4.2
I
()~ CProgramming
v1a10N
9< d is
~ega t i ve" ' nurn);
o
. f ("The number
pri nt
. eve n"' nurn );
9.c 2 ;::::: 0)
dis
%
r
be
num
1· f(nu m
("The
•
pri. ntf
· dd" num ) ;
'
%d is o
er
" he numb
else
-------~
j__ _:pr: in:t~ f(~T= - - - - - - -
else
0
of two numbers
Maximum
,
e have to use the relational operator :::. to c
°llip,
f two numb ers, w
.
To find the maximum o
two numbers.
Q - -- - - - - - - - -
#inc lude<std io . h>
•,oid ma i n ()
(
numbers : ") ;
int a , b ;
prin tf ("\ nEnter t he two
sca nf( "%d%d ", &a , &b) ;
b) intf ( "\ nThe maxi.mum 1·s %d ", a ) ;
i f (a ,pr
else
;
prin tf( "\ nThe maxi·mum is %d ", b )
r)
Maximum of two numbers (using conditional operato
use lhe condi ti ona1 oper ator ?:
To find the maximum of two numbers, we can also
~
-- -- -- ---- -- -- -- -- -- -- -
• - -~J~ <s:J l~ . h>
• "'::. r.
.. :- . . ::, :. -. , , -ax;
, • " " ·i:::i::.er the cw
"·: •~j,: :: • ,~a, ;r1;
numbers : " ) ;
max1m~m 13 %d ",na x);
~
• --
-
-
-.,.... ...-~. 4.L.L:. .~-.
A
I
•
j.,;, ;-x., .._.-..~
-
-
- ,\
'
Programs through Conditional.. .
prograrn
utf.:.11 \
0
4-3
to calc ulate the max imum of three numbers.
maximum of three numbers, we have to use the relational oper
ator > to compare two
.
f
at
a
ume.
...,bers
o find the
otJw
.
g
de< stdi o . h>
#J.· ncll l . ( )
·d rnain
vol.
int a , b , c , max ;
r int f ( 11 \ nEn ter the th ree num bers : •q;
p
f ( 11 %d%d%d ", &a , &b , &c);
s can
if(a >b)
rnax =a ;
else
rnax =b ;
if( c> rnax)
ma x= c ;
pr in t:. f ( \ nThe maxi mum is %ct ", max );
11
~------------------Q
Program ro calc ulate the maxi mum of three numbers
(using conditional operator).
We have seen how to use conditional operator to find
maximum of two numbers . This can be
used in the same way to find the maximum of three num
bers.
Q- - - - - - - - - - - - - - - - - - - -
win c l ude <st:.d i o . h >
vo id main ()
i nt a , b , c , ma x;
pt i nt f ( \n Ente r the t hree num bers : ");
s can f( " ~d%d%d " , &a , &b , &c) ;
ma x -=(a> b)?a : b ;
ma x = (c> max ) ? c : max;
pr1 ntf ( " \n Th e max imum is %d ", max ) ;
11
l
4-4
3.
I
0
l)
UISIOft
C Programming
ing statements
Programs using Loop
+2+3+ ... +n
Sum of first n numbers l
loop statement to start frorn 1
a
e
us
we
,
rs
be
m
nu
n
first
and g0
To calcul ate the sum of
.
0)
ly
al
llpt
to sum (which is initi
0 ~Each number is added
Q------~
#i nc lu de <s td i o .h
vo i d main ()
{
>
in t n, i, s um=O ;
);
th e va lu e of n :"
p ri n tf ( "\ nEnt er
sc an f ( "%d ", & n) ;
i++)
fo r( i= l; i< = n;
su m = su m + i;
numbe rs is
sum of fi rs t %d
he
"T
(
tf
n
ri
P
0
'5
d" , n, sum ) ;
Q
rs
Sum of given n numbe
T
n numbers from the use r. o dot~
pt
ce
ac
to
ve
ha
e
w
,
rs
be
given n num
ch number .is added t~
Ea
.
rs
To calc ulate the sum of
be
m
nu
e
th
pt
ce
ac
d
0
t to count from 1 to n an
we use a loop statemen
.
sum (which is initially 0)
---Q- - - - - - - -
>
# in cl u de <s td i o .h
v o i d m ai n ()
{
= O;
in t n, i , num, sum
nu mbe rs ? ") ;
p ri n tf { "H ow many
sc an £ ( " %d ", &n) ;
i++)
:- :: ,r (i =l ; i< = n;
t h e number : " ) ,·
t
E
er
\
"
n
n
(
tf
~ ri n
);
ss an f( "l d ", &nu m
m;
s ..:~ = su m + nu
,..,.. , .. &,
: , - - · , .....
0
sum);
...,h"" sum is %d ",
.L
-
\
1Jonaf..
Programs through Cond1
.
..j{.~ \
4-5
I of a number
.
ctorlO
ed as n! = I x 2 x 3 x n. This can be done using a loop
lat
lcu
ca
is
r
be
m
nu
a
of
·aJ
....
fO
to n.
,rne fact011
1
. b goes from
wbic
JJl
de <s td i o .h >
•n cl ll
#J. . cl
111
vo J. .
a in ( )
pr od uc t =l ,
i
;
. nt n,
nu m be r:" ) ;
J. . f ( "E nt er th e
0 nt
f (" % d" ' &n.); .
p
.
sc an
, J.++)
J.< =n
· :::1
'
f or ( J.
* i;
pr od uc t= pr od uc t
· %d ", pr od uc t);
t o rJ..a l is
1
f
. tf ("·"\ n Th e ac
prJ.D
0 Cr
nCr::: n! / ( r! x (n -r) !)
To calculate the factorials
we have to use a loop.
g
#i nc l ud e< st di o. h>
vo i d m ai n ()
=1 , rf ac t =l , nm i
i nt n , r , nf ac t
e va lu es of n
p ri nt f (" En te r th
&r ) ;
sc an f( "% d%d " , &n ,
+)
fo r {i =l , i< =n ; i+
nf ac t= n f ac t * i;
+)
fo r {i =l , i< =r ; i+
rf a c t= rf ac t * i;
i+ + )
fo r (i =l , i <= n- r;
us rf ac t
nm in us r fa ct =n m in
sw er is %ct",
pr in tf ( " \ n Th e an
)
nu s rf a ct =l , i
an d r: ") ;
* i;
rf ac t) );
nf ac t / (rf ac t* nm in us
0
4-6 / VISIO~
C Programming
Digit Reve rsing
Example: Input 978 output reverse=879.
To reverse the djgjts of a number. we appIy the % and / operators succe ssive ly tO
get th
one by one till the number becomes 0· For example, if the numb er is 978
.
' We first e di&1~
apply ina 978% 10. Further 978/10 gives 97. 97%10 gives 7
97110
and
gives 9. 9%10 g· get 8 by
0
¾9
9/ 10 gjves 0.
illid
JQ - - - - - - - - - - - - - -
# i nclu de<s t d i o . h>
void main ()
{
int n umb er, re v=O;
prin tf( "Ent er the numb er
scan f( " %d",& numb er);
wh ile (num ber > 0)
");
{
re v = rev* 1 0 +num ber % 1 0 ;
numb er= numb er / 10;
p r intf ( " \ n The re vers e is %d", rev) ;
Table generation tor,n (first l O multiples)
To generate the first 10 multiples of any number, we use a loop
for the multiplier which runs
from 1 to 10 and multiply the number with the multiplier.
Q
#inc lude <std io .h>
void main ()
{
int n , i;
prin tf( "Ent er the va lue o f n ");
scan £ ( " %d" , &n) ;
f or(i =l ; i<= 10; i++)
prin tf( " %d X %d = %ct \n ", n,i, n*i );
_1_ _ _ _ _ _ _
_ __
_ _ _ __ _ _ _ __
__
g
Programs through Ccn<fflicnaJ
_f.!_\
--
4-7
fa number
ctors o
of a number are all the numbers which di vide the number. For lhi~. we have co check
· der 1·s 0, the number is a factor. The loop goes from 1
If th e remam
· · ·
factor
'file ainder after d1v1swn.
we re01
fCl
5
t0 Jl
- 1.
g
de<stdio . h >
•ncltJ
. ()
#l. d rna i n
"ol-
illt n,i;
. tf(" \nE nt er a number :");
prl-n
" %d ", &n) ;
.
scan f(
· -1· i < n; i++)
f o r (i- '
/* * i is a factor** /
if(n % i ==0 )
printf("%d is a factor o f %ct", i , n) ;
_L- ---- ---- ---- ---- ---- ---- ~
GCD of 2 numbers
To calculate the greatest common divisor of two numbers, we repeatedly subtract the smaller
number from the larger till both numbers become equal. The result is the GCD.
Q---------------------#include<s tdio . h >
void main ()
int a , b;
p rint f ( " \ nEnter the numbers : ");
scanf("%d%d ",&a , &b) ;
while( a !=b)
if(a >b )
a =a-b;
else
b=b-a;
}
Printf( " \n\ nThe GCD is %d ", a);
- - - - - - - - - - - - - - - - - - -Q
4-8
I
or;,
C Programming
UISIOft
to accept a number and check whether it is a perfect number (i .e., the
ProQTam
::,
8
1.ltn
factors=number itself).
Gf .
l[I
To find the sum of factors. we repeatedly divide the number n by all numbers fro
1
the remainder is O, it is a factor and we add it to sum. If the sum and· numberareIo. l<h·1-!
eqllaJ, it .· 11
perfect number.
lj .
I
Q - -- - - - - - - - - - - - - -
~
#i nc lude <s tdi o . h>
voi d mai n ()
{
int num, i,sum=0;
printf (" \ nEnter a number
sca nf ("%d" , &num) ;
f or (i =l ; i < num; i++)
; II )
j
{
if (num % i ==0 )
s urn=surn + i ;
/ ** i is a fa ctor** /
if (s um == num)
pri ntf (" %d is Perfec t ", nurn) ;
el se
pri ntf ("%d is not Perfec t ", num) ;
Prime number
To check if a number is prime, we successively divide it by 2 to number -1. If it is divisible ilii
number is not prime. Thus, as soon as we get a Oremainder, we have to break out of the loop.
