r YES and N f o r NO: ” ) ;
answer = scanf ( ”%c” ) ;
// R e a d i n g
character
if ( answer == ’Y ’ ) |( answer == ’ y ’ ) then
// T e s t f o r
answer
disp ( ”My name i s BUSY BEE” ) ;
else
disp ( ”You a r e good f o r n o t h i n g ” )
end
22
Scilab code Exa 4.1cs Case study 1 Inventory report
1 //
Case Study :− C h a p t e r 4 , Page No : 1 0 6
2 //
1 . I nv e nt o ry Report
3
4 ITEMS =4;
5 i =1;
6 printf ( ” [ E n t e r i n s i n g l e l i n e s e p e r a t e d be s p a c e s ] \ n
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
”);
while ( i <= 4)
printf ( ” E n t e r code , q u a n t i t y , and r a t e : ” ) ;
[ code ( i ) , quantity ( i ) , rate ( i ) ]= scanf ( ”%s %d %f ” )
;
i = i +1;
end
printf ( ”INVENTORY REPORT\n ” ) ;
printf ( ”
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n ”
);
printf ( ” Code
Quantity
Rate
V a l u e \n ” )
;
printf ( ”
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n ”
);
total_value =0;
i =1;
while (i <= ITEMS )
value = quantity ( i ) * rate ( i ) ;
printf ( ” %6s %10d %10 . 2 f
%e\n ” , code ( i ) ,
quantity ( i ) , rate ( i ) , value ) ;
total_value = total_value + value ;
i = i +1;
23
23 end
24
printf ( ”
25
26
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\
n”);
printf ( ”
T o t a l V a l u e = %e\n ” ,
total_value ) ;
printf ( ”
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\
n”);
Scilab code Exa 4.2 Whether character is alphabet or digit or special character
//
Example 4 . 2
// The program r e q u e s t s t h e u s e r t o e n t e r a
c h a r a c t e r and d i s p l a y a m e s s a g e on
3 // t h e s c r e e n t e l l i n g t h e u s e r w h e t h e r t h e c h a r a c t e r
i s an a l p h a b e t o r d i g i t ,
4 // o r any o t h e r s p e c i a l c h a r a c t e r .
1
2
5
6
7
8
9
10
11
12
13
14
disp ( ” P r e s s any key ” ) ;
character = scanf ( ”%c” ) ;
//
Reading c h a r a c t e r
if ( isletter ( character ) ) then
// T e s t
for letter
disp ( ” The c h a r a c t e r i s a l e t t e r ” ) ;
elseif ( isdigit ( character ) ) then
// T e s t
for digit
disp ( ” The c h a r a c t e r i s a d i g i t ” ) ;
else
disp ( ” The c h a r a c t e r i s n o t a l p h a n u m e r i c ”
);
end
24
Scilab code Exa 4.2cs Case study 2 Reliability graph
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
//
//
Case s t u d y : c h a p t e r 4
2 . R e l i a b i l i t y Graph
LAMBDA =0.001;
for i =1:27
printf ( ”−−” ) ;
end
printf ( ” \n ” ) ;
for t =0:150:3000
r = exp ( - LAMBDA * t ) ;
R = int32 (50* r +0.5) ;
printf ( ” | ” ) ;
for i =1: R
printf ( ” ∗ ” ) ;
end
printf ( ”#\n ” ) ;
end
for i =1:2
printf ( ” | \ n ” ) ;
end
Scilab code Exa 4.3 Print character in reverse case
1
2
//
Example 4 . 3
// A program t h a t r e a d s a c h a r a c t e r from t h e
k e y b o a r d and t h e n p r i n t i n r e v e r s e
25
3
4
5
6
7
8
9
10
// c a s e , t h a t i s , i f i n p u t i s i n u p p e r c a s e , t h e o u t p u t
w i l l be l o w e r c a s e and v i c e −v e r s a
disp ( ” E n t e r an a l p h a b e t ” ) ;
alphabet = scanf ( ”%c” ) ;
// R e a d i n g
character
if (( ascii ( alphabet ) ) >=97) then
disp ( convstr ( alphabet , ” u” ) ) ;
// R e v e r s e and
display
else
disp ( convstr ( alphabet , ” l ” ) ) ;
// R e v e r s e and
display
end
Scilab code Exa 4.4 Input formatting options
//
Example 4 . 4
// The program i l l u s t r a t e s t h e v a r i o u s o p t i o n s f o r
reading
3 // i n t e g e r s a r e e x p e r i m e n t e d i n t h i s program
1
2
4
5
6 printf ( ” E n t e r t h r e e i n t e g e r numbers \n ” ) ;
7 [n ,a ,b , c ]= mscanf ( ”%d %∗d %d” ) ;
8 disp (c ,b , a ) ;
9 printf ( ” E n t e r two 4− d i g i t numbers \n ” ) ;
10 [n ,x , y ]= mscanf ( ”%2d %4d” ) ;
11 printf ( ’%d %d\n ’ ,x , y ) ;
12
13 printf ( ” E n t e r two i n t e g e r s \n ” ) ;
14 [n ,a , x ]= mscanf ( ”%d %d” ) ;
15 printf ( ’%d %d\n ’ ,a , x ) ;
16
17 printf ( ” E n t e r a n i n e d i g i t number \n ” ) ;
26
18 [n ,p ,q , r ]= mscanf ( ”%3d %4d %3d” ) ;
19 printf ( ’%d %d %d\n ’ ,p ,q , r ) ;
20
21 printf ( ” E n t e r two t h r e e d i g i t numbers \n ” ) ;
22 [n ,x , y ]= mscanf ( ”%d %d” ) ;
23 printf ( ’%d %d \n ’ ,x , y ) ;
Scilab code Exa 4.5 Reading of real numbers
1
2
//
Example 4 . 5
// R e a d i n g o f r e a l numbers ( i n b o t h d e c i m a l p o i n t and
exponential notation
3
4 printf ( ” V a l u e s o f x and y : ” ) ;
5 [x , y ]= scanf ( ” %f %e” ) ; // r e a d i n g x [ d e c i m a l p o i n t ]
6
7
8
9
10
and y [ e x p o n e n t i a l ] from s t a n d a r d i n p u t
printf ( ” \n ” ) ;
printf ( ” x = %f\ ny = %f\n ” ,x , y ) ;
printf ( ” V a l u e s o f p and q : ” ) ;
[p , q ]= scanf ( ” %lg %lg ” ) ;
printf ( ” \ np = %. 1 2 f \ nq = %. 1 2 e \n ” ,p , q ) ;
Scilab code Exa 4.6 Reading of strings
1
2
3
4
//
//
Example 4 . 6
R e a d i n g o f s t r i n g s u s i n g %wc and %ws .
// s c a n f ( ) can o n l y r e a d one c h a r a c t e r a t a t i m e and
r e t u r n i t t o name1 , t h e r e f o r e
27
// m s c a n f ( ) i s u s e d t h a t can r e a d more t h a n one
character
6 printf ( ” E n t e r s e r i a l number and name one : ” ) ;
7 [n , no , name1 ]= mscanf ( ”%d %15c ” ) ;
8 printf ( ”%d %15s \n ” ,no , name1 ) ;
5
9
10 // Read and d i s p l a y a s t r i n g
11 printf ( ” E n t e r s e r i a l number and name two : ” ) ;
12 [ no , name2 ]= scanf ( ”%d %s” ) ;
13 printf ( ”%d %15s \n ” ,no , name2 ) ;
14
15 // Read and d i s p l a y a s t r i n g o f w i d t h 15
16 printf ( ” E n t e r s e r i a l number and name t h r e e : ” ) ;
17 [ no , name3 ]= scanf ( ”%d %15s ” ) ;
18 printf ( ”%d %15s ” ,no , name3 ) ;
Scilab code Exa 4.8 Testing for correctness
1
2
3
4
5
6
7
8
9
10
11
//
Example 4 . 8
// The program i l l u s t r a t e s t h e t e s t i n g f o r
c o r r e c t n e s s o f r e a d i n g d a t a by s c a n f f u n c t i o n
// I n S c i l a b m s c a n f f u n c t i o n i s s i m i l a r t o s c a n f
function of c
printf ( ” E n t e r v a l u e s o f a , b and c ” ) ;
[n ,a ,b , c ]= mscanf ( ”%d %f %c” ) ;
if ( n ==3) then
// T e s t c o n d i t i o n
printf ( ” a = %d b = %f c = %c” ,a ,b , c ) ;
else
printf ( ” E r r o r i n i n p u t . ” ) ;
end
28
Scilab code Exa 4.9 Output of integer numbers
1
2
//
Example 4 . 9
// The program i l l u s t r a t e s t h e o u t p u t o f i n t e g e r
numbers u n d e r v a r i o u s f o r m a t s
3
4 m = int16 (12345) ;
5 n = int32 (987654) ;
6 printf ( ’%d\n ’ ,m ) ;
7 printf ( ’ %10d\n ’ ,m ) ;
8 printf ( ’ %010d \n ’ ,m ) ;
9 printf ( ’%−10d\n ’ ,m ) ;
10 printf ( ’ %10d\n ’ ,n ) ;
11 printf ( ’ %10d\n ’ ,-n ) ;
// i n t e g e r v a r i a b l e m
// i n t e g e r v a r i a b l e n
Scilab code Exa 4.10 Printing a real number
1
2
//
Example 4 . 1 0
// The program i l l u s t r a t e s a l l t h e o p t i o n s o f
p r i n t i n g a r e a l number .
3
4 y =98.7654;
// r e a l number
5 // V a r i o u s o p t i o n s o f p r i n t i n g a r e a l number
6 printf ( ’%7 . 4 f \n ’ ,y ) ;
7 printf ( ’ %f\n ’ ,y ) ;
8 printf ( ’%7 . 2 f \n ’ ,y ) ;
9 printf ( ’%−7.2 f \n ’ ,y ) ;
10 printf ( ’ %07 . 2 f \n ’ ,y ) ;
29
11
12
13
14
15
16
printf ( ’%7 . 2 f \n ’ ,y ) ;
printf ( ’ \n ’ ) ;
printf ( ’ %10 . 2 e \n ’ ,y ) ;
printf ( ’ %12 . 4 e \n ’ ,-y ) ;
printf ( ’%−10.2 e \n ’ ,y ) ;
printf ( ’ %e\n ’ ,y ) ;
Scilab code Exa 4.11 Printing of character and strings
1
2
//
Example 4 . 1 1
// The program i l l u s t r a t e s t h e p r i n t i n g o f c h a r a c t e r
and s t r i n g s .
3
4 x = ’A ’ ;
// x v a r i a b l e h a s c h a r a c t e r
5 name = ’ ANIL KUMAR GUPTA ’ ;
// name v a r i a b l e h a s s t r i n g
6 disp ( ”OUTPUT OF CHARACTERS” ) ;
7 printf ( ’ %c\n%3c\n%5c\n ’ ,x ,x , x ) ;
8 printf ( ’ %3c\n%c\n ’ ,x , x ) ;
9 printf ( ’ \n ’ ) ;
10 disp ( ”OUTPUT OF STRINGS” ) ; // d i s p can a l s o be u s e d
11
12
13
14
15
16
for display
printf ( ’ %s\n ’ , name ) ;
printf ( ’ %20s \n ’ , name ) ;
printf ( ’ %20 . 1 0 s \n ’ , name ) ;
printf ( ’%. 5 s \n ’ , name ) ;
printf ( ’%−20.10 s \n ’ , name ) ;
printf ( ’ %5s\n ’ , name ) ;
30
Chapter 5
Decision Making and Branching
Scilab code Exa 5.1 Ratio
//
Example 5 . 1
// The program r e a d s f o u r v a l u e s a , b , c and d from
t h e t e r m i n a l and
3 // e v a l u a t e s t h e r a t i o o f ( a+b ) t o ( c+d ) and p r i n t s
t h e r e s u l t , i f c−d
4 // i s n o t e q u a l t o z e r o
1
2
5
6
7
8
9
10
11
12
13
14
15
disp ( ” E n t e r f o u r i n t e g e r v a l u e s ” ) ;
a = int8 ( input ( ” a=” ) ) ;
b = int8 ( input ( ” b=” ) ) ;
c = int8 ( input ( ” c=” ) ) ;
d = int8 ( input ( ” d=” ) ) ;
if (c - d ~= 0) then
// E x e c u t e s t a t e m e n t b l o c k
ratio = double ( a + b ) / double (c - d ) ;
printf ( ” R a t i o=%f\n ” , ratio ) ;
end
31
Scilab code Exa 5.1cs Case study 1 range of numbers
1 //
Case Study :− C h a p t e r 5 , Page No : 1 3 9
2 //
1 . Range o f numbers
3
4 sum1 =0;
5 count =0;
6 f =0; value =1;
7 printf ( ” E n t e r numbers [ p r e s s e n t e r a f t e r e a c h number
] : \ n i n p u t a NEGATIVE number t o end \n ” ) ;
8 while ( value )
9
[ value ]= scanf ( ” %f ” ) ; // Read d a t a
10
if ( value <0) then
11
break ;
12
end
13
count = count +1;
14
// C a l c u l a t i n g h e i g h t and l o w e s t v a l u e
15
if ( count ==1) then
16
high = value ;
17
low = value ;
18
elseif ( value > high ) then
19
high = value ;
20
elseif ( value < low ) then
21
low = value ;
22
end
23
// C a l c u l a t e sum
24
sum1 = sum1 + value ;
25 end
26 average = sum1 / count ;
// A v e r a g e c o s t
27 range1 = high - low ;
// Range o f v a l u e s
28 // P r i n t t h e r e s u l t s
29 printf ( ” T o t a l v a l u e s : %d\n ” , count ) ;
32
printf ( ” H i g h e s t −v a l u e s : %f\ nLowest−v a l u e : %f\n ” , high
, low ) ;
31 printf ( ” Range : %f\n Avarage : %f\n ” , range1 , average ) ;
30
Scilab code Exa 5.2 counts the number of boys
//
Example 5 . 2
// The program c o u n t s t h e number o f b o y s whose
w e i g h t i s l e s s t h a n 50Kg
3 // and h e i g h t i s g r e a t e r t h a n 170cm .