Q- -- -- -- ---- -- -- -- -
,: r.c~ude <stdio . h>
·;,:, _ d :nain ()
:~: f.j:nber , i , flag =l ;
1., .c: :-. '::: ( "enter the nurnbe r : " ) ;
.s ::::-c1r.: 1 %d " , &number) ;
:=L ; i<numbe r;i ++)
=~:
• c
11
- ,"L'::I
%i == 0)
I
Program:J lhrougn Cor.d,:h,..~
}
· f(fJ.. 0
4- 9
g====O)
f(" \ n The number is n o t prime " i ;
pLint
se
f(" \ n The number is prime " ) ;
e )..
. nt
J..
p rl·
--- --- --- --- --- --- --- --- -.,~
~
ca1cu1ate
3
sine x using Taylor series
x5
)( + - - ......
". 3!
5!
.
r series to calculate sine(x) has n terms. Each term has a numeraw r and deno~r cr.
the fay 1o
.
.
numerator, we can use the standard function
pow. For the denomm
a10r. we have w
for the
the factorial using a loop. Even terms are subtracted and odd termS are added .
calculate
g
void main ()
fl o at x , sum=O , numer;
i nt i, n , p = l ;
l ong int deno ;
printf( " How many terms: ");
&n l ;
5 canf ( " %d" ,
printf ("e nte r the va lue o f x : " ) ;
s c an f (" %f" , &x ) ;
f or ( i= l ; i<=n ; i++)
{
numer= pow(x ,p ) ; /* calcula te numerat o r* /
/* ca l c ulate denomi nat o r * /
den o = 1 ;
f or(j = l ; j< =p; j++)
deno = deno * j;
if ( i % 2 == 0)
sum=s um - numer I deno ;
e lse
sum= sum + numer I deno ;
p =p + 2 ;
}
p rintf( " \ n The sum is % f" , sum);
~TS
Q
I
(_).,
4-1 0 VIIIOII
C Programming
in th
The Taylor series for cosine( x) is given below. Cosine(x) can be calcula ted
e salh
nl
·
•
.
.
•qe \v
numbe
even
Y
o
s
cont.am
senes
the
that
1s
ce
differen
only
ai,.
sine(x). The
rs as
0
P wer8 . '
factorial.
<11\j
x6 xs
= J - 2! + 41 - 6 ! + g ! - ...
x2 x4
COS X
*, I, %
Progra m to calcula.te multiplication, divisio n and modulu s withou t using
0
Perators .
calcula te x * y, xly and x%y
1,t
#inclu de<s tdio . h>
#in c lude< math . h>
void main ()
int x ,y , i , prod=O , quo=O , rern,s ignx= l , signy =l ;
print f ( "Ente r the value s of x and y " ) ;
scanf ( "%d%d ", &x , &y) ;
if( x<O)
signx=O; / **xi s negat ive** /
if(y<O )
signy =O;
x=abs ( x);
y=abs (y) ;
/ *** ca l cu late x * y *** /
f or(i =l; i<=y; i++)
prod= prod+ x ;
i f(sign x != signy )
prod= -prod ;
/ *** c al c ulate x/ y and x % y *** /
whi l e (x >=y)
I
x=x- y;
quo+ +;
:em=x ;
1f(si gnx' =s1gn y)
quo= quo ;
i: s1gnx ==O)
~• x 1 s negat ive*~ /
r<::m• - x ;
·r. ~f "n nThe produ ct 1s %ct", prod ) ;
Or;, \
Programs through Conditional... 11111011
4-11
quo) ;
11 \n \nThe quotient i s %ct",
•ntf (
ptlf( " \ n \ nThe remainde r i s %ct ", rem);
,int
p
- - - - - - -- - - - - - - - - - - ] ; i
to display the first n terms of the Fibonacci series, i.e., J J 2 3 5 8 ...
ram
g
pro
.
. . th
the fibonacc1 senes 1s e sum of the previous two terms. We initialize fust two
f
0
term
Bach
l and 1 respeeti vel y.
__,._
to
te11,....
,g
cte<stdio . h>
#incll.l
. d main()
vo l
{
· f l= O, f2=1 , f3 ;
int n , 1.,
c irscr () i
rintf( " \ nHow many terms
ps e a..
nf( " %d ", &n). ;
for(i=l ; i< = n ; i++)
; II )
j
f3=fl+f2 ;
printf( " %d\ t ", f2 ) ;
fl=f2 ;
f 2=f3 ;
)
__.!...
l ____________________
.QJ.
To display 'n' lines of Pascal triangle
1
1
1
1
1
1
1
4
6
10
15
1
3
3
5
6
1
2
1
4
10
20
1
5
15
1
6
1
The Pascal triangle contains the coefficients of binomial expansion.
g ____________________
~include <stdio . h>
void main ()
{
inti , j,spaces = 39 , num , n ;
Printf( " \ nHo w ma ny lines : " ) ;
~
~ ,:.. i,; ..... ~
.- ;.~ ~ ~
~k~;...~~...
w.
8fTV71fng
c Pr 09~
c.,J.
,,_ 12
I v,,,o•
. f/ "% d
se a •', = ( " \n
p: 1. -..,_-~1 ;1
iS
\ n
\n ");
T ri a n
'.
< -n ,
"
") ;
a ce s ,
" sp
" % -, s , 1 ;
e sf{
r:.
~1 .n
++ )
es = sp ac
F. ·<-. i ; J
sp ac
J
r- or
{-
-L
;
&n J pA
SC r>
e
T n_i + + !
"
g .1 e
(J.
fo rr j- 0 ,
!
(, f( :; -- 0
4,
I
)
nu m -l ;
j ;
~ (i -j ~ I) /
~ is e
nu m = nu m
f (
" %d
p ri. n t
1)
J. ! ( i --
" nu m ) ;
'
b re a k ;
!
n");
_· _( "
_rt
_. .! ... ~~p_r_i _n
- -- - -----
m
W nt. e a 'C ' p r og ra
I
0
-
----
----
t
E x e r c is e
LO p n n
.c
l n u m b e rs .
tiirst JO n at u ra
e
ic is n eg a ti v
a n u m b er if
d isp la y
,
-~~
------
ra m LO
e ven
Write a 'C ' pr og
g er is o d d o r
te
in
an
er
th
e
h
ra m C h ec k w
J
s y m bo l
W ri te a 'C ' pr og
g = , > o r<
n
i
s
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rs
e
g
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g ra m l o re fa
I 0 0 D iv is ib
W nt e a ·c p ro
o
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m b ers B e tw
o H n d A ll N u
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m
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w ri te a C '
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rn
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\I n ,, ,' C P ro
Arrays and Strings
Unit
1.
introduction
we have used variables to store a single data item in memory However in many
so f ar
·
. t.ons we need to store a large amount of data.
apphca
1
ThUS, we would have to declare and use a large number of variables, which is very inconvenient.
.
Moreover, these variables are independent and unrelated to each other. Many applications
·requ1·re multiple data items to be grouped so that it becomes easy to manipulate them. This can
be done using an array .
Definition
An array is a collection of data items of the same data type
referred to by a common name. Individual data items can be
accessed by the name of the array and an integer called the
'index' or ·~ubscript'. These items occupy contiguous or
Oct. 2016- 2M
What is an Array? Give
an example.
consecuti\'e memory locations.
An arra) i, L1h o called a 'subscripted variable'.
Slngle and Multidlmensional Arrays
An arra) ha\lng only a single subscript is referred to as single subscripted, linear or one-
dimen,1onal .irray.
5-1
5-2
o®
I UIIIOII
C Programming
.
, subscripts is a two-dimensional arr
An array whose elements are specified by l\\ 0
ay (also
cai1
a matrix).
qJ
Conceptually. an array can have any number of dimensions, limited only by the ava·1
table
memory.
2.
Array Declaration
An array has to be declared before it is used in a C program. The declaration tells the c
.
ornp1ler
I
1.
the type of the array,
11.
the name of the array.
lll.
the number of dimensions,
1v.
number of elements in each dimension.
Syntax: data -t ype array - name [ size l ] [ size 2 ] ............ • • [ size n ) ;
•
•
•
data-type specifies the data type of each element of the array.
array-name is a valid C identifier.
[size 1] .. ......... [size n] are the n dimensions of the array size 1. .. ..size n are positive
integers indicating the maximum number of elements in each dimensions.
rr~
Example:
For each dimension, an array subscript begins from Oand has a maximum of size-1.
i nt a [ 10 J;
char name[8 0j ;
In addition, a storage class can also be specified for the array (default is auto).
Example:
s ta t. ic i nt x [ 20] ;
j
//~
/. " t.J/ ., ~l/:
~:rr,r:
,,
one Dimensional Array
1
3.
. ensional array is declared as follows.
one d1rtl
A
array-name(size)
data_type
le . int n [ 1 0 l ;
exarnP .
5-J
WJJIII
'.JEf'A~
t-2, 'J .~
e-am~ cJ ,:;~,
G-tr ~ ~~,..a e:,.'ra
________
p
t
. ·s a declaration of an array n of 10 integers. When an arra, b ~ '.Med. :rce ~, ::-.:= •• .::-:--
rtus
1
•
or allocates a block of memory large enough to slore the enure arra\
reserves
.
number of bytes allocated is:
'[he to tal
rotal bytes = length of array * size of (data-type)
'fhUS, for the above declaration, 20 bytes will be allocated (considering thar an int..eger re:if.!.!:-es .:
bytes) .
I~
Address---+ 1000
e;; GD············· ·······@
1002
1004
101 8
Figure 5.1: One dimensional array
3 .1 Accessing Array Elements
To access a particular array element we have to specify the name of the array folJo,;\·cd by the
index in square braces. The index indicates the particular element we wam to access.
Syntax
array -name [ integer- express ion]
Example:
n[O] refers to the element at position 0, i.e .. the first element.
n[2] refers to the element at position 2 which is the third element.
An integral expression can also be used as a subscript. E.mmple is as follows:
n[5-2]
n[i++]
n[i-2]
n[--i]
n[i+j]
are all valid.