1
2
4
5 count = int (0) ;
6 disp ( ” E n t e r w e i g h t and h e i g h t f o r 10 b o y s ” ) ;
7 for i =1:10
8
[ weight , height ]= scanf ( ” %f %f ” ) ;
9
if ( weight <50& height >170) then
// T e s t f o r
w e i g h t and h e i g h t
count = count +1;
10
11
end
12 end
13 disp ( ”Number o f b o y s w i t h w e i g h t <50 kg and h e i g h t
>170 cm =” ) ;
14 disp ( count )
Scilab code Exa 5.2cs Case study 2 Pay Bill Calculations
1
//
Case Study :− C h a p t e r 5
33
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
//
2 . Pay− B i l l C a l c u l a t i o n s
CA1 =1000;
CA2 =750;
CA3 =500;
CA4 =250;
EA1 =500;
EA2 =200;
EA3 =100;
EA4 =0;
level =1;
while ( level )
printf ( ” E n t e r 0 [ z e r o ] f o r l e v e l t o end ” ) ;
// Read d a t a
level = input ( ” E n t e r l e v e l : ” ) ;
if ( level ==0)
break ;
end
printf ( ” E n t e r j o b number , and b a s i c pay \n ” ) ;
// Read d a t a
[ jobnumber , basic ]= scanf ( ”%d %f ” ) ;
// D e c i d e l e v e l number and c a l c u l a t e p e r k s
select level
case 1 then perks = CA1 + EA1 ;
case 2 then perks = CA2 + EA2 ;
case 3 then perks = CA3 + EA3 ;
case 4 then perks = CA4 + EA4 ;
else
printf ( ” E r r o r i n l e v e l c o d e ” ) ;
return ;
end
house_rent =0.25* basic ;
// C a l c u l a t e g r o s s s a l a r y
gross = basic + house_rent + perks ;
// C a l c u l a t e i n c o m e t a x
if ( gross <=2000) then
incometax =0;
elseif ( gross <=4000)
34
40
incometax =0.03* gross ;
41
elseif ( gross <=5000)
42
incometax =0.05* gross ;
43
else
44
incometax =0.08* gross ;
45
end
46
// Compute t h e n e t s a l a r y
47
net = gross - incometax ;
48
// P r i n t t h e r e s u l t s
49
printf ( ”%d %d %. 2 f \n ” , level , jobnumber , net ) ;
50 end
51 printf ( ”END OF THE PROGRAM” ) ;
Scilab code Exa 5.3 Evaluate the power series
//
Example 5 . 3
// A program t o e v a l u a t e t h e power s e r i e s . I t u s e s
i f . . . e l s e to t e s t accuracy .
3 // e ˆ x=1+x+x ˆ 2 / 2 ! + x ˆ 3 / 3 ! +
+x ˆ n/ n ! ,0 < x<1
1
2
4
5 ACCURACY =0.0001;
6 x = input ( ” E n t e r v a l u e o f x : ” ) ;
7 n =1; term =1; sum1 =1; count = int8 (1) ;
8 while (n <=100)
9
term = term * x / n ;
10
sum1 = sum1 + term ;
11
count = count +1;
12
if ( term < ACCURACY ) then
// T e s t f o r
13
n =999;
14
else
15
n = n +1;
16
end
17 end
18 // P r i n t r e s u l t s
19 printf ( ”Term=%d Sum=%f ” , count , sum1 ) ;
35
accuracy
Scilab code Exa 5.4 Largest of the three numbers
//
Example 5 . 4
// The program s e l e c t s and p r i n t s t h e l a r g e s t o f t h e
t h r e e numbers
3 // u s i n g n e s t e d i f . . . e l s e s t a t e m e n t
1
2
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
disp ( ” E n t e r t h r e e v a l u e s ” ) ;
A = input ( ”A=” ) ;
B = input ( ”B=” ) ;
C = input ( ”C=” ) ;
disp ( ” L a r g e s t v a l u e i s : ” ) ;
if (A > B ) ,
// T e s t f o r l a r g e s t b e t w e e n A
&B
if (A > C ) ,
// T e s t f o r l a r g e s t b e t w e e n A
&C
disp ( A ) ;
else
disp ( C ) ;
end
else
if (C > B ) ,
// T e s t f o r l a r g e s t b e t w e e n C&
B
disp ( C ) ;
else
disp ( B ) ;
end
end
36
Scilab code Exa 5.5 Reads the customer number and power consumed
//
Example 5 . 5
// The program r e a d s t h e c u s t o m e r number and power
consumed and p r i n t s
3 // t h e amount t o be p a i d by t h e c u s t o m e r
1
2
4
5 disp ( ” E n t e r CUSTOMER NO. and UNITS consumed ” ) ;
6 // Read d a t a
7 custnum = input ( ”CUSTOMER NO . : ” ) ;
8 units = input ( ”UNITS : ” ) ;
9 // Use o f e l s e . . . i f l a d d e r
10 // C a l c u l a t e s c h a r g e s
11 if ( units <=200) ,
12
charges =0.5* units ;
13 elseif ( units <=400) ,
14
charges =100+0.65*( units -200) ;
15 elseif ( uints <=600) ,
16
charges =230+0.8*( units -400) ;
17 else
18
charges =390+( units -600) ;
19 end
20 // P r i n t r e s u l t s
21 printf ( ” Customer No : %d C h a r g e s=%. 2 f ” , custnum ,
charges ) ;
Scilab code Exa 5.6 Loan applications and to sanction loans
37
1
2
//
Example 5 . 6
// A program t o
p r o c e s s l o a n a p p l i c a t i o n s and t o
sanction loans .
3
4 MAXLOAN =50000;
5 disp ( ” E n t e r t h e v a l u e s o f p r e v i o u s two l o a n s ” ) ;
6 loan1 = int32 ( input ( ” E n t e r f i r s t l o a n : ” ) ) ;
7 loan2 = int32 ( input ( ” E n t e r s e c o n d l o a n : ” ) ) ;
8 loan3 = int32 ( input ( ” E n t e r t h e v a l u e s o f new l o a n : ” ) ) ;
9 sum23 = loan2 + loan3 ;
10 // C a l c u l a t e t h e s a n c t i o n l o a n
11 if ( loan1 >0) ,
12
sancloan =0;
13
elseif ( sum23 > MAXLOAN ) ,
14
sancloan = MAXLOAN - loan2 ;
15
else
16
sancloan = loan3 ;
17 end
18 // P r i n t t h e r e s u l t s
19 printf ( ” P r e v i o u s l o a n s p e n d i n g : %d %d\n ” , loan1 , loan2 )
;
20 printf ( ” Loan r e q u e s t e d =%d\n ” , loan3 ) ;
21 printf ( ” Loan s a n c t i o n e d =%d\n ” , sancloan ) ;
Scilab code Exa 5.7 square root for five numbers
1
2
3
4
5
6
//
Example 5 . 7
// The program e v a l u a t e s t h e s q u a r e r o o t f o r f i v e
numbers .
count =1;
printf ( ” E n t e r FIVE r e a l v a l u e s \n ” ) ;
while ( count <=5)
38
7
8
9
10
11
x = scanf ( ” %f ” ) ;
if (x <0) then
printf ( ” V a l u e − %d i s n e g a t i v e \n ” , count ) ;
else
y = sqrt ( x ) ;
// C a l c u l a t e s q u a r e
root
printf ( ” %f\ t%f \n ” ,x , y ) ; // P r i n t r e s u l t
end
count = count +1;
12
13
14
15 end
16 printf ( ”End o f c o m p u t a t i o n ” ) ;
39
Chapter 6
Decision Making and Looping
Scilab code Exa 6.1 evaluate the equation
1
2
3
4
5
6
7
8
9
10
11
12
13
//
Example 6 . 1
//A program t o e v a l u a t e t h e e q u a t i o n y=x ˆ n when n i s
a non−n e g a t i v e i n t e g e r .
x = input ( ” E n t e r t h e v a l u e o f x : ” ) ;
n = input ( ” E n t e r t h e v a l u e o f n : ” ) ;
y =1.0; count =1;
// i n t i a l i z a t i o n
// Loop b e g i n s
while ( count <= n )
// T e s t i n g
y=y*x;
count = count +1; // I n c r e a m e n t i n g
end
// End o f l o o p
printf ( ” x = %f ; n = %d ; x t o power n = %f\n ” ,x ,n , y ) ;
Scilab code Exa 6.1cs Case study 1 Table of Binomial Coefficients
40
1
//
Case Study :− C h a p t e r 6 , Page No
:176
2 //
1 . Table o f Binomial
3
4 MAX =10;
5 printf ( ”mx” ) ;
6 for m =0:10
7
printf ( ”%4d” ,m ) ;
8 end
9 printf ( ” \n
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
Coefficients
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n
”);
m =0;
// p r i n t t h e t a b l e o f b i n o m i a l c o e f f i c i e n t s f o r m=10
// Computation u s i n g w h i l e l o o p
while (m <= MAX )
printf ( ”%2d” ,m ) ;
x =0;
binom =1;
while (x <= m )
if ( m ==0| x ==0)
printf ( ”%4d” , binom ) ;
// P r i n t t h e
r e s u l t i . e . binom
else
binom = binom *( m - x +1) / x ; // compute t h e
binomial c o e f f i c i e n t
printf ( ”%4d” , binom ) ;
// P r i n t t h e
r e s u l t i . e . binom
end
x = x +1;
end
printf ( ” \n ” ) ;
m = m +1;
end
printf ( ”
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n
”);
41
Scilab code Exa 6.2 Multiplication table
1
2
//
Example 6 . 2
//A program t o p r i n t m u l t i p l i c a t i o n t a b l e from 1∗1
to 12∗10.
3
4
5 disp ( ”
MULTIPLICATION TABLE
”);
6 disp ( ”−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−” ) ;
7 row =1;
8 while ( row <=12)
// Outer l o o p b e g i n s
9
column =1;
10
while ( column <=10)
// I n n e r l o o p b e g i n s
11
y = row * column ;
12
printf ( ”%4d” ,y ) ;
13
column = column +1;
14
end
15
row = row +1;
16
printf ( ” \n ” ) ;
17 end
Scilab code Exa 6.2cs Case study 2 Histogram
1 //
2 //
3
4 N =5;
5 for n =1: N
Case Study :− C h a p t e r 6
2 . Histogram
42
6
7
8
printf ( ” E n t e r e m p l o y e e s i n Group−%d : ” ,n ) ;
value ( n ) = scanf ( ”%d” ) ;
// Read d a t a i n
t h e a r r a y named v a l u e
printf ( ”%d\n ” , value ( n ) ) ;
// P r i n t number
which i s a t p o s i t i o n n
9 end
10 printf ( ” \n ” ) ;
11 printf ( ”
| \ n”);
12 // Computation u s i n g f o r l o o p and draw a h i s t o g r a m
13 for n =1: N
14
for i =1:3
15
if ( i ==2) then
16
printf ( ” Group−%1d | ” ,n ) ;
17
else
18
printf ( ”
| ”);
19
end
20
for j =1: value ( n )
21
printf ( ” ∗ ” ) ;
22
end
23
if ( i ==2)
24
printf ( ” (%d) \n ” , value ( n ) ) ;
25
else
26
printf ( ” \n ” ) ;
27
end
28
end
29
printf ( ”
| \ n”);
30 end
Scilab code Exa 6.3 Uses a for loop
1
2
//
Example 6 . 3
// The program u s e s a f o r l o o p t o p r i n t t h e ” Power o f
2” t a b l e f o r t h e
43
3
4
5
6
7
8
9
10
11
12
// power 0 t o 2 0 , b o t h p o s i t i v e and n e g a t i v e .
disp ( ”−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−” ) ;
printf ( ” 2 t o power n
n
2 t o power −n\n ” )
;
disp ( ”−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−” ) ;
for n =0:20
// Loop b e g i n s
p =2^ n ;
q =2^ - n ;
printf ( ”%7d
%9d
%10 . 1 2 f \n ” ,p ,n , q ) ;
end
// Loop e n d s
Scilab code Exa 6.3cs Case study 3 Minimum Cost
1 //
Case Study :− C h a p t e r 6
2 //
3 . Minimum C o s t
3
4 for p =0:0.1:10
5
cost =48 -8* p + p ^2;
6
if ( p ==0) ,
7
cost1 = cost ;
8
continue ;
// Use o f c o n t i n u e s t a t e m e n t
9
end
10
11
if ( cost >= cost1 ) ,
12
break ;
// Use o f b r e a k s t a t e m e n t
13
end
14
cost1 = cost ;
15
p1 = p ;
16 end
17 p =( p + p1 ) /2.0;
18 cost =40 -8* p + p ^2;
// Computes t h e c o s t
19 // p r i n t t h e r e s u l t
44
20
printf ( ”MINIMUM COST=%. 2 f AT p=%. 1 f \n ” , cost , p ) ;
Scilab code Exa 6.4 Read the marks and print total marks
//
Example 6 . 4
//A c l a s s o f n s t u d e n t s t a k e an a n n u a l e x a m i n a t i o n
in m subjects .
3 // A program t o r e a d t h e marks o b t a i n e d by e a c h
student in various subjects
4 // and t o compare and p r i n t t h e t o t a l marks o b t a i n e d
by e a c h o f them .
1
2
5
6 FIRST =360; SECOND =240;
7 disp ( ” E n t e r number o f s t u d e n t s and s u b j e c t s ” ) ;
8 [n , m ]= scanf ( ”%d %d” ) ;
9 for i =1: n
10
roll_number = input ( ” E n t e r r o l l n u m b e r : ” ) ;
11
total =0;
12
printf ( ” E n t e r marks o f %d s u b j e c t s f o r ROLL NO
13
14
15
16
17
18
19
20
21
22
23
24
25
%d” ,m , roll_number ) ;
printf ( ” [ E n t e r e a c h i n n e w l i n e ] ” ) ;
for j =1: m
marks = scanf ( ”%d” ) ;
total = total + marks ;
// Compute t h e t o t a l
end
// p r i n t t h e t o t a l marks
printf ( ”TOTAL MARKS =%d” , total ) ;
// T e s t
f o r d i v i s i o n and d i s p l a y i t
if ( total >= FIRST ) ,
disp ( ” F i r s t D i v i s i o n ” ) ;
elseif ( total >= SECOND )
disp ( ” S e c o n d D i v i s i o n ” ) ;
else
45
26
disp ( ” ∗∗∗F A I L ∗∗∗ ” )
27
end
28 end
Scilab code Exa 6.4cs Case study 4 Plotting of two Functions
1
2
//
//
Case Study :− C h a p t e r 6
4 . P l o t t i n g o f two F u n c t i o n s i . e . y1=exp (−ax )
and y2=exp (−ax ˆ 2 / 2 )
3
4 a =0.4;
5 printf ( ”
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
y−−−−−−−>
\n ” ) ;
printf ( ” 0
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\
n”);
for x =0:0.25:4
// E v a l u a t i o n o f f u n c t i o n s
y1 = int32 (50* exp ( - a * x ) +0.5) ;
y2 = int32 (50* exp ( - a *( x ^2) /2) +0.5) ;
// p l o t t i n g when y1=y2
if ( y1 == y2 ) ,
if ( x ==2.5) ,
printf ( ” x | ” ) ;
else
printf ( ” | ” ) ;
end
for i =1:( y1 -1)
printf ( ” ” ) ;
end
printf ( ”#\n ” ) ;
continue ;
end
46
24
25
// P l o t t i n g when y1>y2
26
if ( y1 > y2 ) ,
27
if ( x ==2.5) ,
28
printf ( ” x | ” ) ;
29
else
30
printf ( ” | ” ) ;
31
end
32
for i =1: y2 -1
33
printf ( ” ” ) ;
34
end
35
printf ( ” ∗ ” ) ;
36
for i =1:( y1 - y2 -1)
37
printf ( ”−” ) ;
38
end
39
printf ( ” 0\ n ” ) ;
40
continue ;
41
end
42
43
// P l o t t i n g when y2>y1
44
if ( y2 > y1 ) ,
45
if ( x ==2.5)
46
printf ( ” x | ” ) ;
47
else
48
printf ( ” | ” ) ;
49
end
50
for i =1:( y1 -1)
51
printf ( ” ” ) ;
52
end
53
printf ( ” 0 ” ) ;
54
for i =1:( y2 - y1 -1)
55
printf ( ”−” ) ;
56
end
57
printf ( ” ∗\ n ” ) ;
58
end
59 end
60
printf ( ” | \ n ” ) ;
47
Scilab code Exa 6.5 Use of the break statement
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
//
Example 6 . 5
// Program i l l u s t r a t e u s e o f t h e b r e a k s t a t e m e n t
disp ( ” T h i s program c o m p u t e s t h e a v a r a g e o f s e t o f
numbers ” ) ;
disp ( ” E n t e r v a l u e s and e n t e r a NEGATIVE v a l u e a t t h e
end ” ) ;
sum1 =0;
for m =1:1000
x = scanf ( ” %f ” ) ;
// Read d a t a
if (x <0) then
break ;
//EXIT FROM LOOP
end
sum1 = sum1 + x ;
// Computes sum
end
average = sum1 /( m -1) ; // Computes A v e r a g e
// P r i n t t h e r e s u l t s
printf ( ”Number o f v a l u e s =%d\n ” ,m -1) ;
printf ( ” sum1=%f\n ” , sum1 ) ;
printf ( ” Avarage =%f\n ” , average ) ;
Scilab code Exa 6.6 Evaluate the series
1
2
3
//
Example 6 . 6
// Program t o e v a l u a t e t h e s e r i e s i . e .