5-4 /
0
UISIO~
C Programming
Assigning Values to Array Elements
Values can be assigned to individual elements by using the assignment operator (==).
Syntax
arr a y_ name [ index ]=value
Example:
n[0] = 20:
n[2] = 35;
Entering Data into an Array
In most of the cases, the values are not known in advance. In such cases we can accept th
e da~
from the standard input device (stdin) and store it in the array.
This can be done in the following manner.
The following code accepts ten numbers from the user and places them into the array.
fo r (i =O; i <l O;i++)
print f (" \ n Enter t he value f or p osit i on %ct", i) ;
scanf( " %d", &n (i] ) ;
The value of i goes from Oto 9. Initially i=0 and the scanf statement will cause the integer read
from the keyboard to be stored·at the location (address) of n[0].
This process will be repeated for the entire array.
Warning
C does not perform bound checking for an array, i.e., it does not check for the validity of the
subscript.
This responsibility is of the programmer. Hence the programmer should ensure that the array
length is not exceeded. Otherwise some other data may be overwritten.
Example :
in t n [ 5] ;
f~r(i =O; i <lO ; i++ )
ssanf f"%d ", &n[i]) ;
'f hJ !j c;<)(Je 1i, pcrfeclly valid in a C program.
/ /
hra;; W-rJ :,'J ''11
tLPta \
5-5
oata frorn an Array
eodlng
~
arraY elements can be read (accessed ) from the array u~ing c1 for lo<Jp a, lh"J~·. r
p.ll the
- i<lO ; i++)
(i~o ,
..
for
,
f( " \ n The value at positi o n Id is Id" ' -,
··- .··-.· ,
print
ffl1 .--:::. gram to read 10 integers in an array, display them and calculate their a"eraet "
~/* pro
#incll.l
cte <s tdi o .h>
ma i n ()
int n (l O) '
i,
sum
= O;
fl o at averag e ;
/ * ;,,.cc
ept data and ca l cu l a t e sum* /
. - o · i<lO ; i+ + ) {
for(i '
.
t
f(
" \ n Enter element
prin
%d", i ) ;
s can f( " 9.°c d " ' &n[i)) ;
sum = sum + n [ i ) ; }
/ * Display data* /
f o r (i
=
O; i<lO ; i++)
printf~" \ n Elemen t %dis %d", i , n [i ] ) ;
/* calcula te averag e* /
average = (float)s urn / 10 ;
printf( " \n The average is %f", average ) ;
...;}------------------------ □
An operation cannot be performed on a numeric array as a whole. The operation has to be
performed on individual ele1:1ents.
Examp le: An array cannot be directly copied into another by using the assignme nt operator.
Individual elements of the array have to be copied one-by-one.
{)0
5-6 / u1s1on
C Programming
3.2 Initializing an Array
. • ·a1·zed when it is declared.
Just like ordinary variables, an array can be mitl 1
The entire array or a part of it can be initialized.
·
An array can be initialized by the declarauon
.
f II Owed by an = sign and a list of val
°
Ues en 1
COsed
m braces and separated by commas.
The values are assigned in order, to array elements from SUbscript O.
Examples
1.
int num[S]
(10 , 1s ; 2s , 90 , 100) ;
num
~r rj'fs tfC: tJ
J
10
15
[O]
[1]
[2]
[3]
[4]
Figure 5.2
2.
char c[5] = { 'a', 'b', 'c', 'd', 'e'};
•
During initialization, it is not necessary to specify the array size. The compiler
allocates memory to hold the initialization values.
int num[ ]= { I0, 15, 25, 90, 100};
•
If less number of initialization values are specified, the remaining are initialized to
0.
int a[lO]= I 1, 2, 3};
Here a[O], a[ I] and a[2] are initialized to the specified values and the rest contain 0.
•
If more initializers than the specified number of array elements are specified, the
compiler gives an error.
Arrays and Strings
.
(.,}e \
lttltOII
5-7
s .
~oJT'I~ .
~
~ 1,
;\n example to find the largest and smallest number from an array of n integers.
1fJ1 --- gest and smallest number from array */
t* Laf
~
#).·nclll
.
de<stdio . h>
)110.if1 ()
i nt i,
n,
num[SOJ,
largest , smallest;
·ntf(" \ n Ho w many numbers") ,·
p:r1
f ( " %d", &n) ;
/1'
scan
ept numbers*/
ACC
. tf ( " \ n Array elements
prJ.D
f or(i==O ; i<n ; i++)
scan f(
\ n ");
" ~d"
° ' &num[i]);
/ *Initializat i on * /
1···~:;;:;·····························.......,
~
! How many numbers:
:: _5
I :~ray elements :
:
largest== smallest = num[ OJ ;
/ *Comparing with remaining elements* /
f or(i==l ; i<n ;i++ )
;
l
::
:
j
! -9
120
Q
!
36
if(num(i]
> largest)
largest = num[i] ;
The largest is 120
The smallest is -9
.......................................................
else if(num[ i ]< smallest)
s ma 11 est
= n um [ i ] ;
pr in t f( " \ n The largest is %d" , largest);
p rintf ( " \ n The smallest is %ct", smallest);
Q
5-8
I
{}'9
C programming
UISIOII
rraY·
Reversing an a
►► 2.
ray */
- Reversi.ng ""D ar
'
lnl t*
}E:!,
# in cl u d e< st d
m ai n ()
];
[SO] ' re v[ S O
a
nu m be rs " ) ;
in t 1 , n,
y
an
How m
p ri n tf (" \ n
n) ;
s ca n f ( "%d" , &
\n ");
m be rs *I
/ * A cc ep t nu
ts
en
em
el
y
~ .r ra
p ri n tf ( " \ n .r
;i + + )
fo r( i= O ;i <n
& a[ i] J ;
sc an f( "%d" ,
a rr a y * I
/ *Rev e rs in g
;i + + )
fo r( i =O; i< n
a [i ];
re v fn -i - 1] =
:" ) ;
d a rr a y is
se
er
v
re
he
T
p ri n tf ( " \ n
; i+ + )
fo r (i=O ; i< n
", re v [i] ) ;
p ri n t f ( " %d\n
.
►►
3.
Q
from
d prime numbers
an
en
ev
d/
od
ng
Displayi
*
:-:m-;b~er-s - : : : v ::n:--;a::n:d-:; p:ri~m:-:e:-:n:-:u
/~*:'°iO~d;:;:d/'.ii:e:e
• - :. ~ l ud e < s r d
i o . h>
a[ S O ], fl ag
., .: -, r.,
,,,.. ...-
,... ... F ( "
,_ - · · - -
n
;
How many nu m be rs " ) ;
• ~r~~y
. . . . . . ··;
r;utp~t
i o . r.>
el em en ts
\n " ) ;
mbers : 5
! Array elements :
/ 25
i -9
nu
j How many
i 120
!
!
I
r
/
/
i
!
0
36
j
!
!
is •
reversed array
120
I -9
j
I
I
2s
....................................... :
: .....................
I
1Q
an array
------/-;------
- - - - - -- -
Arrays and Strings
s c an
f( " %d",
&a[ i)) ;
•ng odd- even and prim e* /
· splaY J.
; t Dl· "<n· i++)
(i==O, J.
'
f
or:
if(a [i]% 2== 0)
prin tf( " %d is odd" , a[i] ) ;
prin tf( " %d is even ", a[i] ) ;
flag
=
l ;
/ * chec king prim e* /
for( j=2 ;
j <a[i ] ;
j++)
if(a[ i)%j ==O )
flag= O ; brea k;
if (flag == l.,
prin tf ("%d is prim e", a[i) ) ;
Q
~
4,
To conv ert a decimal number to binary, the decim
al number is successively
divided by 2 till the number becomes Oand the remainder
s of each division are
displayed in the reverse order.
.C, /* Decimal to binary conversion */
#i nc lu de< stdi o .h>
main ()
te!I
unsi gned in t
de c _num , rem( 16] ;
i nt i ;
/ * I nit i a lize rem aind er arra y* /
f or(i = O;i <1 6 ;i++ )
rem ( i) =
o;
pr i n t f ( " \ n Ente r the decim a l numb er : ")
i
scan f( " \% u",& dec_ num );
(),)
I
5-10 UISIOII
i= O;
C Programming
/ * sec i cc
0 th
element o f r emainder
array "
/ * divide numbe ::- succe ssive l y by 2'
wh ile (dec_ ~~m >Ol
rem[i]
=
Output a ..........................
Enter the
···,
number: 25
deciniai
!he binary e .
is: 1100100006u1va1en1
00000o
Output b
dec_ num % 2;
i++;
de c_ num = dec_ num / 2; . }
/*
Di splay remainder array in rever se order* /
l
printf( " \ n The bina ry equivalent is : \ n" ) ;
f o r (i =l S; i> =O; i - - )
.............................................~.~.' ...]
print f (" %d", rem [i ]) ;
►► 5.
To merge two sorted arrays into a third array such that the third is also.
1n the
sorted order.
Al
A2
A3
[J
Enter the
numb~r: 65535 decinia1
,The binary e .
l
is: 1111111~~,~~~e,ni \
10 25 90
9 16 22 26 100
9 10 16 22 25 26 90
100
/* Merging two sorted arrays of m and n elements respectively into a third array*/
#include<stdio . h>
-
main ()
int m, n, i,j , k , al[20) ,a2[20 ],a3[4 0) ;
p rintf( " Enter the length of the fi rst array :" ) ;
scanf (" %d ", &m) ;
/*
Accept elements of first array* /
pr i nt f ("Enter elements in the sorted order
\ n");
fo r(i =O;i <m; i++)
s,:;anf (" %ct", &al [ i)) ;
pr1ntf( " \ n Ent er the length of second array :" ) ;
ssar,f 1"%d",
&n) ;
!
/. " ,J1
~
,,
:;: : / ' ·,, ;
✓-
'iJSU.a
:5-,-, 1
element s o f ses0~d arr~; ·
ccePt
.A
· ot
p rlf or
sc
lG t~~
f( " \ n Enter e l ement s
. · <n ; i++)
( i ""o , 1.
00 f
("
%d "
'
&a2[i]) ;
int o the t hird arr ay as . . 0r.g ~ ::,
.
t-1er-g1.n9
. _ O· k== O;
'
. ,,,, O; J( ( i <rn) & & ( j < n) )
1- .
wn1- l e
1*
/ * In itiali21;-.g •
if(al[i] < a2[j])
a3[k] == al[i) ;
i++;k ++ ;
else
a3[k]
== a2 [ j l ;
j++ ;k++ ;
rrayl has not ended, copy its eleme nts~ ,
/ * if a
whil e (i<m)
a3[k]
==
al [ i];
i++;k++ ;
/ *if array 2 has not ended, copy its elements * /
while ( j <n)
a3[k]
a2 [ j] ;
j++ ; k++ ;
/ *Display merged array* /
printf(" \ n The merged array is \n" ) ;
for (i=O ; i <k ; i++)
1
Printf("%d\ t ", a3 [i )) ;
----- ---- ---- ---- ---Q
5-12
I
CJ.,
UIIIDII
C Programming
:·····················································································································
l
1 Enter the length of the first array : J
!
I
Enter •~ements in the sorted order :
j
I
Enter ttw"°length of the second array:§
Enter e~ements in the sorted order:
1
I
I
I
§
I
f~
I
:
19
l
I
:
19
:
/ The merged array is 3 5 5 7 9 10 15 19 19
,
!
l Output a
I
...............................................................................................................................
4.
Two-Dimensional Arrays
Most often we need to store and perform operations on matrices
and tables. For this, we can use a 2-D array.
A 2-D array has 2 subscripts - one indicating the row and the
other indicating the column.
Example:
,
.......'
,' 1'.
:
' \
~ _<,,.
'Oct. 2017·..:'2u · ,, __,,
Define 2-dimens·1
ona1
•
array. Give exampl e.
int m [ 3 J [ 2 J ;
mis declared as a 2-D array having 3 rows (0-2) and 2 columns (numbered 0-1). The el
emen~
are m[O][O], m[O][l ]...... m[2][1].
4. 1 Initializing the Array
A matrix can be initialized in two ways:
int m(3 ] [ 2)
= {
{1, 2 }
{3 , 4 }
{ 5 , 6}
};
J
Arra;a and St11r.g:,
===
i.Z
..
\
5,-13
{1 ,2, 3 ,4, 5 , 6} ;
m
RowO
rv,eni 0
(/
llJlll
col O col 1
1
2
Row 1
3
4
Row2
5
6
ry Repr esen tation (Physical View)
. linear. Hence even multidimensional arrays are
ytefllorY ·
A 2 _D array can be stored in memory in 2 ways.
·nearlY
•
red ll
;to
11..t"ajor Representation: Elements are stored rowgo\\7 1n
. . e elements of row O are stored first, row 1 next
j,
wise, i. .,
2-0 array can be
stored In memo ry in
2 ways.
15
i.
Row Major
Repres entatio n:
ii. Column Majo r
Repres entatio n
and so on.
Example
Row0
m
u.
Row 1
Aow2
1
2
3
4
s
6
m[O][0]
m[0][1]
m[1 ][0]
m[1 ][1]
m[2][0]
m[2][1]
Cohunn Major Representation: Elements are stored column -wise, i.e., eleme
nts of
column O are stored first, column 1 next and so on.
Column 0
m\
1
m[O][0]
I
Column 1
3
s
2
4
I s I
m[1 ][0]
m[2][0]
m[0][1 I
m[1 ][1]
m[2][1]
Address Calcu lation
Address of any elemen t m[i][j] in a matrix of r rows and c columns can be calcula
ted as
i.
Row major representation: Address of m[i] U]=base address + (i*c+j ) * elemen
t_size
ii.
Column Major: Addres s of m[i] li]=base addres s+ G*r+i) * elemen t_size
0-,
5-14 / UIIIOI
C Programming
4.3 Accepting Matrix Elements
We have already seen how data for one-dimensional array can be accepted from the
User
A similar method is used for a two-dimensional array except tbat we will now ha
·
one for the row subscript and the other for the column.
"' lo,
Ve l\
0
For every value of row subscript, the column subscript has to increment from 0
to nu
columns-1.
illbe1
Examples
The following program illustrates matrix addition.
.. 1.
[CJ
/* Program to add two matrices */
#inc l ude <std i o .h>
ma i n ()
int matl [l 0] (1 0] , mat2[1 0] (1 0] , mat 3 [10 ] [10] , r l , c l , r2 , c 2 , 1· , J. ;
printf( " How many rows and co l umns in mat r ix l ? : " ) ;
s c a nf( "%d%d" , &rl , &c l);
p rint f ("How many r ows and columns in mat rix 2? : " ) ;
sca nf (" %d%d ", &r2 , &c 2) ;
i f ( (rl == r 2)&& (c l==c2 )) / *chec k if they can be added* /
{
printf( " \ n Addi tion pos sibl e \ n") ;
printf (" \ n I nput Mat rix 1 : \ n");
f or (i=O;i <rl;i ++ )
fo r ( j =0 ; j <cl ; j ++ )
/ * r ow subscr ipt */
/ * col umn subsc ript */
sca nf (" %ct", &matl [i ] (j ] );
print f(" \ n Matrix 2 : \ n) ;
for (i = 0; i <r 2 ; it+)
f o r(j=0 ; j <c2 ; i++)
sca nf ( "%d" &mat 2 [ i ] [ j ]) ;
/ *add co rre sponding e lements of both ma trices */
ma t 3 [ i l [ j l
= ma t1 [ i ] [ j ] + mat 2 ( i ] ( j ] ;
, ,..
mi•• '
r es ul t
I
5-1 5
1r ;
;, o i s p la.Y
f(" \ n The s um i s
.
rl-n t
. o · i < rl; i+ + )
P
f o !' (1== '
n ,, I i
for (j= = O;j <c l ; j++ )
;
pr in tf ( " %5 d", ma t 3 [ i ) [ j l)
· tf( " \ n") '·
pr in
Of.
*/
e nd if
}/ *
els e
~~
1
e
("\ n Ad dit i on not pos s i· b l \ ::" ) ;
P rin tf
. *I
/ * end ma in
. :!!: ,
................
r·..;~~;~~·~····...........................................................................................
ix 1? :
HoW many rows and column~ in ma~
rix 2? :
HoW many rows and columns in mat
i,,,i
.
I
~
~
Addition not possible.
, Output b
i
rix 1?:
How many,;cws agd columns in mat
rix 27 :
How many~ows a~d columns in mat
Addition possible
· l Input Matrix 1
:·,•
.
I
,
I
:.!,
i,.!
I
I
Input Ma J 2
ii
1
The sum
i1
l
I
I
:
.!
············ ····························································································. ............................
:
5~ 16
I
V"
VIIIOII
.,. 2.
c Programming
Transpose of a matrix
Transpose is a matrix with the rows and columns of the original matrix interchanged .
.QI /*Transpose of a
•in2lude<s t d i o . h >
:r.a.1n ( J
1.nt ml[lO] [10 ], tr a ns [ lO ] [1 0 ] , i , j , r , c ;
2lrsc r() ;
print f ( " \ n Ho w ma ny r o ws and c olumns i n t he matri x
:" ) ;
scan f ( " %d%d", &r , &c ) ;
")i
print f(" \ n Ent er the e l e ments
/* r ro ws* /
f o r ( i=O ; i< rl ; i++)
/ * c c o lumns * /
f o r(j=O ; j<c ; j ++ )
I
scan f ("%d" , &ml [ i] [j ] ) ;
trans[j ] [i]= ml[i ) [j] ;
\ n" ) ;
pr.1ntf (" \ n The trans p ose i s
/ * c r ows * /
f o r(i=O ; i<c ; i++)
/ * r columns * /
for(j=O ; j < r ; j ++ )
printf( " %5d" , tra n s( i ] [j ] 1 ;
pri n t f(" \ n " ) ;
·································································•·...
.
Output
Hvfl _man; rows and columns in the matrix: ~
... rte·
g
•rs e'em ents
-7
-4
:
I
,..~ ''af S.½$':: 5.
7
4
~
2
i
.................................... ···································--···j
Cbeeking uppe
r/\ower
Arrays and Stri
ngs
triangular and sy
mmetric
O~\
"""
5-17
tt 3·
matrix.
. matrix is suc
h
th
at ati1G1 = aG1U1
J1111letfl
. ln an upl)er trian
f. s~ . d. Cgon
gular matri x all
a
l
are 0.
elements below
in ia
U\e¢ -
p01, i , j
, r , c , f l ag
in t a l 1 01
;
t" \ n Ho w
many rows
p r int f
a n d column
s i.n th e m
at r ix : n ) ;
f l " %d%d " ,
&
r
, &c) ;
scan
. f l" \ n
En t e r t h e
p rJ.n t
e l e me n ts
") ;
fo r ti.=0 ;i .
<r ; i+ + )
o
f O f \'J·- '· J' < c ; j ++ )
s c a n f \, 11 9.cd " &al i. 1 lj 1 ) ;
nag = 1 ;
0
'
: or (i = 0 i i<
r ; i+ + )
f o r l j =0 i j<
c ; j++)
i fl al ii l j
i \ =a l j i l i
1l
l f la g =0 ;
b re a k ; }
if ( fl a g = =
l )
p r in tf l " 'I
he mat r i x
is symme t r
ic") ;
e ls e
p r i:Yc.f t " T
h e ma tr ix
is n o t sym
roetric u) ;
f la g = l ;
fo r( i=0 ; i<
r ; i - +)
:o r ( j=0 ; j<
i ; j+ + )
if ta ll .i l j
i \ =0)
lf la g =0 ;
b re a k ; )
i f l f la g ==
l)
p r in tf t " T
Y tr
h e m a tr ix
is uppe - ia n g ula r " ) ;
e ls e
P r in tf t " T
e r t r ia n g u
h e ma t r ix
la r " ) ;
is no t upp
5-18
I
().,
VISION
C Programming
for (i =0; i <r ; i++)
for( j=0 ; j>i ;j ++ )
if ( a [ i l [ j l ! =0)
(fla g =0; brea k ; )
i f ( fla g== l)
lar" ) ;
prin tf(" The mat r ix is l ower tri angu
els e
ngu lar " ) ;
pri ntf( " The mat rix is not l ower tria
retu rn 0;
5.