// 1/1−x = 1+x+xˆ2+x ˆ 3 + . . . . . + x ˆn
48
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
x = input ( ” I n p u t v a l u e o f x : ” ) ;
LOOP =100; ACCURACY =0.0001;
sum1 =0; term =1; flag =0;
// Computation u s i n g f o r l o o p
for n =1: LOOP
sum1 = sum1 + term ;
if ( term <= ACCURACY ) then
flag =1;
break ;
end
term = term * x ;
// Read v a l u e o f x
// I n i t i a l i z a t i o n
// T e s t f o r a c c u r a c y
end
// P r i n t t h e r e s u l t s
if ( flag ==1) then
printf ( ” EXIT FROM LOOP\n ” ) ;
printf ( ” Sum =%f ; No . o f t e r m s =%d” , sum1 , n ) ;
else
disp ( ”FINAL VALUE OF N I S NOT SUFFICIENT TO
ACHIEVE DESIRED ACCURCY” ) ;
24 end
Scilab code Exa 6.7 Use of continue statement
1
2
3
4
5
6
7
//
Example 6 . 7
// The program i l l u s t r a t e t h e u s e o f c o n t i n u e
statement
disp ( ” E n t e r 9 9 9 9 t o STOP” ) ;
count =0;
negative =0;
while ( count <=100)
49
8
number = input ( ” E n t e r a number : ” ) ;
9
if ( number ==9999) then
10
break ;
//EXIT FROM THE LOOP
11
end
12
if ( number <0) ,
13
disp ( ”Number i s n e g a t i v e ” ) ;
14
negative = negative +1;
15
continue ;
// SKIP REST OF LOOP
16
end
17
sqrot = sqrt ( number ) ; //COMPUTE SQUARE ROOT
18
printf ( ”Number = %f\n ” , number ) ;
19
printf ( ” S q u a r e r o o t = %f ” , sqrot ) ;
20
count = count +1;
21 end
22 //PRINT RESULTS
23 printf ( ”Number o f i t e m s done = %d\n ” , count ) ;
24 printf ( ” N e g a t i v e i t e m s = %d\n ” , negative ) ;
25 disp ( ”END OF DATA” ) ;
50
Chapter 7
Arrays
Scilab code Exa 1.cs Case study 1 Median of list of numbers
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
//
//
Case Study : C h a p t e r 7 , Page No : 2 1 0
1 . Median o f l i s t o f numbers
N =10;
disp ( ” E n t e r t h e number o f i t e m s ” ) ;
n = scanf ( ”%d” ) ;
// R e a d i n g i t e m s i n t o a r r a y a
printf ( ” I n p u t %d v a l u e s [ One a t a t i m e ] \ n ” ,n ) ;
for i =1: n
a ( i ) = scanf ( ” %f ” ) ;
end
// S o r t i n g b e g i n s
for i =1: n -1
for j =1: n - i
if ( a ( j ) <= a ( j +1) )
t=a(j);
a ( j ) = a ( j +1) ;
a ( j +1) = t ;
else
continue ;
end
51
22
23
24
25
26
27
28
29
30
31
32
33
34
end
end // s o r t i n g e n d s
// C a l c u l a t i o n o f median
if ( n /2==0) then
median1 =( a ( n /2) + a ( n /2+1) ) /2.0;
else
median1 = a ( n /2+1) ;
end
// P r i n t i n g
for i =1: n
printf ( ” %f ” ,a ( i ) ) ;
end
printf ( ” \ nMedian i s %f\n ” , median1 ) ;
Scilab code Exa 2.cs Case study 2 Calculation of standard deviation
1 //
Case Study : Chapter −7
2 //
2. Calculation of standard deviation
3
4 MAXSIZE =100;
5 sum1 =0; n =0; sumsqr =0;
6 disp ( ” I n p u t v a l u e s : i n p u t −1 t o end ” ) ;
7 for i =1: MAXSIZE
8
value ( i ) = scanf ( ” %f ” ) ; // E n t e r i n g v a l u e s i n t h e
a r r a y named v a l u e
if ( value ( i ) == -1)
break ;
end
sum1 = sum1 + value ( i ) ;
n = n +1;
9
10
11
12
13
14 end
15 mean1 = sum1 / n ;
16 for i =1: n
// Computes mean
52
17
deviation = value ( i ) - mean1 ;
18
sumsqr = sumsqr + deviation ^2;
19 end
20 variance1 = sumsqr / n ;
21 stddeviation = sqrt ( variance1 ) ;
22
23
24
25
// Computes
standard deviation
// P r i n t i n g i t e m s , Mean and S t a n d a r d d e v i a t i o n
printf ( ”Number o f i t e m s : %d\n ” ,n ) ;
printf ( ”Mean : %f\n ” , mean1 ) ;
printf ( ” S t a n d a r d d e v i a t i o n : %f\n ” , stddeviation ) ;
Scilab code Exa 3.cs Case study 3 Evaluating a Test
1 //
Case Study : Chapter −7
2 //
3 . Evaluating a Test
3
4 STUDENTS =3;
5 ITEMS =5;
6 // R e a d i n g o f c o r r e c t a n s w e r s
7 printf ( ” I n p u t key t o t h e i t e m s \n ” ) ;
8 for i =1: ITEMS
9
key ( i ) = read ( %io (1) ,1 ,1 , ’ ( a ) ’ ) ; // Read d a t a u s i n g
10
read function
// key ( i )=s c a n f ( ” %c ” ) ; I t can be u s e d t o r e a d
data
11 end
12 // E v a l u a t i o n b e g i n s
13
14 for student =1: STUDENTS
15
// R e a d i n g s t u d e n t s r e s p o n s e s and c o u n t i n g
16
17
c o r r e c t ones
count =0;
printf ( ” \ n I n p u t r e s p o n s e s o f s t u d e n t −%d” , student
53
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
);
for i =1: ITEMS
response ( i ) = scanf ( ”%c” ) ;
end
correct = zeros (1 , ITEMS )
// Commented c o d e can be u s e d t o r e p l a c e a b o v e l i n e i
. e . c o r r e c t =z e r o s ( 1 , ITEMS )
// f o r i =1:ITEMS
//
c o r r e c t ( i ) =0;
// end
for i =1: ITEMS
if ( response ( i ) == key ( i ) ) then
count = count +1;
correct ( i ) =1;
end
end
// P r i n t i n g o f r e s u l t s
printf ( ” S t u d e n t −%d\n ” , student ) ;
printf ( ” S c o r e i s %d o u t o f %d\n ” , count , ITEMS ) ;
printf ( ” R e s p o n s e t o t h e i t e m s b e l o w a r e wrong \n ”
);
n =0;
for i =1: ITEMS
if ( correct ( i ) ==0)
printf ( ” %d” ,i ) ;
n = n +1;
end
end
if ( n ==0) then
printf ( ” NIL\n ” ) ;
end
end
54
Scilab code Exa 4.cs Case study 4 Production and sales analysis
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
//
//
Case Study : Chapter −7
4 . P r o d u c t i o n and s a l e s a n a l y s i s
// I n p u t Data
disp ( ” E n t e r p r o d u c t s m a n u f a c t u r e d w e e k w i s e ” ) ;
disp ( ”M11 , M12,−−,M21 , M22,−− e t c ” ) ;
for i =1:2
for j =1:5
M (i , j ) = scanf ( ”%d” ) ;
end
end
disp ( ” E n t e r p r o d u c t s s o l d w e e k w i s e ” ) ;
disp ( ” S11 , S12 ,−−, S21 , S22 ,−− e t c ” ) ;
for i =1:2
for j =1:5
S (i , j ) = scanf ( ”%d” ) ;
end
end
disp ( ” E n t e r c o s t o f e a c h p r o d u c t ” ) ;
for j =1:5
C ( j ) = scanf ( ”%d” ) ;
end
// V a l u e s m a t r i c e s o f p r o d u c t i o n and s a l e s
for i =1:2
for j =1:5
Mvalue (i , j ) = M (i , j ) * C ( j ) ;
Svalue (i , j ) = S (i , j ) * C ( j ) ;
end
end
// T o t a l v a l u e o f w e e k l y p r o d u c t i o n and s a l e s
for i =1:2
Mweek ( i ) =0;
Sweek ( i ) =0;
for j =1:5
Mweek ( i ) = Mweek ( i ) + Mvalue (i , j ) ;
Sweek ( i ) = Sweek ( i ) + Svalue (i , j ) ;
55
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
end
end
// Monthly v a l u e o f p r o d u c t w i s e p r o d u c t i o n and s a l e s
for j =1:5
Mproduct ( j ) =0;
Sproduct ( j ) =0;
for i =1:2
Mproduct ( j ) = Mproduct ( j ) + Mvalue (i , j ) ;
Sproduct ( j ) = Sproduct ( j ) + Svalue (i , j ) ;
end
end
// Grand t o t a l o f p r o d u c t i o n and s a l e s v a l u e s
Mtotal =0; Stotal =0;
for i =1:2
Mtotal = Mtotal + Mweek ( i ) ;
Stotal = Stotal + Sweek ( i ) ;
end
// ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
// S e l e c t i o n and p r i n t i n g o f i n f o r m a t i o n r e q u i r e d
// ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
disp ( ” F o l l o w i n g i s t h e l i s t o f t h i n g s you r e q u e s t
f o r ”);
59 disp ( ” E n t e r a p p r o p r i a t e number and p r e s s r e t u r n key ”
);
60
61
62
disp ( ” 1 . V a l u e m a t r i c e s o f p r o d u c t i o n and s a l e s ” ) ;
disp ( ” 2 . T o t a l v a l u e o f w e e k l y p r o d u c t i o n and s a l e s ” )
;
63 disp ( ” 3 . P r o d u c t i o n w i s e monthly v a l u e o f p r o d u c t i o n
and s a l e s ” ) ;
64 disp ( ” 4 . Grand t o t a l v a l u e o f p r o d u c t i o n and s a l e s ” ) ;
65 disp ( ” 5 . E x i t ” )
66
67
68
69
70
number =0;
while (1)
// B e g i n i n g o f w h i l e l o o p
number = input ( ”ENTER YOUR CHOICE : ” ) ;
56
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
if ( number ==5) then
disp ( ”Good Bye ” ) ;
break ;
end
select number
// V a l u e M a t i c e s
case 1 then
disp ( ”VALUE MATRIX OF PRODUCTION” ) ;
for i =1:2
printf ( ”Week (%d) \ t ” ,i ) ;
for j =1:5
printf ( ”%7d” , Mvalue (i , j ) ) ;
end
printf ( ” \n ” ) ;
end
disp ( ”VALUE MATRIX OF SALES” ) ;
for i =1:2
printf ( ”Week (%d) \ t ” ,i ) ;
for j =1:5
printf ( ”%7d” , Svalue (i , j ) ) ;
end
printf ( ” \n ” ) ;
end
// Weekly A n a l y s i s
case 2 then
disp ( ”TOTAL WEEKLY PRODUCTION AND SALES” ) ;
disp ( ”
PRODUCTION
SALES” ) ;
disp ( ”
−−−−−−−−−−
−−−−−−” ) ;
for i =1:2
printf ( ”Week (%d) \ t ” ,i ) ;
printf ( ”%7d\ t%9d \n ” , Mweek ( i ) , Sweek ( i ) ) ;
end
// P r o d u c t w i s e A n a l y s i s
case 3 then
disp ( ”PRODUCTWISE TOTAL PRODUCTION AND SALES
”);
disp ( ”
PRODUCTION
SALES” ) ;
disp ( ”
−−−−−−−−−−
−−−−−−” ) ;
57
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
for i =1:5
printf ( ” P r o d u c t (%d) \ t ” ,i ) ;
printf ( ”%7d\ t%7d \n ” , Mproduct ( i ) , Sproduct
(i));
end
// Grand T o t a l s
case 4 then
disp ( ”GRAND TOTAL OF PRODUCTION AND SALES” ) ;
printf ( ” T o t a l p r o d u c t i o n = %d\n ” , Mtotal ) ;
printf ( ” T o t a l s a l e s = %d\n ” , Stotal ) ;
// D e f a u l t
else
printf ( ”Wrong c h o i c c e , s e l e c t a g a i n \n ” ) ;
end // End o f s e l e c t
end // End o f w h i l e
disp ( ” E x i t from t h e program ” ) ;
Scilab code Exa 7.1 Sum of squares of 10 numbers
//
Example : 7 . 1
// W r i t e a program u s i n g s i n g l e −s u b s c r i p t e d v a r i a b l e
to evaluate :
3 // sum o f s q u a r e s o f 10 numbers . The v a l u e s x1 , x2 , . . .
a r e r e a d from t h e t e r m i n a l .