Multidimensional Arrays
. However common programs do not ne
Multidimensional arrays have more than one subscript
to use arrays more than 3 dimensions.
Exa mple :
int rn[3) (4) (2) ;
arrays having 4 rows and 2 columns each.
Here m is an an·ay consisting of 3 two-dimensional
col 0 col 1
~
2D array 2
.
2D array 1
2D array 0
.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Figure 5.3
17
18
row O
19
20
row 1
21
22
row 2
23
24
row 3
:ed
Arrays and Slt1ngs
.. \
VlllOI
()
;
5-19
Memory Map
20 array 0
--.i.--
2D array 1 - - 1M4---
11
1002
14
1008
I
2D arra12 ~
8 1920 2122232:4
•
1014
Figure 5.4
Apf. 2018- 2M
Limitations of an Array·
6.
What are limitations of
Array?
The compiler uses static memory allocation for an array, i.e., we have to specify the array
size in advance. It is not possible to increase or decrease the array size at runtime.
l.
2.
Elements cannot be inserted into an array.
3.
we cannot delete elements into an array.
4.
If the number of elements to be stored is not known in advance, there may be memory
wastage if an array of large size is specified.
s.
If a small array size is specified, there may not be enough memory to place all elements.
6.
C does not perform bound checking on an array, i.e., it does not check for the validity of
the array subscript. Hence, if the array range is exceeded, some other data may ge~
overwritten.
7.
Arrays and Functions
Arrays can be passed to a function in two ways.
1.
ii.
Element by element
Passing the entire array
When the elements are passed to the function, their value is copied into the corresponding
fun ct·10n parameter and cannot
· .
be modified by the function
I
5-20
I
(.}0
UIIIOII
C Programming
· ai,·ay can be passed
An entire
as an argument tO the function. The function gets coIUplete aec
to the original array rather than a copied array.
Before we see how it
n~e.
7. l
can
ess
be done, it is important to underStaod th e importance of th
ea
~
Name of the Array
Consider the declaration
int n[lO) ;
The name of the array is n. The name n without any subscript refers to the starting add
.
ress or
base address of the array in memory, i.e., it refers to the memory locat10n of the Qth element of
the array.
Thus, n is eq uivalent to & n[O] . It is a constant pointer and cannot be changed ('
I.e.,
incremented/ decremented, etc.)
When we want to pass the array n to a function sum_of_elements, the call wi ll be given as
surn_ o f_elements(n) ;
This makes the complete array available to the function and the function can make modifications
to the original a.ITay.
This is so because by specifying only the array name, we are in effect sending the base address
of the array to the function.
Thus, the name of the array can be used effectively as any other pointer to the array in accessing
the array elements. This is illustrated below.
Example
/*Illustrate array name as a pointer to the array */
_ ·.:
'Cl • '
51
i fi!i;~~~:~:~~:~ l
= ( 1 0 , 2 0 , 3 0 , 4 0 , 5 0 };
r c::iins %d \n ", (al+i),
~
(al+i)) ;
.:.~ 65494
65496
i 65498
1 65500
contains 20
contains 30
contains 40
contains 50
:i,
i
i
i.......................................................
Arrays and Strings
{)e \
UIIIOII
5-21
contains the base address, al+O points to the 0th element
and *(al +O) gives the valu e of
·oce a 1
s1
Ul elernent.
weO
.
wes ha11
7,2
.
see rnore about arrays and pomters m
the next section.
passing the Array Element By Element
'fhe arraY
'[be f un
elements can-be passed one-by-one to the function.
ction thus gets acce ss only to one element at a time and
cannot modify this valu e.
le as f ollows:
ExamP
main ()
int n[S) ={l 0 , 20 , 30 , 40 , 50) ;
vo id d ispl ay( int) ;
/ * func tion prot otyp e* /
int. i ;
f or( i=O ; i<S ; i++)
ctis play (n[i ));
vo id disp lay( int x )
p r intf ( " %d " , x ) ;
7.3 Passing the Entire Array
In orde r to pass the entire array , we just have to send
the name of the array to the function.
Since the nam e contains the base address, we are
effectively sending the entire array . The
function gets access to the original array (since it has
the base address). It can mod ify the array
contents.
{)0
5-22
I u1s1on
C Programming
Eram ple is as follows :
ma i n ()
int n[S] = {10 , 20,3 0,4 0, 50} ;
/ *Fun ction decl arat ion* /
vo id mod ify(i nt b[S) ) ;
modi fy(n ) ;
void mod ify(i nt b[S] )
int i;
for( i=0 ; i <S ; i++)
b[i)
= b[i) * 2 ;
the array n, they are changed to
In this exam ple, the function modifies the contents of
20,40,60,80 and 100.
LNote I
l
e b stores the base address of n
b is not a new array created but it is the same as n becaus
(Covered in detail in the next chapter).
7.4 Passing Dimensional Array to Function
l array to a function to find the
The following program illustrates passing of an one dimen siona
largest number from the array.
C
/*Passing an array to a function */
~~nc lude <std io . h>
-~L
array (SO] , i,n ;
:r.: ... arge st ( int x [] , int n) ;
/ * func tion prot otyp e * /
~r.~ :~I" ~ How many elem e nt s? :" )
·1a • ..ies • /
f-, _; · : " - :np' r ·;al Jes
n" );
'
►
Arrays and S1rings
(_}£ \
VISIOII
5-23
• / n ·i++)
( i;:::: O'• ).' '
f0'
n.d" &array[ i]);
f ( ,, "6 '
00
sc
. o n call and display result * /
I * ftlnct J.
f( " \n Largest number= %d", largest (arra y , n )J ;
rint
.
P
est(int x[] , int n)
a
r
g
l
iI1t
1arge = x[ O] ;
int j'
·-1
for ( J - '· j<n ; j++)
/ * initializ e
Tl
/ * compare remainin g eleme n r.s " /
if(x [j] > large)
large = x[j];
return
( large) ;
_1-,;,__---------------~
,..................................
1
l
l
!
output
\ How many elements? : §.
\ Input values: 100
:
0
\
:90
\
i
48
1
:
1
1
2000
!
l Largest number= 2000
j
i.......................................................................................................................:
Note: The function header could also be written as
i nt large s t ( i n t
* x,
int n)
In both the cases, i.e., int x[] and int *x, x means a pointer to an int. The loop inside the function
could also be written as
for(i=l ; j<n ; j ++)
if (* (x + j )> l a rge)
large = *( x+ j) ;
I
00
5-24 UISIOII
C Programming
In the above program, we have passed a 1D array to the function. In the same way
, We c
an Pa
a 2D array to the function as shown below.
81
Program to accept, multiply and display matrices using functions.
Example
Q/*Illustration of 2D Arrays and functions */
#i nclude <s tdio . h>
ma i n( )
int ver i fy(int x , int y) ;
vo id readmat( i nt x ( l 0 ] (10), int r , int c ) ;
vo i d multma t ( int x( l 0) [l 0] , int y[l 0 ] [1 0) , int z [1 0 ) [10 ) ;
vo id dispmat(in t m[ l 0 ] [1 0] , int r , int c) ;
i nt a [ l 0 ] [10 ], b[l0] [10] , c[l0 , 10], rl, r2 , c l , c2 ;
pr i ntf( " \ n Number o f r ows a nd co l umns i n matrix A:" ) ;
sca n f( " %d %d", &r l , &cl) ;
prin t f (" \ n Number of r ow s a nd co l umns in matr i x B:" ) ;
scan f (" %d%d " , &r2 , &c2) ;
if(verify (c l , r2) == 1 )
printf( " \ n Mul tipli c ati on Possible \ n" ) ;
printf(" \ n Input Matr i x A \ n" ) ;
readmat(a , r l ,c l ) ;
pri ntf(" \ n Input Mat rix B \ n " ) ;
readmat(b , r2,c2) ;
/ *Multip ly A and B givi ng C */
mul tmat(a,b , c ,r l , cl , c2) ;
/ *
display result* /
printf(" \ n The resultant matr ix is \ n " ) ;
11spmat(c , rl, c2) ;
} /*
end if * /
J
•
►
Arrays and Stflngs
ns o f A must be equa l t o r o ws
).
. tf{ " \ n Mul tipl icat ion not .Pos sibl e" '
. tf(C olum
pr 1n
R
• _
1r.
O• \ 5-25
lftllO I
n u J ,·
pr1n
a dma t ( int x [ 1 0 l [ 1 0 J , int r, int c )
re
void
int i, j;
for(i === O;i< r;i++ )
for(j === O;j< c;j++ )
scan f ( " %d",
int ver ify( i nt x,
&x [
i
l [ j)) ;
int y)
retur n(x= ===y ) ;
}
vo1'd
disp mat (int m[l OJ (1 0 ] ,
i nt i,
int r, int c)
j;
f o r ( i= O; i < r;i+ +)
f o r (j = O; j <c ; j++)
p rin tf("% 5d" , m[i] ( j]);
p r in tf ( " \ n");
(1 0] ,
vo i d mult mat (int x [ l O] ( 1 0 ], int y [ lO]
int z (10] (10] , int rl , int cl , int c2)
int i , j , k ;
for( i=O ; i< rl,i+ +)
for( j = O; j<c2 ;j++ )
z[i) (j]= O;
for( k = O; k <cl ; k++)
z[i] (j] = z(i] [ j] + x [ i ] [ k l
*
y[k ] ( j] ;
- -- - -- - - - - - - - - - - - - -Q.
5-26/ VI~~
C Programming
Output a
Number of rows and columns is matrix A : 2
~umber of rows and columns in matrix B : g
ol~m~s of A must be equal to rows in B
Multrplrcation not possible.
g
Output b
Number of rows and columns in matrix A : 2
Number of rows and columns in matrix B : 2Multiplication possible
Input Matrix A
g
1 1
1
Input Matrix B -
2
-
g J 1
4
5
The resultant ~atrix
6
1
i;
8 2
10 13 3
.