1
2
4
5 // R e a d i n g v a l u e s i n t o a r r a y
6 disp ( ”ENTER 10 REAL NUMBERS[ Each i n n e w l i n e ] ” ) ;
7 total =0;
8 for i =1:10
9
x ( i ) = input ( ” ” ) ;
10
total = total + x ( i ) ^2; // Computation o f t o t a l
11 end
12 // P r i n t i n g o f x ( i ) v a l u e s and t o t a l
58
13 for i =1:10
14
printf ( ” x ( %2d ) =%5 . 2 f \n ” ,i , x ( i ) ) ;
15 end
16 printf ( ” T o t a l =%. 2 f ” , total ) ;
Scilab code Exa 7.2 Count the number of students
1
2
3
4
5
6
7
//
Example : 7 . 2
// Given b e l o w i s t h e l i s t o f marks o b t a i n e d by a
c l a s s o f 50 s t u d e n t s i n an
// a n n u a l e x a m i n a t i o n . 43 65 51 27 79 11 56 61 82 09
25 36 07 49 55 63 74 81 49
// 37 40 49 16 75 87 91 33 24 58 78 65 56 76 67 45 54
36 63 12 21 73 49 51 19 39
// 49 68 93 85 59
// W r i t e a program t o c o u n t t h e number o f s t u d e n t s
b e l o n g i n g to each o f
// f o l l o w i n g g r o u p s o f marks
:0 −9 ,10 −19 ,20 −29 ,.......100.
8
9
// T h i s program c o m p u t e s f o r 10 s t u d e n t s . We c o u l d
compute f o r 50 s t u d e n t s by
10 // c h a n g i n g MAXVAL=50.
11
12 MAXVAL =10; COUNTER =11;
13 disp ( ” I n p u t Data [ Marks o f 10 s t u d e n t s ] ” ) ;
14 group1 = zeros (1 ,11) ;
15 // R e a d i n g and c o u n t i n g
16 for i =1: MAXVAL
17
// R e a d i n g o f v a l u e s
18
value ( i ) = input ( ” ” ) ;
19
// C o u n t i n g f r e q u e n c y o f g r o u p s
20
a = int16 (( value ( i ) /10) ) ;
59
21
if ( a ==0) then
22
group1 ( a +1) = group1 ( a +1) +1;
23
else
24
group1 ( a +1) = group1 ( a +1) +1;
25
end
26
27 end
28 // P r i n t i n g o f f r e q u e n c y t a b l e
29 printf ( ” Group
Range
F r e q u e n c y \n ” ) ;
30 for i =0: COUNTER -1
31
if ( i ==0) ,
32
low =0;
33
else
34
low = i *10;
35
end
36
if ( i ==10) ,
37
high =100;
38
else
39
high = low +9;
40
end
41
printf ( ”%2d %8d t o %3d %5d\n ” ,i +1 , low , high ,
group1 ( i +1) ) ;
42 end
Scilab code Exa 7.3 Compute and print
1
2
3
4
5
6
//
Example : 7 . 3
// W r i t e a program u s i n g two d i m e n s i o n a l a r r a y t o
compute p r i n t f o l l o w i n g
// i n f o r m a t i o n from t h e t a b l e o f d a t a d i s c u s s e d :
// ( a ) T o t a l v a l u e o f s a l e s by e a c h g i r l .
// ( b ) T o t a l v a l u e o f e a c h i t e m s o l d
// ( c ) Grand t o t a l o f a l l s a l e s o f a l l i t e m s by a l l
60
girls .
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
MAXGIRLS =4; MAXITEMS =3;
frequency = zeros (1 ,5) ;
disp ( ” I n p u t d a t a ” ) ;
// R e a d i n g v a l u e s and c o m p u t i n g g i r l t o t a l
disp ( ” E n t e r v a l u e s , one a t a t i m e ” ) ;
for i =1: MAXGIRLS
girl_total ( i ) =0;
for j =1: MAXITEMS
value (i , j ) = scanf ( ”%d” ) ;
girl_total ( i ) = girl_total ( i ) + value (i , j ) ;
end
end
// Computing i t e m t o t a l
for j =1: MAXITEMS
item_total ( j ) =0;
for i =1: MAXGIRLS
item_total ( j ) = item_total ( j ) + value (i , j ) ;
end
end
// Computing g r a n d t o t a l
grand_total =0;
for i =1: MAXGIRLS
grand_total = grand_total + girl_total ( i ) ;
end
// P r i n t i n g o f r e s u l t
disp ( ”GIRLS TOTALS” ) ;
for i =1: MAXGIRLS
printf ( ” S a l e s g i r l (%d)=%d\n ” ,i , girl_total ( i ) ) ;
end
disp ( ”ITEM TOTALS” ) ;
for j =1: MAXITEMS
printf ( ” Item (%d)=%d\n ” ,j , item_total ( j ) ) ;
end
printf ( ” Grand T o t a l=%d\n ” , grand_total ) ;
61
Scilab code Exa 7.4 Multiplication table
1
2
//
Example : 7 . 4
// W r i t e a program t o compute and p r i n t a
m u l t i p l i c a t i o n t a b l e f o r numbers 1 t o 5
3
4 ROWS =5; COLUMNS =5;
5 disp ( ”MULTIPLICATION TABLE” ) ;
6 printf ( ” ∗ | ” )
7 for j =1: COLUMNS
8
printf ( ”%4d” ,j ) ;
9 end
”);
10 disp ( ”
11 for i =1: ROWS
12
printf ( ”%1d | ” ,i ) ;
13
for j =1: COLUMNS
14
product (i , j ) = i * j ;
// C a l c u l a t e t h e
product
15
printf ( ”%4d” , product (i , j ) ) ;
16
end
17
printf ( ” \n ” ) ;
18 end
// P r i n t t h e p r o d u c t
Scilab code Exa 7.5 Popularity of various cars
1
2
//
Example : 7 . 5
//A s u r v e y t o know t h e p o p u l a r i t y o f f o u r c a r s (
Ambassador , f a i t , D o l p h i n and
62
// M a r u t i ) was c o n d u c t e d i n f o u r c i t i e s ( Bombay ,
C a l c u t t a , D e l h i and Madras ) .
4 // Each p e r s o n s u r v e y e d was a s k e d t o g i v e h i s c i t y
and t y p e o f c a r he was u s i n g .
5 // W r i t e a program t o p r o u c e a t a b l e s h o w i n g t h e
popularity of various cars in
6 // f o u r c i t i e s .
3
7
8
9
frequency = zeros (5 ,5) ;
printf ( ” For e a c h p e r s o n , e n t e r t h e c i t y c o d e [ B , C , D,M
] \ n”);
10 printf ( ” f o l l o w e d by t h e c a r c o d e [ 1 , 2 , 3 , 4 ] . \ n ” ) ;
11 printf ( ” E n t e r t h e l e t t e r X 0 ( z e r o ) t o i n d i c a t e end . \ n
”);
12
13 // T a b u l a t i o n b e g i n s
14 for i =1:99
15
[n , city , car ]= mscanf ( ”%c %d” ) ;
16
if ( city == ’X ’ ) then
17
break ;
18
end
19
select city
20
case ’B ’ then frequency (1 , car ) = frequency (1 , car )
21
22
23
+1;
case ’C ’ then frequency (2 , car ) = frequency (2 , car )
+1;
case ’D ’ then frequency (3 , car ) = frequency (3 , car )
+1;
case ’M’ then frequency (4 , car ) = frequency (4 , car )
+1;
end
24
25
26 end
27
28 // T a b u l a t i o n c o m p l e t e d and P r i n t i n g b e g i n s
29 disp ( ”
POPULATORY TABLE” ) ;
30 printf ( ”
\n ”
63
);
31 printf ( ” C i t y
\n ” ) ;
32 printf ( ”
Ambasseador
fait
Dolphin
Maruti
\n ”
);
33
34 for i =1:4
35
select i
36
case 1 then printf ( ” Bombay ” ) ;
37
case 2 then printf ( ” C a l c u t t a ” ) ;
38
case 3 then printf ( ” D e l h i
”);
39
case 4 then printf ( ” Madras ” ) ;
40
end
41
for j =1:4
42
printf ( ”%8d” , frequency (i , j ) ) ;
43
end
44
printf ( ” \n ” ) ;
45 end
46 printf ( ”
”);
47
//
P r i n t i n g ends
64
Chapter 8
Character Arrays and Strings
Scilab code Exa 1.cs Case study 1 Counting words in a text
1
Csae s t u d y : Chapter −8 , Page No
//
:253
2 //
1 . C o u n t i n g words i n a t e x t
3
4 characters =0; words =0; lines1 =0;
5 printf ( ”
KEY IN THE TEXT. \ n ” ) ;
6 printf ( ”GIVE ONE SPACE AFTER EACH WORD. \ n ” ) ;
7 printf ( ”WHEN COMPLETED, ENTER end \n ” ) ;
8
9 l= ’ ’ ;
10 while ( l ~= ’ end ’ )
11
l = read ( %io (1) ,1 ,1 , ’ ( a ) ’ ) ; // R e a d i n g a l i n e o f
12
13
14
15
16
17
text
if ( l == ’ end ’ ) then
break ;
end
line =[ ascii ( l ) ];
values of line l
len = length ( line ) ;
line
for i =1: len
// Array o f a s c i i
// compute l e n g t h o f
65
// a s c i i v a l u e o f ’ ’ ( i . e . s p a c e ) i s 32
if ( line ( i ) ==32) then
words = words +1;
// Count t h e number
o f words
end
18
19
20
21
22
23
end
lines1 = lines1 +1;
// Count t h e number
of l i n e s
characters = characters + len ; // Count t h e number
of characters
24
25 end
26 // P r i n t i n g r e s u l t s
27 printf ( ”Number o f l i n e s = %d\n ” , lines1 ) ;
28 // Number o f l i n e s a r e added t o words b e c a u s e
word o f e a c h
29 // u n c o u n t e d a s
occurence of
30 printf ( ”Number
31 printf ( ”Number
last
l i n e remains−
words a r e i n c r e m e n t e d a t t h e
space .
o f words = %d\n ” , words + lines1 ) ;
o f c h a r a c t e r s = %d\n ” , characters ) ;
Scilab code Exa 2.cs Case study 2 Processing of a customer list
1
//
2
3
4
5
6
7
//
Csae s t u d y : Chapter −8 , Page No
:253
2 . Processing of a customer l i s t
CUSTOMERS =10;
printf ( ”
I n p u t names and t e l e p h o n e numbers \n ” ) ;
printf ( ” [ Names must have F i r s t , S e c o n d and L a s t n a m e
] \ n”);
8 for i =1: CUSTOMERS
9
// Read d a t a
66
10
11
12
[ first_name ( i ) , second_name ( i ) , surname ( i ) ,
telephone ( i ) ]= scanf ( ”%s %s %s %s” ) ;
// C o n v e r t i n g f u l l name t o surname w i t h i n i t i a l s
l1 = length ( surname ( i ) ) ;
// Compute
l e n g t h o f surname a t i
name ( i ) = strncpy ( surname ( i ) , l1 ) ;
name ( i ) = strcat ([ name ( i ) , ’ , ’ ]) ;
dummy (1) = part ( first_name (i ,1) ,1) ;
name ( i ) = strcat ([ name ( i ) , dummy ]) ;
name ( i ) = strcat ([ name ( i ) , ’ . ’ ]) ;
dummy (1) = part ( second_name (i ,1) ,1) ;
name ( i ) = strcat ([ name ( i ) , dummy ]) ;
13
14
15
16
17
18
19
20 end
21 // A l p h a b e t i c a l o d e r i n g o f s u r n a m e s
22 for i =1: CUSTOMERS
// Outer l o o p
23
begins
for j =2: CUSTOMERS - i +1
// I n n e r l o o p
begins
k = strcmp ( name (j -1) , name ( j ) ) ;
if (k >0) then
24
25
26
27
// Swaping names
28
l1 = length ( name (j -1) ) ;
29
l2 = length ( name ( j ) ) ;
30
dummy = strncpy ( name (j -1) , l1 ) ;
31
name (j -1) = strncpy ( name ( j ) , l2 ) ;
32
l3 = length ( dummy ) ;
33
name ( j ) = strncpy ( dummy , l3 ) ;
34
35
// Swapping t e l e p h o n e numbers
36
l3 = length ( telephone (j -1) ) ;
37
l4 = length ( telephone ( j ) ) ;
38
dummy = strncpy ( telephone (j -1) , l3 ) ;
39
telephone (j -1) = strncpy ( telephone ( j ) , l4 ) ;
40
telephone ( j ) = strncpy ( dummy , l3 ) ;
41
end
42
end // I n n e r l o o p e n d s
43 end
// Outer l o o p e n d s
67
44 // P r i n t i n g a l p h a b e t i c a l l i s t
45 disp ( ”CUSTOMER LIST IN ALPHABETICAL ORDER” ) ;
46 for i =1: CUSTOMERS
47
printf ( ”%−20 s \ t %−10 s \n ” , name ( i ) , telephone ( i ) ) ;
48 end
Scilab code Exa 8.1 Read a series of words
1
2
//
Exampple 8 . 1
// W r i t e a program t o r e a d a s e r i e s o f words from
terminal using scanf function .
3
4 // Read d a t a u s i n g s c a n f f u n c t i o n
5 disp ( ” E n t e r t e x t : ” )
6 [ word1 , word2 , word3 , word4 ]= scanf ( ”%s %s %s %s” ) ;
7 // P r i n t i n g t h e r e s u l t s
8 printf ( ” word1 = %s\ nword2 = %s\n ” , word1 , word2 ) ;
9 printf ( ” word3 = %s\ nword4 = %s\n ” , word3 , word4 ) ;
Scilab code Exa 8.2 Read a line of text
//
Example 8 . 2
// W r i t e a program t o r e a d a l i n e o f t e x t c o n t a i n i n g
a s e r i e s of−
3 // words from t h e t e r m i n a l .
1
2
4
5
6
7
disp ( ” E n t e r t e x t . P r e s s <Return > a t end ” ) ;
line = read ( %io (1) ,1 ,1 , ’ ( a ) ’ ) ;
// Read a l i n e
disp ( line ) ;
// D i s p l a y l i n e
68
Scilab code Exa 8.3 Copy one string into another
//
Example 8 . 3
// W r i t e a program t o copy one s t r i n g i n t o a n o t h e r
and c o u n t t h e number
3 // o f c h a r a c t e r s c o p i e d .
1
2
4
5
6
7
8
9
10
11
12
13
14
15
// Read d a t a u s i n g s c a n f f u n c t i o n
disp ( ” E n t e r a s t r i n g : ” )
[ string2 ]= scanf ( ”%s” ) ;
// Read s t r i n g
l = length ( string2 ) ;
// Compute t h e l e n g t h
string1 = ’ ’ ;
// s t r i n g 1 i s empty
for i =1: l
string1 = string1 + part ( string2 , i ) ;
end
// P r i n t i n g t h e r e s u l t s
printf ( ” %s\n ” , string1 ) ;
printf ( ” Number o f c h a r a c t e r s = %d\n ” ,l ) ;
Scilab code Exa 8.4 Display the string under various format specifications
//
Exampple 8 . 4
// W r i t e a program t o s t o r e t h e s t r i n g ” U n i t e d
Kingdom ” i n t h e a r r a y c o u n t r y −
3 // and d i s p l a y t h e s t r i n g u n d e r v a r i o u s f o r m a t
specifications .
1
2
4
69
5
6
7
8
9
10
11
12
13
14
15
16
17
country = ’ U n i t e d Kingdom ’ ;
printf ( ” \n ” ) ;
printf ( ” ∗ 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 ∗ \ n ” ) ;
printf ( ”−−−−−−−−\n ” ) ;
printf ( ” %15s \n ” , country ) ;
printf ( ” %5s\n ” , country ) ;
printf ( ”%15 . 6 s \n ” , country ) ;
printf ( ”%−15.6 s \n ” , country ) ;
printf ( ”%15 . 0 s \n ” , country ) ;
printf ( ”%. 3 s \n ” , country ) ;
printf ( ”%s\n ” , country ) ;
printf ( ”−−−−−−−−\n ” ) ;
Scilab code Exa 8.5 Program using for loop
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
//
Example 8 . 5
// W r i t e a program u s i n g f o r l o o p t o p r i n t t h e
f o l l o w i n g output :
//
C
//
CP
//
....
//
....
//
CProgrammimg
//
CProgrammimg
//
....
//
....
//
CPr
//
CP
//
C
string1 = ’ CProgramming ’ ;
printf ( ” −−−−−−−−−−−−−\n ” ) ;
f= ’ ’ ;
70
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
for i =1:12
f = f + part ( string1 , i ) ;
printf ( ” | %−13 s | \ n ” ,f ) ;
end
printf ( ”|−−−−−−−−−−−−−|\n ” ) ;
for j =0:11
s= ’ ’ ;
for i =1:12 - j
s = s + part ( string1 , i ) ;
end
printf ( ” | %−13 s | \ n ” ,s ) ;
end
printf ( ” −−−−−−−−−−−−−” ) ;
// f o r c =0:11
//
d=c +1;
//
m p r i n t f ( ” | %−12.∗ s | \ n ” , d , s t r i n g 1 ) ;
// end
// d i s p(”−−−−−−−−−−−−−−−−−−−−−”) ;
// f o r c =11: c −1:0
//
d=c +1;
//
p r i n t f ( ” | %−12.∗ s | \ n ” , d , s t r i n g 1 ) ;
// end
// d i s p(”−−−−−−−−−−−−−−−−−−−−−−”) ;
Scilab code Exa 8.6 Print the alphabet set a to z and A to Z
//
Example 8 . 6
// W r i t e a program which would p r i n t t h e a l p h a b e t s e t
a to z A to Z in decimal −
3 // c h a r a c t e r form .
1
2
4
5 for c =65:122
71
6
7
8
9
10
11
12
if (c >90& c <97) then
continue ;
// T e r m i n a t e c u r r e n t
iteration
end
c1 = ascii ( c ) ;
// C o n v e r t a s c i i v a l u e t o
character
printf ( ” | %4d − %c \ ” , c , c 1 ) ;
end
p r i n t f ( ” |\ n ” )
Scilab code Exa 8.7 Concatinate the three parts into one string
//
Example 8 . 7
// The name o f e m p l o y e e s o f an o r g a n i z a t i o n a r e
s t o r e d i n t h r e e a r r a y s namely−
3 // f i r s t n a m e , s e c o n d n a m e and l a s t n a m e .
4 // W r i t e a program t o c o n c a t i n a t e t h e t h r e e p a r t s
i n t o one s t r i n g c a l l e d name .
1
2
5
6
7 // S t o r e t h e name i n t h e t h r e e a r r a y s
8 first_name =[ ’VISWANATH ’ ];
9 second_name =[ ’PRATAP ’ ];
10 last_name =[ ’ SINGH ’ ];
11
12 // C o n c a t i n a t e t h r e e p a r t s i n t o one
13 name =[ first_name second_name last_name ];
14 // P r i n t t h e r e s u l t
15 for i =1:3
16
printf ( ”%s ” , name ( i ) ) ;
17 end
18 // S t a t e m e n t b e l o w can a l s o be u s e d t o p r i n t t h e
result
72
19
// d i s p ( name ) ;
Scilab code Exa 8.8 Compare whether strings are equal
//
Example 8 . 8
// s1 , s 2 and s 3 a r e t h r e e s t r i n g v a r i a b l e s . W r i t e a
program t o r e a d two s t r i n g −
3 // c o n s t a n t s i n t o s 1 and s 2 and compare w h e t h e r t h e y
a r e e q u a l o r not , j o i n −
4 // them t o g e t h e r . Then copy c o n t e n t s o f s 1 t o
v a r i a b l e s 3 . At t h e end program−
5 // s h o u l d p r i n t a l l t h r e e v a r i a b l e s and t h e i r
lengths
1
2
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
// Read d a t a
printf ( ” E n t e r two s t r i n g c o n s t a n t s \n ” ) ;
[ s1 s2 ]= scanf ( ”%s %s” ) ;
// Comparing two s t r i n g s
x = strcmp ( s1 , s2 ) ;
if x ~=0 then
printf ( ” S t r i n g a r e n o t e q u a l \n ” ) ;
// C o n c a t i n a t e two s t r i n g s s 1 and s 2
s1 = strcat ([ s1 , s2 ]) ;
else
printf ( ” S t r i n g a r e e q u a l \n ” ) ;
end
l1 = length ( s1 ) ;
// Coping s 1 t o s 3
s3 = strncpy ( s1 , l1 ) ;
// f i n d i n g l e n g t h o f s t r i n g s
73
26 l2 = length ( s2 ) ;
27 l3 = length ( s3 ) ;
28 // Output
29 printf ( ” s 1 = %s\ t l e n g t h = %d c h a r a c t e r s \n ” ,s1 , l1 ) ;
30 printf ( ” s 2= %s\ t l e n g t h = %d c h a r a c t e r s \n ” ,s2 , l2 ) ;
31 printf ( ” s 3= %s\ t l e n g t h = %d c h a r a c t e r s \n ” ,s3 , l3 ) ;
Scilab code Exa 8.9 Sort a list of names in alphabetical order
1
2
//
Example 8 . 9
// W r i t e a program t h a t would s o r t a l i s t o f names i n
alphabetical order .
3
4 ITEMS =5;
5 // R e a d i n g t h e l i s t
6 printf ( ” E n t e r names o f %d i t e m s \n ” , ITEMS ) ;
7 i =1;
8 while (i <= ITEMS )
9
string1 ( i ) = scanf ( ”%s” ) ;
10
i = i +1;
11 end
12 // S o r t i n g b e g i n s
13 for i =1: ITEMS
// Outer l o o p b e g i n s
14
for j =2: ITEMS - i +1
// I n n e r l o o p b e g i n s
15
k = strcmp ( string1 (j -1) , string1 ( j ) )
16
if (k >0) then
17
// Compute l e n g t h and Exchange o f
18
19
20
21
22
contents
l1 = length ( string1 (j -1) ) ;
l2 = length ( string1 ( j ) ) ;
dummy = strncpy ( string1 (j -1) , l1 ) ;
string1 (j -1) = strncpy ( string1 ( j ) , l2 ) ;
l3 = length ( dummy ) ;
74
23
string1 ( j ) = strncpy ( dummy , l3 ) ;
24
end
25
end // I n n e r l o o p e n d s
26 end
// Outer l o o p e n d s
27 // S o r t i n g c o m p l e t e d
28 disp ( ” A l p h a b e t i c a l l i s t ” ) ;
29 for i =1: ITEMS
30
printf ( ”%s\n ” , string1 ( i ) ) ;
31 end
75
Chapter 9
User Defined Functions
Scilab code Exa 1.cs Case study 1 Calculation of Area under a Curve
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
//
Case Study : Chapter −9[ p a g e no : 3 1 0 ]
//
1 . C a l c u l a t i o n o f Area u n d e r a Curve
funcprot (0) ;
// g l o b a l v a r i a b l e s
global start_point
global end_point ;
global total_area ;
global numtraps ;
function []= input1 ()
global start_point ;
global end_point ;
global total_area ;
global numtraps ;
total_area =0;
start_point = input ( ” E n t e r l o w e r l i m i t : ” ) ;
end_point = input ( ” E n t e r u p p e r l i m i t : ” ) ;
numtraps = input ( ” E n t e r number o f t r a p e z o i d s : ” ) ;
endfunction
function [ total_area ]= find_area (a ,b , n )
global total_area ;
base =( b -1) / n ;
// b a s e i s l o c a l
76
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
variable
lower = a ;
// l o w e r i s l o c a l
variable
for lower = a :( lower + base ) :( b - base )
h1 = function_x ( lower ) ;
h2 = function_x ( lower + base ) ;
total_area = total_area + trap_area ( h1 , h2 , base ) ;
// t o t a l a r e a i s c a l c u l a t e d
end
endfunction
function [ area ]= trap_area ( height_1 , height_2 , base )
area =0.5*( height_1 + height_2 ) * base ;
// a r e a
is local variable
endfunction
function [ x ] = function_x ( x )
x =( x ^2) +1;
endfunction
// c a l l i n g f u n c t i o n s
disp ( ”AREA UNDER CURVE” ) ;
input1 () ;
// c a l l i n g i n p u t 1
() function
39 total_area = find_area ( start_point , end_point , numtraps )
; // c a l l i n g f i n d a r e a ( ) f u n c t i o n
40 printf ( ”TOTAL AREA = %f ” , total_area ) ;
Scilab code Exa 9.1 Multiple functions
//
Example 9 . 1
// W r i t e a program w i t h m u l t i p l e f u n c t i o n s t h a t do
n o t communicate−
3 // d a t a b e t w e e n them .
1
2
4
77
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
// F u n c t i o n 1 : p r i n t l i n e ( )
funcprot (0) ;
function []= printline ()
// c o n t a i n s no argument
for i =1:35
printf ( ”%c” , ’− ’ ) ;
end
printf ( ” \n ” ) ;
endfunction
// F u n c t i o n 2 : v a l u e ( )
function []= value ()
// c o n t a i n s no argument
principal = input ( ” P r i n c i p a l amount ? ” ) ;
inrate = input ( ” I n t e r e s t r a t e ? ” ) ;
period = input ( ” P e r i o d ? ” ) ;
sum1 = principal ;
year =1;
// Computation u s i n g w h i l e l o o p
while ( year <= period )
sum1 = sum1 *(1+ inrate ) ; // c a l c u l a t e s
p r i n c i p a l amount a f t e r c e r t a i n y e a r s
year = year +1;
end
printf ( ”%8 . 2 f %5 . 2 f %5d %12 . 2 f \n ” , principal ,
inrate , period , sum1 ) ;
endfunction
// C a l l i n g
functions
printline () ;
value () ;
printline () ;
Scilab code Exa 9.2 Include arguments in function calls
1
2
//
Example 9 . 2
// Modify Example 9 . 1 t o i n c l u d e a r g u m e n t s i n
78
function calls .
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
funcprot (0) ;
function []= printline ( ch )
// f u n c t i o n w i t h
argument ch
for i =1:52
printf ( ”%c” , ch ) ;
end
printf ( ” \n ” ) ;
endfunction
function []= value (p ,r , n )
// f u n c t i o n w i t h
argument p , r , n
sum1 = p ;
year =1;
while ( year <= n )
sum1 = sum1 *(1+ r ) ;
year = year +1;
end
printf ( ” %f\ t%f \t%d\ t%f \n ” ,p ,r ,n , sum1 ) ;
endfunction
printf ( ” E n t e r p r i n c i p a l amount , i n t e r e s t r a t e , and
p e r i o d \n [ E n t e r i n s i n g l e l i n e s e p e r a t e d by s p a c e
] ”);
[ principal , inrate , period ]= scanf ( ” %f %f %d” ) ; // r e a d
from s t a n d a r d i n p u t
// C a l l i n g
functions
printline ( ’ z ’ ) ;
value ( principal , inrate , period ) ;
printline ( ’ c ’ ) ;
Scilab code Exa 9.3 Return result
1
//
Example 9 . 3
79
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
// Modify Example 9 . 2 ’ s f u n c t i o n v a l u e ( ) t o r e t u r n
r e s u l t and e x t e n d v e r s a t i l i t y o f t h e f u n c t i o n
// p r i n t l i n e by t a k i n g t h e l e n t h o f l i n e a s an
argument
function []= printline ( ch , len )
for i =1: len
printf ( ”%c” , ch ) ;
end
printf ( ” \n ” ) ;
endfunction
function [ amount ]= value (p ,r , n )
// r e t u r n s amount
sum1 = p ;
year =1;
while ( year <= n )
sum1 = sum1 *(1+ r ) ;
year = year +1;
end
amount = sum1 ;
endfunction
printf ( ” E n t e r p r i n c i p a l amount , i n t e r e s t r a t e , and
p e r i o d \n [ E n t e r i n s i n g l e l i n e s e p e r a t e d by s p a c e ]
”);
[ principal , inrate , period ]= scanf ( ” %f %f %d” ) ;
// C a l l i n g
functions
printline ( ’ ∗ ’ ,52) ;
amount = value ( principal , inrate , period ) ;
printf ( ” %f\ t%f \t%d\ t%f \n ” , principal , inrate , period ,
amount ) ;
printline ( ’= ’ ,52) ;
Scilab code Exa 9.4 Computes x raised to the power y
1
//
Example 9 . 4
80
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
// W r i t e a program w i t h f u n c t i o n power t h a t c o m p u t e s
x r a i s e d t o t h e power y .
funcprot (0) ;
function p = power (x , y )
p =1.0;
// x t o power 0
if (( y >=0) ) then
while ( y )
// c o m p u t e s p o s i t i v e p o w e r s
p=p*x;
y =y -1;
end
else
while ( y )
// c o mp u t e s n e g a t i v e p o w e r s
p=p/x;
y = y +1;
end
end
endfunction
disp ( ” E n t e r x , y : ” ) ;
disp ( ” [ E n t e r i n s i n g l e l i n e s e p e r a t e d by s p a c e ] ” ) ;
[x , y ]= scanf ( ”%d %d” ) ;
// i n p u t u s i n g s c a n f
function
// c a l l i n g power ( ) f u n c t i o n and p r i n t i n g i t s o u t p u t
printf ( ”%d t o t h e power %d i s %f ” ,x ,y , power (x , y ) ) ;
Scilab code Exa 9.5 Calculate standard deviation of an array values
//
Example 9 . 5
// W r i t e a program t o c a l c u l a t e s t a n d a r d d e v i a t i o n o f
an a r r a y v a l u e s .