.
•·········•·············............................. .............................................................................J
8.
String
r--::::----...!.: 4
A string is an array of characters terminated by a special
character called NULL character ('\O').
Strings in C are enclosed within double quotes.
Example: "Welcome to C" is a string and it is stored in memory
~pr.18,0ct.16- 2M ',. __
Define string with
example.
Oct., Apr. 17 - 2M ·
Define string. Give
example .
·
as:
w
1000
e
C
1002
j
0
1004
m
e
1006
t
1008
C
0
1010
\I
10
Each character is stored in 1 byte as its ASCII code. Since the string is stored as an array, ii
possible to manipulate individual characters using either subscript or pointer notation.
f.rra; s ar.1J Stnr~
fl•\
11S1U
5--27
ring and Initializing Strings
.1
6
'flee
0
·og is a
. ng n
st! J.
8'!
d
-
ame[length) ;
•
.-rllines the maximum number of characters in the string
~d~~J•-
file }etlg
tes:
c,,:ClrrlP
P
fher
. .
character array, 1t 1s eclared as follows.
.
8 s~
51
~
eclO
a r cit Y [ 1 o)
.
;
cl:1 r name [ 2 O) ;
cna.,,. message [ 80) ;
cha .I,.
ays to initialize strings.
.
e are ewo w
char city [] = { 'p', 'u', 'n', 'e', ' \ 0 ' } ;
j,
Bowever, C offers a beue·r way to initialize strings.
char ci· ty [] ="Pune" ,·
,·
The compiler automatically stores the null character at the end of the string .
the following two declarations. But are valid however there is a distinction between the
.
consider
tWO,
\
I
,'
char amessg [] ="C programming language";
cha r *pmessg=" C programming language ";
/ *arra y"' /
/ *poi nt.er~
amessg is an array big enough to hold the sequence of characters and '\O'.
Even if the characters are later changed, amessg will always refer to the same storage.
On the other hand, pmessg is a pointer, initialized to point to a string constant.
pmessg may be later made to point elsewhere.
The result is undefined if we try to modify the string contents.
D
12
l is
amessg \ C programming language \0 \
.,_____.. C programming language \0
Figure 5.5
I
00
5-28 u1s1on
C Programming
8. 2 String Inpu t/ Output
Similarly the functions printf and scanf can also be used wi th th
e format specifier o/,
.
os. 1'h
function does not allow a strino with embedded spaces. gets()
allows a stnng to co
e Sc~.
e
.
ntajI)
" I(
We shall see an example of usino the printf function to display
chara
cters of a string. Spac.
,
e
1n th •s.
of a kite.
e Sha1x;
Although this can be obtained by different methods, the easieSt
of them is to use th
.
e fo11n .
features provided by the printf functi.on.
attin~
If the string is abed, the required output is
aa
abab
abcabc
abcdabcd
abcabc
abab
aa
It]
~ ;*~'7: D~i-=-sp~l:""'a_y_s".""
tr".in
""-g-:i-n""'t:.'he---:sh_a_p_e_o-::-f-a":"'ki"."t-"e --::*'."/':" - - #inc lude <std io . h>
main ()
-------------......
{
char s[80 ) ;
puts ( " Plea se ente r the strin g" ) ;
gets (s) ;
I * Uppe r tr iang le * /
for( i=l ;i <=st rlen (s) ; i++)
{
prin tf( " %40 . *s "i , s) ;
/ * Disp lays i c hara cter s fr om stri ngs
in a tota l o f 40
cha ra c ters .De faul t is righ t just ifi cati
on* /
prin tf(" %-40 . *s ", i , s) ;
/*D isp lays i char acte rs left justi fied
*/
i
I , , Lowe r Tria ng le * /
for( i=st rle n(s) -1; i> =l; i--)
prin tf( " %4 0 .*s" , i , s ) ;
pr:n tf("% - 40 . *s ", i,s) ;
_,;,___ __ __ ____ __ __ __ __ _ g
~
(
Arrays and St"r.gs
stondar
d Library String Functions
prov1·des a large number of functions in the header tile string•h for the han dlir.c0
,'
, 1at1guage
(>f
.
/
,triJlg
5
nly used functions are:
·
,.,i0st corn,rno
rh8
()·
.
.
This funct10n returns an mteger corresponding to the number of characters in the
strJen .
.
•fied stnng.
spec1
:'
size_t strlen (char * s)
syntax,:
, --
. - - - - - - - ----
,
1 \
,'
Apr. 2016- 2M
Give syntax and use of
strlenO and strcat(}.
char str[20] ;
gets (str);
printf("%d", strlen(str)) ;
Example:
,.
If the user enters C language, the output will be 10.
. This function is used to concatenate Uoin) two strings It concatenates a copy of
ii,
strcat() ·
.
·
the second string to the first and returns the first. The second remains unchanged.
Syntax:
·
char * strc at(char * s1 , char* s2 )
Example:
c har s1[2 0 )= "Pune ",s 2[2 0 ] =" Murnbai ";
st rcat (s l, s2) ;
puts( sl ) ;
p uts (s2 ) ;
The output is:
PuneMumbai
Mumbai
· ·u"
I '
strcmp(): This function is used to compare two strings. It returns an integer.
which is -ve
~
if string 1<string 2.
0 if string l is equal to string 2
+ve if string 1 > string 2.
Syntax:
Example:
i nt strcmp(cha r * s1, char • s 2) ;
Oct. 2016- 2M
What is the prototype and
use of strcmp() ?
c har sl [lO)=" ABC", S2[10]="abc ";
p rintf(" %d" , strcrnp (s1 ,s 2)) ;
The output will be - ve. Since "ABC" is less than "abc" because the ASCII value of 'A'
is<ASCII value of 'a'.
1%2~~
The fu nction strcmpi is used to compare two strings ignoring the case.
5-30
I o.,
VIIIOII
C Programming
.
iv.
t of string 2 to string 1 and retu
ms string
.
strcpy: This function copies the conten s
l. \
original contents of string are lost.
Syntax:
Example:
r *s2)
char *strcpy (char *sl , c h a
20 ) =" Mumb a i. " ;
c har s1[ 20) ="Pune", ? 2 [
strcpy(sl , s2 ) ;
puts (sl);
ii. strlen
puts(s2) ;
iii. tolower
iv. strrev
V. strcpy
The output will be Mumbai
Mumbai
The following are other string handling functions:
Name
Prototype
Description
strlen
size_t strlen(char • s)
Returns the length of the string (numbe~
characters excluding the NULL character)o
strcpy
char • strcpy(char *d, char *s)
Copies the contents of string s to d and r~
urns Pointer
to d
strcat
char • strcat(char* s1 ,char *s2)
Concatenates a copy of s2 to s 1 and terminat
es s1
with a null. Returns s1.
strcmp
int strcmp(char *s 1,char *s2)
Compares s1 and s2 and returns ,
-ve if s1 is less than s2
0 if s1 is equal to s2
+ve ifs 1 is greater than s2
strcmpi
int strcmpi(char *s1 , char· s2)
Compares s1 and s2 ignoring the case and returns
similar results as strcmp.
strlwr
char *strlwr(char *s)
Converts a string pointed to by s to lowercase.
strupr
char *strupr(char *s)
Converts a string pointed to by s to uppercase.
strncat
char • strncat(char *s1,char *s2, int
n)
Concatenates first n characters of s2 to s1 and returns 7
s1. s2 is unchanged.
I
r--
I
strre,
- char 'strrev(char 's)
Reverses the string s and returns the reversed string.
str~rir
char 'strchr(char ·s,char ch)
Returns a pointer to the first occurrence of character
ch in string s.
l,; 1 f6l/
i:.11ar ·sustr(chars·1 ,char *s2)
Returns a pointer to the first occurrence of s2 in s1 .
Returns null of no match 1s found .
.
.
-
r:.har
.. ,;i' a,
. strsetlchar .s.char ch)
~
Sets all characters in pointed to by s to the value ol
ch
• :.tm;;.et 'char 's char ch, int n) / Sets the first n characters of string s to the value?' ch
Arrays and Strings
5tr
mp(char•s1 ,char*s2,int n)
...1 crTlP
•
char
•strdup(char *s)
• strchr(char •s, int c)
char
Compares first n characters of s1 and s2. Returns<O
if s1 is less than s2,0 if they are the same, >0 if s1 is
greater than s2 .
Duplicates a string at another location and returns
NULL if space could not be allocated or returns a
pointer to the storage location containing duplicated
string.
strdLlP
char
5-31
Conv~rts a string pointed to by s into an integer,
returning the result. Similarly there is atol and atof.
i(char •s)
int a to
int strnc
Oe \
VJIIOI
Returns a pointer to the last occurrence of character c
in string s, NULL if not found.
--i
by
Searches s1 for tokens that are separated
delimiters specified in s2. Returns the pointers to the
first character of first token in s1 .
~---------t---.::...______::. .:. . _:,. : .__::.:_________
•strtok(char •s1 ,char *s2)
strtOI<
. tf and sscanf
• tf is the same as printf except that the output is written into a string rather than
sprtr"I
.
spr1n
. layed on the output device. The return value is equal to the number of characters
• d1SP
actually placed into the array.
The string is terminated with '\0' and it must be long enough to hold the result.
Prototype:
int sprint f ( char * buf, char * format,
Example:
cha r
s ( 80 l ;
spri ntf(str,
arg_l ist ) ;
" %s %d %f", "Hello" , 2, 5.0);
This will result in the data. Hello 2 5.0 to be put into the string str.
•
sscanf is identical to scanf except that data is read from the array pointed to by buf rather
than stdin. The return value is equal to the number of fields that were actually assigned
values.
Prororype:
int sscanf (cha r *buf , char format , arg_list)
Example :
char s[20] ;
1.nt nl , n2 ;
sscan f ( " hello 1 0
20 " , " %s %d %d ", s, &nl,
&n2);
Thi<; assigns hello to strings and 10 and 20 to nl and n2 respectively.
5-32 / v1?o'11
C Programmmg
10. Array of Strings
•
· al array. This is often required in applicatio
An array of strinos is a two d1mens10n
ns dealin
~
a list of names, etc.
lw
%
An array of strings can be initialized.
Example:
11
char c i t i es ( 4 ] [ l 0 ] = ( "Pune , "Mumbai", "Delh i ", " Ch e nnai " );
They are stored as:
O
2
3
4
5
6
\0
cities [OJ
p
u
n
e \0
cities [1]
M
u
m
b
a
i
cities [2]
D
e
I
h
i
\0
cities [3)
C
h
e
n
n
a
i
7
89
\0
In the following program, we will accept 'n' names and sort them alphabetically:
Q Ii;*;";ni71'us."'.': :t:rat::-e::-:s'."'a:=-:arr
n = a:-:-y'."'o:~f~s=tr1-:-.n:-:g:-:-s-:*-;
/ - - - - - - - - - - -- - --------..
#include<s t dio . h>
main ()
{
char names [ 2 0) [ 3 0 ) , temp [ 30) ;
int i , j , n ;
print f (" Ho w many na mes ? : " ) ;
sca n f (" %d" , &n) ;
I* Accept names */
for(i =0;i<n ; i t + )
/ *list o f 2 0 na mes */
{
p r intf("Ente r name %d ," i+l) ;
gets (list [i)) ;
/* li s t [ i ] i s the name o f the Fh st ri ng */
)
I* Sor ting* /
f o r(i=0 ; i<n-1 ; i++)
for(j=i+l ; j<n ; j ++ ) ;
i f(strcmp(name s[i] , names[j]> 0)
~tr c py(temp , names[i]) ;
','"u:py /names[i), names(j]);
/ ,) 6 \
Afra; s ar.d S:r ".•;p
l'IIJH \
5-33
strcpy (n ames [j] , temp ) ;
sort e d List r /
}
·sp l aY
/ "' DJ.
" The s o rted name s are" ) ;
p\JtS (
. <n . i++ )
'
r ( i === Oi i
fo
(names [ i]) ;
-=-
pt.1tS
- - - - - - - - - - - - - - - - - - - - - - - - - :'ir:
11.
implementation without Standard Library Functions
d pointers are very closely related, since a string is an array the name of the srrirnz is a
strings an
·
ointer
to
the
string.
In
this
section
we
will
see
some
functions
on
strines.
constant P
...
Find the length of the string.
g
int 1ength(cha r *s)
int count=O ;
while (*s! = ' \ 0 ' )
count++;
s++ ;
re turn count ;
main ()
char str[80] ;
gets (str) ;
Printf( "The length i s %d" , lengt h(s tr )) ;
-}- - - - - - - - - - - - - - - - - - - - - Q
5-34
I
00
UISIOII
C Programming
st
Copy one string to another and return the copied ring.
.. 2.
Q -- -- -- -- -- -- -- -- -
char* scopy (char * s2, cha r* s1)
char *t=s2 ;
while (*s1 ! = ' \ 0 ' )
{ *s2=* s1 ;
s1++;
s2++;
*s2= ' \ 0 '; I termi nate copie d s tring * /
}
retur n t ; / * return copied strin g* /
main ( J
{
c har sourc e[20] , targe t(20) ;
gets( sou r c e) ;
puts( scopy (targe t , sou rc e )) ;
Convert a string to uppercase and return the converted string.
.. 3.
Q- -- -- -- -- -- -- -- ----
c har * suppe r(cha r *s )
c har *t =s;
while (*s ! = ' \ 0 ')
{
if (islow er (*s))
t o upper (*s) ;
s ++;
retur n t ;
J
main ()
I
c har s tr ( 8 0 ] ;
g ets (str) ;
pu t s (suppe r ( st r));
~--- -- ---- -- -- -- -- --,
V
Arrays and Stnngs
5-35
111~•1 \
d progra
~m
:..:.::...
s_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
soive~
~
1
-w--'-r:i~t·e a C program to find the largest number from 3 x 3 t .
.
ma nx.
~g
Jtit"lcl ll
cte<std io . h>
!1lait1 ()
I
. nt m, O , c , d , mat r ix [ 3 ] [ 3 ] , maximum ;
l. .
f ( " Enter the elemen ts of matrix \ n " ) ;
rJ..n t
P
-o ,· c<3 ; c++)
f or (c-
,' 1 \
I
I
...--_-A_pr
___2_0_1_8___4--,/>-
j
\
I
.{
for (d= O;
d<3 ; d++)
scanf( " %d ",
&matrix [ c][d)) ;
!.,
~a x imum= matri x[ O] [OJ ;
for(c = O; c<3 ; c ++)
for (d= O;
d<3; d++)
f
~E~he elemenffi of
, 1
45
56
I 2
56
48
:
l 10
15
98
Maximum element
l matrix is 98
l
1.,,
,
l
:
l
in
j
l
~
...................
................... ........-- ·-----~
~
-
if(mat ri x[c] [d) >maximu m)
max imum = matrix [c ] [d);
printf ( " Maxirnu m elemen t in matrix is l d \ n ", maximum) ;
return O;
....:........:....___ ___ ___ ___ ___ ___ ___ _ Q
~
2.
Trace the output and justify:
main()
{
static char str[ ] = "Malayalam";
char * s;
s = str + 8;
while(s >= str)
{
printf("o/oc", *s);
s- - ;
,
--- '
,' 1 \
______:, j'
'
,
Apr. 2018- 4M ,_ -:
5 3 / ()4,
- s VIIIOII
C Programming
51>/1aion
str [
M
/
a
I
I
/
I
Y
/
a
a
str(?]
str[6]
str[S]
str[4]
str[3]
str[2]
str(1]
str[O]
I
I
I
a
str[9]
str(8]
I
m
I
~
· · '==~Lr+ 8 poims to last locati on. i.e. str[8]
A\
111
while loop character is displayed by every iteration and s is decrememed by one lo
Ourpur:
.,. J.
,
ma layalaM
:'
Trace the output and justify:
ma111( J
Apr. 2018-
(Apr.20 18)
I
char* a(5 ] = {"GOOD", "BAD", "UGLY". "WICKED", "NICE");
prim!'(" % s\n", al0JJ;
prilllf( " '½ <;\n ", ·• ( a+ 2));
print! ('''½ c\n", * (a[2 1+ 2)):
pnntf( " 1k ..,\n", * af3]);
pnnlff'% c\n", ... (a[J j + 2));
Solurw11
Arr.t) of ..,tnng..,
1:-.
g1vi:n as
a[OI
a[ 11
! G I O I O I D I \0 :
B
D \0
A
1
I
a[2]
L1
G L
af 3] W
af➔ ]
Ju• ur
.
cation
GOOD
I
( ,I 'r
Y.f( Yf [)
(
~
----K
E
C E
\0
C
_...
Y 1 \0 '
D \0
_,
,
1 '-:
4M'. __ ,,
Arrays and Strings
~
write a
Oe \
UIIIOII
5-37
'C' program for multiplication of p x q matrix.
4.
, ... --,,
,_________, , 1 \
\
Oct. 2017- 4M' .. ___ .,
p , q ' c , d , k , sum=0 ;
n
. t rn ,· r s t' [ l O] [ 10) , seco nd (l0J [ 1 0) , multiply [ l 0J (10).'
1n
,int fJ. ( " En ter n o . o f r o ws and columns o f fi r st matri x \ n " ) ;
print f ,, %d %d " ' &m, &n ) ;
scanf ( ( " En ter elements o f first mat r ix \ n " ) ;
nr.i.ntf . c<m ; c +T )
)
(c::: O,
rf o r ::: O; d < n ; d ++
f o r (d ( " %d " , &f i rst ( c l [d ] ) ;
f
\ ri" ) ;
scan
. t: f ( ,, w~nte r no . o f r o ws and columns o f second matrix
po n f ( " %d %d ", &p , &q ) ;
scan
\ ")
·
b
,
· f ( n ! ==p)
e mu 1 t1plied with each other n ;
1 . _ f ( " :-h e mat. rice s c a n t.
o r1 nc.
e lse
r·" - -&( "E ~ ~ e r e l e me n t. s o : second mat rix \ n " ) ;
·· P-c ++ )
.: - y (~-.
..._ - ....,, ,• c<p ;
,
; : ; ('.:i==v ; d< q ; d T+ )
d] ) ;
.
econd[c]
&s
",
~d
"
'
sea .. -&
: c.:-(:: ==
;
::<~ ;
c -- )
: :.r ~ = 8 ; k <p ; k -- )
s~~
= surn- :~ !" st.'.:: J [k l
~
seco nd [k ] [d] ;
p:::..:"'-:1 " ?:-duct. o : the matri c es : \ n " ) ;
.
! (~c
;
~,r ; c -- )
: :: d= ; d<q ; d· - )
p ! :. :1 t. !: " ~ j t. " , r:-:.: : t. .:. p l y [ c ) [ d ) ) ;
,
I
{),:,
5-38 / UISIOII
C Programming
······················································································································1.
Outpur
Enter no. of rows and columns of first matrix
23
Enter elements of first matrix
159785
Enter no. of rows and columns of second matrix
32
Enter elements of second matrix
459873
Product of the matrices:
112 72
135 114
• ••• • • ••• ••••• • • • • •••• •• •• ••••• •• • •• • • ••••••••• • •••••• • ••••••• • • • • •• •••••• •• • • • ..... ... . . ... . .. .. . ...... . . . ...... . . . . 1
» 5.
Trace the output and justify:
--, '
,' 1',
,
main()
I
I
I
{
r - - - - - - -_.:_I
\
char s[]= "Happy New Year";
printf("%s", s);
printf("\n%c", s[3]);
printf("\n%c", s[8]);
Solution
String s is stored as,
s[O]
s[1]
s[2]
s[3]
s[4]
s[5]
s[6]
printf("%s", s) displays - Happy New Year
~[3] - p
1., !8] - w
Output · Happy New Year
p
w
s[7]
I
Oct. 2017- 4M '~--·
s[8]
s[9]
s[10]
s[11]
s[12]
s[13]
s[1 4]
Write a program to concatenate two strings.