3 // Array e l e m e n t s a r e r e a d from t e r m i n a l . Use
f u n c t i o n s to c a l u l a t e −
4 // s t a n d a r d d e v i a t i o n and mean
1
2
81
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
funcprot (0) ;
// p a s s i n g a r r a y named ’ v a l u e ’ t o f u n c t i o n s t d d e v a t
’a ’
function [ std ]= std_dev (a , n )
sum1 =0;
x = mean1 (a , n ) ;
// c a l l i n g
mean1 ( ) f u n c t i o n
for i =1: n
sum1 = sum1 +( x - a ( i ) ) ^2;
std = sqrt ( sum1 / double ( n ) ) ;
// c o mp u t e s
standard deviation
end
endfunction
function [ x ]= mean1 (a , n )
sum1 =0;
for i =1: n
sum1 = sum1 + a ( i ) ;
x = sum1 / double ( n ) ;
// x c o n t a i n
mean v a l u e
end
endfunction
SIZE = int8 (5) ;
// s i z e o f
array
printf ( ” E n t e r %d f l o a t v a l u e s ” , SIZE ) ;
for i =1: SIZE
value ( i ) = input ( ” ” ) ;
// e n t e r i n g
values in the array
end
printf ( ” Std . d e v i a t i o n i s %f ” , std_dev ( value , SIZE ) ) ;
// c a l l i n g s t d d e v ( ) f u n c t i o n
Scilab code Exa 9.6 Sort an array
82
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
//
Example 9 . 6
// W r i t e a program t h a t u s e s a f u n c t i o n t o s o r t an
array of i n te ge rs .
funcprot (0) ;
function [ x ]= sort (m , x )
// P a s s i n g an a r r a y i .
e . marks t o f u n c t i o n s o r t ( )
for i =1: m
// i r e p e s e n t s number
of passes
for j =2: m - i +1
// j r e p r e s e n t s
number o f c o m p e r i s i o n i n e a c h p a s s
if ( x (j -1) >= x ( j ) ) then
t = x (j -1) ;
x (j -1) = x ( j ) ;
x(j)=t;
end
end
end
endfunction
marks = int16 ([40 ,90 ,73 ,81 ,35]) ; // c r e a t i n g an a r r a y
named marks o f 5 i n t e g e r s
disp ( ” Marks b e f o r e s o r t i n g ” ) ;
disp ( marks ) ;
x = sort (5 , marks ) ;
// c a l l i n g s o r t ( )
function
disp ( ” Marks a f t e r s o r t i n g ” ) ;
disp ( x ) ;
Scilab code Exa 9.7 Autometic variables
//
Example 9 . 7
// W r i t e a m u l t i f u n c t i o n how a u t o m e t i c v a r i a b l e s work
.
3 funcprot (0) ;
1
2
83
4 function []= function1 ()
5
m = int32 (10) ;
// L o c a l V a r i a b l e
6
disp ( m ) ;
// F i r s t Output
7 endfunction
8 function []= function2 ()
9
m = int32 (100) ;
// L o c a l V a r i a b l e
10
function1 () ;
// C a l l i n g f u n c t i o n 1 ( )
11
disp ( m ) ;
// S e c o n d Output
12 endfunction
13 function []= funcmain ()
14 m = int32 (1000) ;
15 function2 () ;
// c a l l i n g f u n c t i o n 2 ( )
16 disp ( m ) ;
// T h i r d o u t p u t
17 endfunction
18 funcmain ()
// c a l l i n g f u n c m a i n ( ) f u n c t i o n
Scilab code Exa 9.8 Global variables
1
2
3
4
5
6
7
8
9
10
11
12
13
14
//
Example 9 . 8
// W r i t e a m u l t i f u n c t i o n t o i l l u s t r a t e t h e p r o p e r t i e s
of global variables .
funcprot (0) ;
function [ x ]= fun1 ()
global x ;
x = x +10;
// g l o b a l x
endfunction
function [ x ]= fun2 ()
x =1
// L o c a l x
endfunction
function [ x ]= fun3 ()
global x ;
x = x +10;
// g l o b a l x
endfunction
84
15
16
17
18
19
20
21
global x ;
x =10;
printf ( ” x=%d\n ” ,x )
// c a l l i n g f u n 1 ( ) , f u n 2 ( ) , f u n 3 ( ) f u n c t i o n s
printf ( ” x=%d\n ” , fun1 () ) ;
printf ( ” x=%d\n ” , fun2 () ) ;
printf ( ” x=%d\n ” , fun3 () ) ;
Scilab code Exa 9.16 Factorial of a number using recursion
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
//
T o p i c 9 . 1 6 RECURSION
//
Page no . 288
// W r i t e a program t o c a l c u l a t e f a c t o r i a l o f a number
using recursion
function [ fact1 ]= factorial1 ( n )
fact1 = -1
if (n <0) then
disp ( ” P l e a s e e n t e r p o s i t i v e v a l u e [ i . e . 0 o r
g r e a t e r than 0 ] ”);
return ;
// Q u i t s t h e
current function
end
if (( n ==0) |( n ==1) ) then
fact1 =1;
else
fact1 = n * factorial1 (n -1) ;
// r e c u r s i v e c a l l
to f a c t o r i a l 1 ()
end
endfunction
n = input ( ” E n t e r number : ” ) ;
// c a l l i n g f a c t o r i a l 1 ( ) f u n c t i o n i n s i d e p r i n t f ( )
printf ( ” F a c t o r i a l o f %d = %d” ,n , factorial1 ( n ) ) ;
85
86
Chapter 10
Structures and Unions
Scilab code Exa 1.cs Case study 1 Book Shop Inventory
1
//
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
//
Case s t u d y : Chapter −10 , Page No
:341
Book Shop I n v e n t o r y
funcprot (0) ;
// D e f i n i n g f u n c t i o n s
function [ string1 ]= get1 ()
string1 = read ( %io (1) ,1 ,1 , ’ ( a ) ’ ) ;
endfunction
function [ i ] = look_up ( table , s1 , s2 , m )
for i =1: m
x = strcmp ( s1 , table ( i ) . title ) ;
y = strcmp ( s2 , table ( i ) . author ) ;
if x ==0 & y ==0 then
return i ;
// Book f o u n d
end
end
i = -1; // Book n o t f o u n d
endfunction
// C r e a t e s a r r a y o f s t r u c t u r e s
87
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
book =[ struct ( ’ a u t h o r ’ , ’ R i t c h e ’ , ’ t i t l e ’ , ’C Language ’ ,
’ p r i c e ’ ,45.00 , ’ month ’ , ’ May ’ , ’ y e a r ’ ,1977 , ’
p u b l i s h e r ’ , ’ PHI ’ , ’ q u a n t i t y ’ ,10)
struct ( ’ a u t h o r ’ , ’ Kochan ’ , ’ t i t l e ’ , ’ Programming
i n C ’ , ’ p r i c e ’ ,75.50 , ’ month ’ , ’ J u l y ’ , ’ y e a r ’
,1983 , ’ p u b l i s h e r ’ , ’ Hayden ’ , ’ q u a n t i t y ’ ,5)
struct ( ’ a u t h o r ’ , ’ Balagurusamy ’ , ’ t i t l e ’ , ’ BASIC ’
, ’ p r i c e ’ ,30.00 , ’ month ’ , ’ J a n u a r y ’ , ’ y e a r ’
,1984 , ’ p u b l i s h e r ’ , ’TMH ’ , ’ q u a n t i t y ’ ,0)
struct ( ’ a u t h o r ’ , ’ Balagurusamy ’ , ’ t i t l e ’ , ’COBOL ’
, ’ p r i c e ’ ,60.00 , ’ month ’ , ’ December ’ , ’ y e a r ’
,1988 , ’ p u b l i s h e r ’ , ’ M a c m i l l a n ’ , ’ q u a n t i t y ’
,25)
];
n = size ( book ) ;
no_of_records = n (1) ;
response = ’ ’ , a =1;
while (( response == ’Y ’ | response == ’ y ’ ) | a ==1)
// Read d a t a
printf ( ” E n t e r t i t l e and a u t h o r name a s p e r t h e
l i s t : \ n”);
printf ( ” T i t l e : \n ” ) ;
title1 = get1 () ;
printf ( ” Author : \ n ” ) ;
author1 = get1 () ;
// C a l l i n g i n d e x ( ) f u n c t i o n and
// P a s s i n g s t r u c t u r e book t o f u n c t i o n l o o k u p ( )
index = look_up ( book , title1 , author1 , no_of_records )
;
// I f book f o u n d t h e n p r i n t t h e book d e t a i l
o t h e r w i s e not in l i s t
if index ~= -1 & index then
// Book f o u n d
printf ( ”% s %s” , book ( index ) . author , book (
index ) . title ) ;
printf ( ”% . 2 f %s” , book ( index ) . price , book (
index ) . month ) ;
printf ( ”% d %s\n ” , book ( index ) . year , book (
index ) . publisher ) ;
88
44
45
46
47
48
49
50
51
52
53
quantity = input ( ” E n t e r number o f c o p i e s : ” ) ;
if quantity < book ( index ) . quantity then
printf ( ” C o s t o f %d c o p i e s = %. 2 f \n ” , quantity ,
book ( index ) . price * quantity ) ;
else
printf ( ” R e q u i r e d c o p i e s n o t i n s t o c k \n ” ) ;
end
else
printf ( ” Book n o t i n l i s t \n ” ) ;
end
printf ( ” \nDo you want any o t h e r book ? ( YES/NO) : ” )
;
response = get1 () ; a =2;
54
55 end
56 printf ( ”
Thank you .
Good Bye ” ) ;
Scilab code Exa 10.1 Define a structure type
//
Example 1 0 . 1
// D e f i n e a s t r u c t u r e type , s t r u c t p e r s o n a l t h a t would
c o n t a i n p e r s o n name ,−
3 // d a t e o f j o i n i n g and s a l a r y . W r i t e a program t o
r e a d t h i s i n f o r m a t i o n from
4 // k e y b o a r d and p r i n t same on t h e s c r e e n .
1
2
5
6
7
8
9
funcprot (0) ;
function [ ]= struc (n ,d ,m ,y , s )
// D e f i n i n g s t r u c t u r e members
personal = struct ( ’ name ’ ,n , ’ day ’ ,d , ’ month ’ ,m , ’ y e a r
’ ,y , ’ s a l a r y ’ ,s ) ;
10
person = personal ;
11
// A c c e s s i n g s t r u c t u r e members
12 printf ( ”
%s %d %s %d %. 2 f ” , person . name , person . day ,
89
13
14
15
16
17
18
person . month , person . year , person . salary ) ;
endfunction
disp ( ” I n p u t v a l u e s [ Name day month y e a r and s a l a r y ] ” )
;
// R e a d i n g d a t a
[ name , day , month , year , salary ]= scanf ( ”%s %d %s %d %f ” )
;
// C a l l i n g f u n c t i o n s t r u c ( )
struc ( name , day , month , year , salary ) ;
Scilab code Exa 10.2 Comparison of structure variables
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
//
Example 1 0 . 2
// W r i t e a program t o i l l u s t r a t e t h e c o m p a r i s o n o f
structure variables .
function []= class ()
// D e f i n i n g s t r u c t u r e s
student1 = struct ( ’ number ’ ,111 , ’ name ’ , ’ Rao ’ , ’ marks
’ ,72.50) ;
student2 = struct ( ’ number ’ ,222 , ’ name ’ , ’ Raddy ’ , ’
marks ’ ,67.00) ;
student3 = struct ( ’ number ’ ,[] , ’ name ’ ,[] , ’ marks ’
,[]) ;
student3 = student2 ;
if ( student3 == student2 ) ,
// L o g i c a l o p e r a t i o n
on s t r u c t u r e s
disp ( ” S t u d e n t 2 and s t u d e n t 3 a r e same ” ) ;
printf ( ” %d %s %f ” , student3 . number , student3 .
name , student3 . marks ) ;
else
disp ( ” S t u d e n t 2 and s t u d e n t 3 a r e n o t same ” ) ;
end
90
16
17
18
endfunction
// c a l l i n g f u n c t i o n c l a s s
class ()
Scilab code Exa 10.3 Calculate the subject wise and student wise totals
//
Example 1 0 . 3
// W r i t e a program t o c a l c u l a t e t h e s u b j e c t −w i s e and
s t u d e n t −w i s e t o t a l s
3 // and s t o r e them a s a p a r t o f t h e s t r u c t u e .
1
2
4
5
6
7
8
// D e f i n i n g a r r a y o f s t r u c t u r e s
student =[ struct ( ’ s u b 1 ’ ,45 , ’ s u b 2 ’ ,67 , ’ s u b 3 ’ ,81 , ’ t o t a l
’ ,0)
struct ( ’ s u b 1 ’ ,75 , ’ s u b 2 ’ ,53 , ’ s u b 3 ’ ,69 , ’ t o t a l
’ ,0)
struct ( ’ s u b 1 ’ ,57 , ’ s u b 2 ’ ,36 , ’ s u b 3 ’ ,71 , ’ t o t a l
’ ,0)
];
total = struct ( ’ s u b 1 ’ ,0 , ’ s u b 2 ’ ,0 , ’ s u b 3 ’ ,0 , ’ t o t a l ’ ,0) ;
9
10
11
12 // C a l c u l a t e t h e s t u d e n t −w i s e and s u b j e c t −w i s e t o t a l s
13 for i =1:3
14
student ( i ) . total = student ( i ) . sub1 + student ( i ) . sub2
+ student ( i ) . sub3 ;
total . sub1 = total . sub1 + student ( i ) . sub1 ;
total . sub2 = total . sub2 + student ( i ) . sub2 ;
total . sub3 = total . sub3 + student ( i ) . sub3 ;
total . total = total . total + student ( i ) . total ;
15
16
17
18
19 end
20 // P r i n t i n g s t u d e n t −w i s e t o t a l s
21 printf ( ”STUDENT
TOTAL\n ” ) ;
22 for i =1:3
91
23
printf ( ” s t u d e n t (%d)
total ) ;
%d\n ” ,i , student ( i ) .
24 end
25 // P r i n t i n g s u b j e c t −w i s e t o t a l s
26 printf ( ”SUBJECT
TOTAL\n ” ) ;
27 printf ( ” %s
%d\n ” ,” S u b j e c t 1 ” , total . sub1 ) ;
28 printf ( ” %s
%d\n ” ,” S u b j e c t 2 ” , total . sub2 ) ;
29 printf ( ” %s
%d\n ” ,” S u b j e c t 3 ” , total . sub3 ) ;
30 // P r i n t i n g g r a n d t o t a l
31 printf ( ” Grand T o t a l = %d” , total . total ) ;
Scilab code Exa 10.4 Array member to represent the three subjects
//
Example 1 0 . 4
// R e w r i t e t h e program o f Example 1 0 . 3 t o u s i n g an
a r r a y member t o r e p r e s e n t
3 // t h e t h r e e s u b j e c t s .