1
strl[25] ,
str2[25] ;
.
' == o, J == 0
'
iI1t l
. f{ " \ n Enter first string :" ) ;
print
ets(s tr l) ;
g ·ie(strl[i· 11.- ' \0 ' )
cna r
.
Apr. 2017 - 4'/
whl
i++ ;
while( st r2 [ J') I. == ' \0 ' )
s trl [i] == str2[j) ;
j++;
i+ +;
s t rl[i) == ' \0 ';
printf ( " \n concate nated string is %s ", strl) ;
return 0 ;
...:..--------- ---------- ---------- --g
r,==-
.. 7.
Trace the output and justify
main()
,,- 1-- \
...
I
-r--A-pr-._20_1_7__-4--~ -
char sl [] = "Kolkotta";
char s2[] = "Pune";
strcpy (sl, s2);
printf("o/os", sl);
...
-j
I
5-40
o~
I u1s1on
c Programming
Solurion
Output: Pune
strcpy(s 1. s:2) copies second string into first string.
:. Output is Pune .
.. 8.
Trace the output and justify
main()
---
r - - - - - - 1: 1\
I
'
I
\
I
Oct. 2016 - M,, __ ,,
4
char string[] = "HELLO WORLD";
int m;
for(m = 0; string [m] ! = '\0'; m++)
if((m % 2) = = 0)
printf("o/o c" , string [m]);
Solution
In this program all odd numbers alphabets in the string are skipped •
Output is HLOWRD
Trace the output and justify
main()
{
int c, d;
char string []= "A BC D";
printf("\n\n ");
for(c = O; c <= 11; c++)
{
d= c +l ;
printf("o/o-12,* "s\n", d, string);
}
printf("\n");
for( c = 11 ; c >= O; c- - )
{
\
Oct. 2016-4M ', . . ,
I
d== C+ l ;
printf( "- 12,* "s\n", d, string);
} . f("\n")·
print
'
·on
Sol,,tz
· f statement m
· b
rogram , pnnt
oth loop is not proper k-. .
,
•
ai ven P
. .
~a use n prr,·.iue
r ,l, o ~ ar.:..: .::
1" we r,
d string out of one 1s mteger and other is characte b
w _ entifier ,
.
.
r ut comro1~tnngder-,;-::?=' ·. . ::-::
or 1d .
d
o/os
for
stnng
also
control
stnng
fonnat
is not given p
nt an
rope rl ".
~~1
.
~ 10,
Trace the output and justify:
#includ e<strin g.h>
#includ e<ctyp e.h>
int rnain(void)
:1
,______ ____\
,1•
Apr. 2016- -!.I.'· - - - ,
int length i;
char string[ ] = "This is A string";
length = strlen(s tring);
for(i = 0, i < length; i + 1)
{
string[i ] = to lower(string[i]) ;
}
printf(" o/os\n", string);
getch() ;
return 0 ;
Solution
The given code will display "this is a string" (i. e. all characters of given string will be displaye
d
in lower case).
strlen() functio n will give the length of string and toLower0 function converts the characte
r to
lower case. In for loop each character of string is converted to lower case using tolower
O
function . So the output.
5-42
I
(},,
UISIDn
C Programmmg
Exercises
A.
Short answer questions:
I.
What i-. row major representation of an array?
List an)' four standard string library functi ons.
3.
What is the difference between array and ordinary variable?
What i-. the use of strrev and strcmpi ?
5.
Define row major and column major.
6.
Li-.t -.tring manipulation fu nctions.
7.
Define string. Which character is used to temlinate the string?
8.
What i-. an array? Give an example.
9.
Define '> lring with an example.
10
Whai 1·s 1hc difference between the S's in these two expressions - int num [5] ·, nurn [5 :
.
I I .!
II
f:.xplarn strlen() and strcpy( ).
12.
Define array. How is it declared?
What 1.., the Prototype and use of strcmp()?
1-l.
What are the advantages of Array?
15
\\ hat 1.., ..,tnng > State at least -l function!) of ,tring with c;ymax
16
What t'> -.,1r,tr( l function ·>
17
l::x pl..1111 ~pnntl anJ ,..,cant \.\- 1th example.
H.
Long anliwers questions:
/~ Marl.
Deline arr,1} Hl \\ 2D arra)" are repre~enteJ m memor}'
\\ me
1 (
prol!ntm
h>
,h.Lept IO number, from u,er. ,tore the-.,e numtier.., m ,111 JIT:J~ Jr.
m.ille I number ffl)ffi ,IITcl)
Arrays and Strings
111~•11 \
5-43
'C' pro gram to calculate length of string w·th
.
1
· out usmg standard library function
.
. an arra y? Wh at are the limitations of array?
What is
..
wnt. e •C' pro gram to calculate sum of non-diagonal e1
.
ements of n 'i'. n matnx.
. d the outp ut
Write a
5.
f10
6-
rnain O
cha r S 1[] = "FY BCA SYBCA TYBCA";
cha r S2[ 20];
cha r S3[ 20];
sscanf (S l ,"% S %S" , S 1. S2, S3);
printf< "o/oS, %S, %S", Sl, S2, S3);
7.
Explain memory representation of two-dimensio
nal array.
Explain U!,e of followin g standard library func
tion s: strlen(), strcmp(), strcatO, strcpy(),
strre v( ).
9.
Write a 'C' program to print arra y elements in reve
rse
order.
10.
Write a 'C' program to calculate sum of majo
r and minor diagonal elements of m x n
matri x.
11.
Write a 'C' prog ram to check if given string is palin
dro_me or not.
12.
Write a 'C' pwg ram to displa) transpose of a matr
ix.
13.
Write
:i
program to :iccept two strings and print common
characters
from both the strings.
5.4 4
I
('.'./@
UISI OII
C Programm ing
ed in Previous Exams
sk
.
tions a
Ques
PJMffftffl
What are lim itat ion s of array?
1.
2.
De fine stri ng with exa mpl e.
3.
mpl e.
Define 2-d ime nsio nal array. Gi ve exa
5.
De fine st1ing. Give exa mpl e.
arrays in memory?
Wh at is array? Ho w to represent 2D-
6.
Wh at is an array? Give an example.
4.
8.
mp()?
Wh at is the prototy pe and use of strc
strcat().
Gi ve syn tax and use of strlen() and
9.
ensional array.
De fin e array. Giv e exa mp le of one-dim
7.
O•Mtatffl
num ber fro m 3 x 3 ma trix .
Wr ite a C pro gra m to find the largest
1.
Tra ce the out put and jus tify :
main()
{
"UGLY", "W ICK ED ", "NI CE "};
cha r* a[S] = {"G OO D", "BA D",
pri nt[ ("% s\n ", a[O]);
pri nt[ ("% s\n ", *(a + 2));
pri nt[ ("% c\n", * (a[ 2] + 2));
pri nt[( "% s\n", * a[3 ]);
pri nt[ ("% c\n ", * (a[ 3] + 2)) ;
}
3.
Trace the out put and justify:
ma in( )
' sta tic cha r str[ ] = ''M alayalam" ;
char* s;
str + 8;
while( s >= str)
>=
( pn mf r %c", ~s);
1,,- - ,
(Apr. 20
y
. a 'C' program for multiplication f
O p )( q lllatrix
. write
J
·
the output and justify:
frace
rnainO
Arrays and St .
nngs
o.\ 5.45
v11101
~
IQ_ct. 201Zl
{ bars[ ] = "Happy New Year";
" s)·
crintf( "m
'
-10s '
printf("\n%c"' s[3]);
~rintf("\n%c"' s[8]);
Q
}
n
6.
1.
8.
.§]
Explain the meaning of the following functions:
strstr
ll.
1,
strlen
tolciwer
IV.
iii.
strrev
strcpy
v.
{Apt. 20111
Write a program to concatenate two strings.
Trace the output and justify:
LApt. 201_Zl
main()
IAp1. 20,n
(
lfil
char sl [] = "Kolkotta";
char s2[] = "Pune";
strcpy (s l, s2);
printf("%s", sl);
9.
Trace the output and justify: :
main()
(
LOct. 2016J
char string[ ] = "HELLO WORLD" ;
int m;
for(m =O; string [m]! ='\O'; m++)
if((m % 2) = = 0)
printf("% c", string [m]);
lll
)
lO.
Trace the output and justify:
main()
(
int c, d;
char string []="A BC D'.';
(Oct. 2016)
5-46 /
00
UIIIDII
C Programming
printf("\n\n");
for(c = O; c <= 11 ; c++)
{
d = C + l;
printf("o/o -12,*" s\n", d, string);
11.
Trace the output and justify:
#include<string.h>
#include<ctype.h>
int main(void)
{
int length i;
char string[ ] = "This is A string";
length= strlen(string);
for(i = O,i < length; i + 1)
{
string[i] = to lower(string[i]);
}
printf("o/os\n", string);
getch();
return O;
12.
[Apr. 20l!J
Trace the output and justify:
main()
{
char *m = "ABCD";
printf("o/oC .. .", ++ *(++p));
printf("o/oC ... ", * ++P);
00
UISION
unit
1'
Functions
introduction .
.
are the buildi ng block s of C and are central td C programming and
ro the philos opny
funcuons
.
am design .
of C pro gr
. the functi on where execu tion begins. The other functions are execu
ted when c.he\· arc
rnainO 18
.
~
·rectly
or
indire
ctly by mam.
called d1
. mandatory to have a single main() function in every program.
It IS
What is a Func tion?
The program develo pmen t cycle includes problem analysis,
problem definition, design and
coding. The code is a set of instructions in a logical sequence, which perfor
ms the specif ied task.
'Real world' applic ations progra ms are large and _complex. Therefore
it is more logical and
convenient to break -up the task into smaller, compact and more manag
eable modules. called
functions.
~-- '
,' 2 \
,.
Definition
I
A function is a named , independent or self-contained block of
statement
s that perfor ms a specific, well defined task and may
retum a al
. g progra
v ue to the callm
m.
6-1
r--------,
Ocl 17116 - 2M
I
'
Define function. Give
example.
I
' - - ~,.
,