1
2
4
5
6
7
8
9
10
11
12
13
14
15
16
17
// D e f i n i n g a r r a y o f s t r u c t u r e s and a r r a y w i t h i n
structure
student (1) =[ struct ( ’ sub ’ ,[45 67 81] , ’ t o t a l ’ ,0) ];
student (2) =[ struct ( ’ sub ’ ,[75 53 69] , ’ t o t a l ’ ,0) ];
student (3) =[ struct ( ’ sub ’ ,[57 36 71] , ’ t o t a l ’ ,0) ];
total = student ;
for i =1:3
total . sub ( i ) =0;
end
total . total =0;
// C a l c u l a t e t h e s t u d e n t −w i s e and s u b j e c t −w i s e t o t a l s
for i =1:3
for j =1:3
student ( i ) . total = student ( i ) . total + student ( i )
. sub ( j ) ;
92
18
19
20
total . sub ( j ) = total . sub ( j ) + student ( i ) . sub ( j ) ;
end
total . total = total . total + student ( i ) . total ;
Grand t o t a l
//
21 end
22 // P r i n t i n g s t u d e n t −w i s e t o t a l s
23 printf ( ”STUDENT
TOTAL\n ” ) ;
24 for i =1:3
25
printf ( ” s t u d e n t (%d)
%d\n ” ,i , student ( i ) .
total ) ;
26 end
27 // P r i n t i n g s u b j e c t −w i s e t o t a l s
28 printf ( ”SUBJECT
TOTAL\n ” ) ;
29 for j =1:3
30
printf ( ” s u b j e c t −(%d)
%d\n ” ,j , total . sub ( j )
);
31 end
32 // P r i n t i n g g r a n d t o t a l
33 printf ( ” Grand T o t a l = %d” , total . total ) ;
Scilab code Exa 10.5 structure as a parameter to a function
//
Example 1 0 . 5
// W r i t e a s i m p l e program t o i l l u s t r a t e t h e method o f
s e n d i n g an e n t i r e
3 // s t r u c t u r e a s a p a r a m e t e r t o a f u n c t i o n .
1
2
4
5
6
7
8
9
10
funcprot (0) ;
// D e f i n i n g f u n c t i o n s
function [ item ]= update ( product ,p , q )
product . price = product . price + p ;
product . quantity = product . quantity + q ;
item = product ;
93
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
endfunction
function [ value ] = mul ( stock )
value = stock . price * stock . quantity ;
endfunction
// C r e a t e s s t r u c t u r e i t e m
item = struct ( ’ name ’ , ’XYZ ’ , ’ p r i c e ’ ,25.75 , ’ q u a n t i t y ’
,12) ;
// Read d a t a
printf ( ” I n p u t i n c r e m e n t v a l u e s : ” ) ;
printf ( ” p r i c e i n c r e m e n t and q u a n t i t y i n c r e m e n t \n ” )
;
[ p_increment , q_increment ]= scanf ( ” %f %d” ) ;
// C a l l i n g u p d a t e ( ) and mul ( ) f u n c t i o n s
// P a s s i n g s t r u c t u r e i t e m t o f u n c t i o n s u p d a t e ( ) and
mul ( )
//−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
item = update ( item , p_increment , q_increment ) ;
value = mul ( item ) ;
//−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
// P r i n t i n g R e s u l t s
printf ( ” Updated v a l u e s o f i t e m s \n ” ) ;
printf ( ”Name
: %s\n ” , item . name ) ;
printf ( ” P r i c e
: %f\n ” , item . price ) ;
printf ( ” Q u a n t i t y
: %d\n ” , item . quantity ) ;
printf ( ” V a l u e o f i t e m = %f\n ” , value ) ;
94
Chapter 12
File Management in C
Scilab code Exa 12.1 Read data from keyboard and write it to a file
//
Example 1 2 . 1
// W r i t e a program t o r e a d d a t a from k e y b o a r d , w r i t e
i t t o a f i l e c a l l e d INPUT ,
3 // a g a i n r e a d t h e same d a t a from t h e INPUT f i l e and
d i s p l a y i t on t h e s c r e e n .
1
2
4
5 warning ( ’ o f f ’ ) ;
6 disp ( ” Data I n p u t ” ) ;
7
8 // Open t h e f i l e INPUT
9 f1 = mopen ( ’INPUT . t x t ’ , ’w ’ ) ;
10 // Get c h a r a c t e r from k e y b o a r d
11 c = read ( %io (1) ,1 ,1 , ’ ( a ) ’ ) ;
12 mfprintf ( f1 , ’ %s ’ ,c ) ;
13
14 // c l o s e t h e f i l e i n p u t
15 mclose ( f1 ) ;
16
17
18 disp ( ” Data Output ” ) ;
19 // Reopen t h e f i l e INPUT
95
20 f1 = mopen ( ’INPUT . t x t ’ , ’ r ’ ) ;
21
txt = mgetl ( f1 ) ;
22
printf ( ”%s” , text ) ;
23 // c l o s e t h e f i l e i n p u t
24 mclose ( f1 ) ;
Scilab code Exa 12.2 Read and write odd and even numbers
//
Example12 . 2
//A f i l e named DATA c o n t a i n s a s e r i e s o f i n t e g e r
numbers . Code a program
3 // t o r e a d t h e s e numbers and t h e n w r i t e a l l ’ odd ’
numbers t o a f i l e t o be
4 // c a l l e d ODD and a l l ’ even ’ numbers t o a f i l e t o be
c a l l e d EVEN .
1
2
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
warning ( ’ o f f ’ ) ;
// I n p u t numbers i n t h e DATA. t x t f i l e
printf ( ” C o n t e n t s o f DATA f i l e \n ” ) ;
f1 = mopen ( ’DATA. t x t ’ , ’ wt ’ ) ;
for i =1:30
number ( i ) = scanf ( ”%d” ) ;
if ( number ( i ) == -1)
break ;
end
mfprintf ( f1 , ’%d\n ’ , number ( i ) ) ;
end
mclose ( f1 ) ;
f2 = mopen ( ’ODD. t x t ’ , ’ wt ’ ) ;
f3 = mopen ( ’EVEN . t x t ’ , ’ wt ’ ) ;
f1 = mopen ( ’DATA. t x t ’ , ’ r t ’ ) ;
// Read numbers from DATA f i l e
96
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
EOF = length ( number ) ;
i =1;
even =0;
odd =0;
while (i < EOF )
[n , number ]= mfscanf ( f1 , ” %d” )
if ( pmodulo ( number ,2) ==0)
mfprintf ( f3 , ’%d\n ’ , number ) ;
even = even +1;
else
mfprintf ( f2 , ’%d\n ’ , number ) ;
odd = odd +1;
end
i = i +1;
end
mclose ( f1 ) ;
mclose ( f2 ) ;
mclose ( f3 ) ;
// W r i t e odd numbers i n t h e ODD. t x t f i l e
f2 = mopen ( ’ODD. t x t ’ , ’ r t ’ ) ;
printf ( ” \ n C o n t e n t s o f ODD f i l e \n ” ) ;
i =1;
while (i <= odd )
[n , number ]= mfscanf ( f2 , ”%d” )
printf ( ”%4d” , number ) ;
i = i +1;
end
// W r i t e e v e n numbers i n t h e EVEN . t x t f i l e
f3 = mopen ( ’EVEN . t x t ’ , ’ r t ’ ) ;
printf ( ” \ n C o n t e n t s o f EVEN f i l e \n ” ) ;
i =1;
while (i <= even )
[n , number ]= mfscanf ( f3 , ”%d” )
printf ( ”%4d” , number ) ;
i = i +1;
end
// c l o s e t h e f i l e s
mclose ( f2 ) ;
97
61
mclose ( f3 ) ;
Scilab code Exa 12.3 Read and write data to and from the file INVENTORY
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
//
Example12 . 3
//A program t o open a f i l e named INVENTORY and
s t o r e in i t the f o l l o w i n g
// d a t a : Item name
Number
Price
Quantity
//
AAA−1
111
17.50
115
//
BBB−2
125
36.00
75
//
C−3
247
31.75
104
// Extend t h e program t o r e a d t h i s d a t a from t h e f i l e
INVENTORY and d i s p l a y
// i n v e n t o r y t a b l e w i t h t h e v a l u e o f e a c h i t e m .
disp ( ” I n p u t f i l e name ” ) ;
filename = scanf ( ”%s” ) ;
// Read f i l e name t h a t i s ’
INVENTORY’
fp = mopen ( filename , ’w ’ ) ; // Open f i l e i n w r i t e mode ,
fp i s f i l e d e s c r i p t o r
disp ( ” I n p u t i n v e n t o r y d a t a ” ) ;
disp ( ” Item name
Number
Price
Quantity ”);
for i =1:3
// r e a d d a t a from t e r m i n a l
[n , item ( i ) , number ( i ) , price ( i ) , quantity ( i ) ]= mscanf (
”%s %d %f %d” ) ;
// w r i t e d a t a t o t h e f i l e
mfprintf ( fp , ’ %s\t%d\t% . 2 f \t%d\n ’ , item ( i ) , number ( i
) , price ( i ) , quantity ( i ) ) ;
end
mclose ( fp ) ;
// c l o s e t h e f i l e
fp = mopen ( filename , ’ r ’ ) ; // open f i l e i n r e a d mode
98
23
24
25
26
27
28
29
30
31
disp ( ” Item name
Number
Price
Quantity
Value ”
);
for i =1:3
// Read d a t a from t h e f i l e ’INVENTORY’
[n , item , number , price , quantity ]= mfscanf ( fp , ”%s %d
%f %d” ) ;
value = price * quantity ; // Computes v a l u e
// P r i n t i n g o f t h e d a t a
printf ( ’
%s
%7d %8 . 2 f %8d %11 . 2 f \n ’ , item ,
number , price , quantity , value ) ;
end
mclose ( fp ) ;
Scilab code Exa 12.4 Error handling in file operations
1
2
//
Example12 . 4
// W r i t e a program t o i l l u s t a t r e e r r o r h a n d l i n g i n
f i l e operations .
3
4 warning ( ’ o f f ’ ) ;
5 fp1 = mopen ( ’TEST ’ , ’w ’ ) ;
6
7
8
9
10
11
12
13
14
15
16
// Open f i l e i n w r i t e mode ,
fp1 i s f i l e d e s c r i p t o r
for i =10:10:100
// w r i t e d a t a t o t h e f i l e
mfprintf ( fp1 , ’%d\n ’ ,i ) ;
end
mclose ( fp1 ) ;
disp ( ” I n p u t f i l e name ” ) ;
filename = ’ a ’ ;
while ( filename ~= ’ ’ )
filename = scanf ( ”%s” ) ;
// E r r o r h a n d l i n g
try
99
17
18
19
20
21
22
23
fp2 = mopen ( filename , ’ r ’ ) ;
if ( fp2 >0) ,
break ;
// T e r m i n a t e s t h e l o o p i f
f i l e e x i s t or opened
end
catch
// M e s s a g e s t o be d i s p l a y e d when e r r o r
occured
printf ( ”Can n o t open f i l e . \ n ” ) ;
printf ( ” Type f i l e name a g a i n . \ n ” ) ;
end
24
25
26
27 end
28 // Code b e l o w r u n s w h i l e t h e r e i s no e r r o r
29 for i =1:20
30
number = mfscanf ( fp2 , ”%d” ) ; // Read d a t a from
31
f i l e ’TEST ’
if meof ( fp2 ) then
// T e s t f o r end o f
file
printf ( ”Ran o u t o f d a t a ” ) ;
break ;
else
printf ( ”%d\n ” , number ) ;
// p r i n t s t h e d a t a
end
32
33
34
35
36
37 end
38 mclose ( fp2 ) ;
Scilab code Exa 12.5 use of function ftell or mtell and fseek or mseek
1
2
//
Example12 . 5
// W r i t e a program t h a t u s e s f u n c t i o n
and f s e e k ( mseek ) .
3
100
f t e l l ( mtell )
4
5
6
7
8
9
warning ( ’ o f f ’ ) ;
// Open f i l e ’RANDOM’ i n w r i t e mode , f p i s f i l e
descriptor
fp = mopen ( ’RANDOM ’ , ’w ’ ) ;
c = read ( %io (1) ,1 ,1 , ’ ( a ) ’ ) ;
mfprintf ( fp , ’ %s ’ ,c ) ;
// w r i t e d a t a t o t h e f i l e
printf ( ”Number o f c h a r a c t e r s e n t e r e d = %d\n ” , mtell (
fp ) ) ;
mclose ( fp ) ;
10
11
12 // Open f i l e ’RANDOM’ i n r e a d mode
13 fp = mopen ( ’RANDOM ’ , ’ r ’ ) ;
14 n =0;
15 while ( meof ( fp ) ==0)
16 // n i s t h e o f f s e t from o r i g i n i n number o f b y t e s .
17 // The new p o s i t i o n i s a t t h e s i g n e d d i s t a n c e g i v e n
18
19
20
21
22
23
24
25
26
27
28
29
30
31
by n b y t e s from t h e b e g i n n i n g
mseek (n , fp , ’ s e t ’ ) ;
// P r i n t t h e c h r a c t e r and i t s p o s t i o n
printf ( ” P o s i t i o n o f %c i s %d\n ” , ascii ( mget (1 , ’ c ’
, fp ) ) , mtell ( fp ) ) ;
n = n +5;
end
n =0;
// I n i t i a l o f f s e t
while ( mtell ( fp ) >1)
//New p o s i t i o n i s a t t h e s i g n e d d i s t a n c e g i v e n by n
b y t e s from t h e end
mseek (n , fp , ’ end ’ ) ;
// P r i n t t h e c h a r a c t e r s from t h e end
printf ( ”%c” ,( ascii ( mget (1 , ’ c ’ , fp ) ) ) ) ;
n =n -1;
end
mclose ( fp ) ;
101
Scilab code Exa 12.6 Append additional items to the file INVENTORY
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
//
Example 1 2 . 6
// W r i t e a program t o append a d d i t i o n a l i t e m s t o t h e
f i l e INVENTORY
// c r e a t e d i n Example 1 2 . 3 and p r i n t t h e t o t a l
contents of the f i l e .
funcprot (0) ;
warning ( ’ o f f ’ ) ;
function [ item ] = append ( product , fp )
printf ( ” Item name : \ n ” ) ;
product . name = scanf ( ”%s” ) ;
printf ( ” Item number : . \ n ” ) ;
product . number = scanf ( ”%d” ) ;
printf ( ” Item p r i c e \n ” ) ;
product . price = scanf ( ” %f ” ) ;
printf ( ” Q u a n t i t y : \ n ” ) ;
product . quantity = scanf ( ”%d” ) ;
// W r i t e d a t a i n t h e f i l e
mfprintf ( fp , ’ %s %d %. 2 f %d\n ’ , product . name ,
product . number , product . price , product .
quantity ) ;
item = product ;
endfunction
// C r e a t i n g s t r u c t u r e
item = struct ( ’ name ’ , ’ 0 ’ , ’ number ’ , ’ 0 ’ , ’ p r i c e ’ , ’ 0 ’ , ’
q u a n t i t y ’ , ’ 0 ’ );
// Read f i l e name t h a t i s ’INVENTORY’
disp ( ” Type f i l e name ” ) ;
filename = scanf ( ”%s” ) ;
// Open f i l e i n append mode , f p i s f i l e d e s c r i p t o r
fp = mopen ( filename , ’ a+ ’ ) ;
b =0; response = -1;
// Read d a t a
while ( response ==1| b ==0)
item = append ( item , fp ) ; // c a l l i n g append ( )
function
printf ( ” Item %s appended . \ n ” , item . name ) ;
102
31
printf ( ”Do you want t o add a n o t h e r i t e m \ ( 1 f o r
YES/0 f o r NO) ? ” ) ;
response = scanf ( ”%d” ) ;
b =1;
32
33
34 end
35 n = mtell ( fp ) ;
// p o s i t i o n o f l a s t c h a r a c t e r
36 mclose ( fp ) ;
37
38 // Opening f i l e i n t h e r e a d mode
39 fp = mopen ( filename , ’ r ’ ) ;
40 while ( mtell ( fp ) < n -2)
41
// r e a d d a t a from t e r m i n a l
42
[g , item . name , item . number , item . price , item . quantity
]= mfscanf ( fp , ”%s %d %f %d” ) ;
// P r i n t Data t o s c r e e n
printf ( ’ %s %7d %8 . 2 f %8d\n ’ , item . name , item . number
, item . price , item . quantity ) ;
45 end
46 mclose ( fp ) ;
43
44
103
Chapter 13
Dynamic Memory Allocation
and linked Lists
Scilab code Exa 1.cs Case study 1 Insertion in a sorted list
1
//
Case Study : C h a p t e r : 1 3 , Page No
.:434
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
//
1. Insertion in a sorted
list
funcprot (0) ;
// C r e a t e t h e l i s t
function [ List ]= create ( list1 )
global List ;
// C r e a t e t h e c u r r e n t node
list1 . number = input ( ” I n p u t a number ( Type −999 t o
end ) ; ” ) ;
if list1 . number == -999 then
list1 . next = NULL ;
list1 . add = NULL ;
else
list1 . add = list1 . add +1;
list1 . next = NULL ;
List ( i ) = list1 ;
if ( i ==1) then
104
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
else
List (i -1) . next = List ( i ) . add
end
i = i +1;
create ( list1 ) ; // C r e a t e t h e n e x t node
end
return ;
endfunction
function []= print1 ( list1 )
if list1 ( i ) (1) . next ~= NULL then
printf ( ”%d−−−>” , list1 ( i ) (1) . number ) ; // P r i n t
c u r r e n t item
i = i +1;
if list1 ( i ) (1) . next == NULL then
printf ( ”%d” , list1 ( i ) (1) . number ) ;
end
print1 ( list1 ) ; // Move t o n e x t i t e m
end
return ;
endfunction
function [ List ]= insert ( list1 )
global List ;
x = input ( ” I n p u t number t o be i n s e r t e d : ” ) ; // Read
t h e number
// f i n d t h e l o c a t i o n s o t h a t number c o u l d be
placed in sorted order
while ( list1 ( i ) (1) . next ~= NULL )
if ( list1 ( i ) (1) . number >= x ) then
break ;
end
i = i +1;
end
key = i ;
// I n s e t i o n a t end
if ( list1 ( i ) (1) . next == NULL & list1 ( i ) (1) . number <
x ) then
list1 ( i +1) (1) . number = x ;
105
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
list1 ( i +1) (1) . add = i +1;
list1 ( i +1) (1) . next = NULL ;
list1 ( i ) (1) . next = list1 ( i +1) (1) . add ;
List = list1 ;
return ;
end
i =1;
while ( list1 ( i ) (1) . next ~= NULL )
i = i +1;
end
j = i +1;
// Key node f o u n d and i n s e r t new node o r i t e m
while ( list1 ( i ) (1) . add ~= key )
list1 ( i +1) (1) . number = list1 ( i ) (1) . number ;
i =i -1;
end
list1 ( i +1) (1) . number = list1 ( i ) (1) . number
list1 ( i ) (1) . number = x ;
list1 ( j ) (1) . add = j ;
list1 ( j ) (1) . next = NULL ;
list1 (j -1) (1) . next = list1 ( j ) (1) . add ;
List = list1 ;
endfunction
global List ;
NULL =0; i =1;
// C r e a t e t h e s t r u c t u r e i . e . node
node = struct ( ’ number ’ ,0 , ’ add ’ ,0 , ’ n e x t ’ ,0) ;
head = node ;
// C a l l i n g t h e f u n c t i o n s
printf ( ” I n p u t a s o r t e d ( a s c e n d i n g ) l i s t ” ) ;
List = create ( head ) ;
printf ( ” \ n O r i g i n a l L i s t : ” ) ;
print1 ( List ) ;
List = insert ( List ) ;
printf ( ” \nNew L i s t : ” ) ;
print1 ( List ) ;
106
Scilab code Exa 2.cs Case study 2 Building a Sorted List
1
//
Case Study : C h a p t e r : 1 3 , Page No
.:438
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
//
2. Building a Sorted List
funcprot (0) ;
// C r e a t e t h e l i s t
function [ List ]= create ( list1 )
global List ;
// C r e a t e t h e c u r r e n t node
list1 . number = input ( ” I n p u t a number ( Type −999 t o
end ) ; ” ) ;
if list1 . number == -999 then
list1 . next = NULL ;
list1 . add = NULL ;
else
list1 . add = list1 . add +1;
list1 . next = NULL ;
List ( i ) = list1 ;
if ( i ==1) then
else
List (i -1) . next = List ( i ) . add
end
i = i +1;
create ( list1 ) ; // C r e a t e t h e n e x t node
end
return ;
endfunction
function []= print1 ( list1 )
if list1 ( i ) (1) . next ~= NULL then
107
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
printf ( ”%d−−−>” , list1 ( i ) (1) . number ) ; // P r i n t
c u r r e n t item
i = i +1;
if list1 ( i ) (1) . next == NULL then
printf ( ”%d” , list1 ( i ) (1) . number ) ;
end
print1 ( list1 ) ; // Move t o n e x t i t e m
end
return ;
endfunction
// S o r t i n g o f t h e numbers i n t h e l i s t
function [ List ]= insert_sort ( list1 )
global List ;
j =1;
while ( list1 ( j ) (1) . next ~= NULL )
i =1;
while ( list1 ( i ) (1) . next ~= NULL )
if ( list1 ( i ) (1) . number > list1 ( i +1) (1) .
number ) then
temp = list1 ( i ) (1) . number ;
list1 ( i ) (1) . number = list1 ( i +1) (1) .
number ;
list1 ( i +1) (1) . number = temp ;
end
i = i +1;
end
j = j +1;
end
List = list1 ;
endfunction
global List ;
NULL =0; i =1;
// C r e a t e t h e s t r u c t u r e i . e . node
node = struct ( ’ number ’ ,0 , ’ add ’ ,0 , ’ n e x t ’ ,0) ;
head = node ;
// C a l l i n g t h e f u n c t i o n s
List = create ( head ) ;
108
64
65
66
67
68
printf ( ” \ n O r i g i n a l L i s t : ” ) ;
print1 ( List ) ;
List = insert_sort ( List ) ; // S o r t t h e l i s t
printf ( ” \ n A f t e r s o r t i n g : ” ) ;
print1 ( List ) ;
Scilab code Exa 13.3 Create a linear linked list
//
Example 1 3 . 3
// W r i t e a program t o c r e a t e a l i n e a r l i n k e d l i s t
interactively
3 // and p r i n t t h e l i s t and t o t a l number o f i t e m s i n
the l i s t .
1
2
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
funcprot (0) ;
NULL =0; i =1;
// C r e a t e t h e l i s t
function [ List ]= create ( list1 )
global List ;
// C r e a t e t h e c u r r e n t node i n t h e l i s t
list1 . number = input ( ” I n p u t a number ( Type −999 t o
end ) ; ” )
if list1 . number == -999 then
list1 . next = NULL ;
list1 . add = NULL ;
else
// C r e a t e t h e n e x t node i n t h e l i s t
list1 . add = list1 . add +1;
list1 . next = NULL ;
List ( i ) = list1 ;
if ( i ==1) then
else
109
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
List (i -1) . next = List ( i ) . add
end
i = i +1;
create ( list1 ) ; // C a l l c r e a t e ( ) f u n c t i o n
end
return ;
endfunction
// F u n c t i o n t o p r i n t t h e numbers o f l i s t
function []= print1 ( list1 )
if list1 ( i ) (1) . next ~= NULL then
printf ( ”%d−−−>” , list1 ( i ) (1) . number ) ; // P r i n t
c u r r e n t item
i = i +1;
if list1 ( i ) (1) . next == NULL then
printf ( ”%d” , list1 ( i ) (1) . number ) ;
end
print1 ( list1 ) ; // Move t o n e x t i t e m
end
return ;
endfunction
// F u n c t i o n t o c o u n t t h e number o f i t e m s i n t h e l i s t
function []= count ( list1 )
global c ;
if list1 ( i ) (1) . next == NULL then
return ;
else
i = i +1;
c=i;
count ( list1 ) ;
end
return ;
endfunction
// C r e a t e t h e s t r u c t u r e i . e . node
node = struct ( ’ number ’ ,0 , ’ add ’ ,0 , ’ n e x t ’ ,0) ;
head = node ;
global List ;
// C a l l i n g t h e f u n c t i o n s
List = create ( head ) ;
110
60 print1 ( List ) ;
61 global c ;
62 c =1;
63 count ( List ) ;
64 // P r i n t t h e t o t a l number o f i t e m s
65 printf ( ” \nNumber o f i t e m s = %d” ,c ) ;
Scilab code Exa 13.4 Insert the item before the specified key node
//
Example 1 3 . 4
// W r i t e a f u n c t i o n t o i n s e r t a g i v e n i t e m b e f o r e a
s p e c i f i e d node known a s
3 // key node .
1
2
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
funcprot (0) ;
// C r e a t e t h e l i s t
function [ List ]= create ( list1 )
global List ;
// C r e a t e t h e c u r r e n t node
list1 . number = input ( ” I n p u t a number ( Type −999 t o
end ) ; ” ) ;
if list1 . number == -999 then
list1 . next = NULL ;
list1 . add = NULL ;
else
list1 . add = list1 . add +1;
list1 . next = NULL ;
List ( i ) = list1 ;
if ( i ==1) then
else
List (i -1) . next = List ( i ) . add
end
111
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
i = i +1;
create ( list1 ) ; // C r e a t e s t h e n e x t node
end
return ;
endfunction
// F u n c t i o n t o i n s e r t t h e i t e m b e f o r e t h e s p e c i f i e d
key node
function [ List ]= insert ( list1 )
x = input ( ” V a l u e o f new i t e m ? ” ) ;
printf ( ” V a l u e o f key i t e m ? ( B e f o r e which you want
to i n s e r t ?) ”);
key = scanf ( ”%d” ) ;
while list1 ( i ) (1) . next ~= NULL
i = i +1;
end
j = i +1;
// Find t h e key node and i n s e r t t h e new node
while ( list1 ( i ) (1) . number ~= key )
list1 ( i +1) (1) . number = list1 ( i ) (1) . number ;
i =i -1;
if ( i ==0) then
printf ( ” Item n o t Found ” ) ;
return ;
end
end
list1 ( i +1) (1) . number = list1 ( i ) (1) . number
list1 ( i ) (1) . number = x ; // I n s e t t h e new node
b e f o r e t h e key node
list1 ( j ) (1) . add = j ;
list1 ( j ) (1) . next = NULL ;
list1 (j -1) (1) . next = list1 ( j ) (1) . add ;
List = list1 ;
endfunction
// F u n c t i o n t o p r i n t t h e numbers o f l i s t
function []= print1 ( list1 )
if list1 ( i ) (1) . next ~= NULL then
printf ( ”%d−−−>” , list1 ( i ) (1) . number ) ; // P r i n t
c u r r e n t item
112
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
i = i +1;
if list1 ( i ) (1) . next == NULL then
printf ( ”%d” , list1 ( i ) (1) . number ) ;
end
print1 ( list1 ) ; // Move t o n e x t i t e m
end
return ;
endfunction
global List ;
NULL =0; i =1;
// C r e a t e t h e s t r u c t u r e i . e . node
node = struct ( ’ number ’ ,0 , ’ add ’ ,0 , ’ n e x t ’ ,0) ;
head = node ;
// C a l l i n g t h e f u n c t i o n s
List = create ( head ) ;
printf ( ” \ n O r i g i n a l L i s t : ” ) ;
print1 ( List ) ;
List = insert ( List ) ;
printf ( ” \nNew L i s t : ” ) ;
print1 ( List ) ;
Scilab code Exa 13.5 Delete a specified node in the list
1
2
3
4
5
6
7
8
9
//
Example 1 3 . 5
// W r i t e a program / f u n c t i o n t o d e l e t e a s p e c i f i e d
node .
funcprot (0) ;
// C r e a t e t h e l i s t
function [ List ]= create ( list1 )
global List ;
// C r e a t e t h e c u r r e n t node
list1 . number = input ( ” I n p u t a number ( Type −999 t o
113
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
end ) ; ” ) // s c a n f ( ”%d” ) ;
if list1 . number == -999 then
list1 . next = NULL ;
list1 . add = NULL ;
else
list1 . add = list1 . add +1;
list1 . next = NULL ;
List ( i ) = list1 ;
if ( i ==1) then
else
List (i -1) . next = List ( i ) . add
end
i = i +1;
create ( list1 ) ; // C r e a t e t h e n e x t node
end
return ;
endfunction
// F u n c t i o n t o p r i n t t h e numbers o f l i s t
function []= print1 ( list1 )
if list1 ( i ) (1) . next ~= NULL then
printf ( ”%d−−−>” , list1 ( i ) (1) . number ) ; // P r i n t
c u r r e n t item
i = i +1;
if list1 ( i ) (1) . next == NULL then
printf ( ”%d” , list1 ( i ) (1) . number ) ;
end
print1 ( list1 ) ; // Move t o n e x t i t e m
end
return ;
endfunction
// F u n c t i o n t o d e l e t e t h e s p e c i f i e d node
function [ List ]= delet ( list1 )
key = input ( ” V a l u e o f i t e m number t o be d e l e t e d ? ” )
; // Read v a l u e o f key
// Find and d e l e t e t h e key node
while ( list1 ( i ) (1) . number ~= key ) then
if list1 ( i ) (1) . next == NULL then
114
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
printf ( ” Item n o t f o u n d i n t h e l i s t ” ) ;
return ;
end
i = i +1;
end
while ( list1 ( i ) . next ~= NULL )
list1 ( i ) (1) . number = list1 ( i +1) (1) . number ;
i = i +1;
end
list1 (i -1) (1) . next = NULL ;
List = list1 ;
endfunction
global List ;
NULL =0; i =1;
// C r e a t e t h e s t r u c t u r e i . e . node
node = struct ( ’ number ’ ,0 , ’ add ’ ,0 , ’ n e x t ’ ,0) ;
head = node ;
// C a l l i n g t h e f u n c t i o n s
List = create ( head ) ;
printf ( ” \ n O r i g i n a l L i s t : ” ) ;
print1 ( List ) ;
List = delet ( List ) ;
printf ( ” \ n A f t e r d e l e t i o n L i s t i s : ” ) ;
print1 ( List )
115