Topic 4: Control structures
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Control Structures Overview
Conditions
if Statement
Nested if Statements
switch Statement
Common Programming Errors
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CPS125
1
Control Structures
• Using for control the flow of the execution
in a program or function.
• They are a combination of the individual
instructions into a single logical unit with
one entry point and one exit point.
Three kinds of control instructions are:
• Sequence
• Selection
• Repetition
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Control Structures
Sequence flow:
• A compound statement is used to specify sequential
flow
{
statement 1;
statement 2;
…..
Control flow
statement n;
}
• A function body consist of a single compound statement
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Conditions
• An expression that is either false
(represented by 0) or true (usually
represented by 1)
For example :
rest_heart_rate > 75
• Conditions establish criteria for executing
or skipping a group of statements
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Relational and Equality Operators
variable relational-operator variable/constant
variable equality-operator variable/constant
Examples :
x <= 0
salary < MIN_SALARY
x>Y
dependents >= 5
item == 3 (It is not the assignment operator ‘=‘ )
num != SENTINEL
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Logical Operators
• Three logical operators && (and), || (or)
and ! (not) can be used to form logical
expressions:
Examples:
n >= 0 && n <= 100
0<= n && n <= 100
! ( 0 <= n && n <= 100 )
0 > n || n > 100
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Logical operators
• && (and)
op1 op2
op1 && op2
------------------------------------T
T
T
T
F
F
F
T
F
F
F
F
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Logical operators
• || (or)
op1 op2
op1 || op2
------------------------------------T
T
T
T
F
T
F
T
T
F
F
F
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Logical operators
• ! (not)
op1
! op1
-------------------T
F
F
T
• The result of logical expression is always 0
or 1. C accepts any nonzero value as a
representation of true
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Operator Precedence
Operator
Precedence
--------------------------------------------------------function calls
highest
! + - & (unary operators)
*/%
+< <= >= >
== !=
&&
||
=
lowest
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Operator Precedence
-2 – 1 * 2
=> -4
( 2 < 3 || 2 > 3 ) && 2 > 1
1 (true)
2 < 3 || 2>3 && 2 >1
=>1(true)
!0 || ( 4.0 + 2.0 >= 3.0 – 2.0)
=> 1(true)
!(0 || ( 4.0 + 2.0 >= 3.0 – 2.0))
=> 0(false)
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Short-Circuit Evaluation
• For && and || C uses the short-circuit
evaluation. It means stopping the
evaluation of the logical expression as
soon as its value can be determined
Examples:
! 1 && ( 5+ 7 >= 3 )
1 || ( 6+ 4 < 7 )
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Examples
• Range checking
x > 2 && y > 2 but not using x && y > 2
• Character comparison
‘9’ >= ‘0’
1(true)
‘Z’ == ‘z’
0(false)
‘a’ <= ch && ch <=‘z’ true if ch is a lowercase letter
‘a’ <= ‘A’
system dependent
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Logical Assignment
• Example 1:
int age, senior_citizen;
scanf(“%d”, &age);
senior_citizen = (age >= 65);
• Example 2:
int even, n;
scanf(“%d”, &n);
even = ( n %2 == 0 )
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Complementing a Condition
• The condition item == sent
complemented as !(item == sent ) or
item != sent
• DeMorgan’s Theorem
age > 25 && ( s == ‘S’ || s == ‘D’)
complemented as
age <=25 || ( s != ‘S’ && s != ‘D’)
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15
THE if STATEMENT
if ( mark > 50 )
printf ( “pass”);
else
printf(“fail”);
F
T
mark> 50
Display
“fail”
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Display
“pass”
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THE if STATEMENT
if ( mark == 50 )
mark = mark + 1;
if ( mark > 50 )
printf ( “pass”);
else
printf(“fail”);
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T
F
mark==50
mark = mark + 1
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THE if STATEMENT
• if statement (One Alternative)
if ( condition )
statementT;
• if statement ( Two Alternatives)
if ( condition )
statementT;
else
statementF;
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if ( mark == 50 )
mark = mark + 1;
if ( mark > 50 )
printf ( “pass”);
else
printf(“fail”);
if ( mark >= 90 )
printf (“outstanding\n” );
printf (“Enter another mark”);
It is not depending on the last if statement
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Errors, when using if statement
• if mark == 100
printf (“highest mark \n” );
• if (mark = 100)
printf (“highest mark \n” );
• if (mark == 100);
printf (“highest mark \n” );
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if statements with compound
statements
• Computing the increase in fruit fly population
if ( pop_today > pop_yesterday ) {
growth = pop_today – pop_yesterday;
growth_pct = 100.0 * growth /pop_yesterday;
printf ( “ The growth percentage is %.2f \n”
, growth_pct);
}
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if statements with compound
statements
• Keeping records of safety rating of the fleet cars
if (ctri <= MAX_SAFE_CTRI) {
printf(“Car #%d: safe\n”, auto_id);
safe = safe + 1;
} What is the result when omitting this brace
else {
printf(“Car #%d: unsafe\n”, auto_id);
unsafe = unsafe + 1;
}
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if statements with compound
statements
if (condition)
{
statements;
}
else
{
statements;
}
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Tracing an if statement
if (x > y) {
/* Switch x and y */
temp = x;
/* Store old x in temp */
x = y;
/* Store old y in x */
y = temp;
}
Stat.
x
y
temp
-------------------------------------------------------------12.5
5.0
?
If (x>y) {
temp = x;
12.5
x = y;
5.0
y = temp;
12.5
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Nested if Statements
if (expression1)
statement1;
else
if (expression2)
statement2;
else
statement3;
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Nested if statements
if (x > 0)
num_pos = num_pos + 1;
else
if (x < 0)
num_neg = num_neg + 1;
else
num_zero = num_zero + 1;
• Sequence of if statements: (It is less readable and less efficient
because all conditions are always tested)
if (x > 0)
num_pos = num_pos + 1;
if (x < 0)
num_neg = num_neg + 1;
if (x == 0)
num_zero = num_zero + 1;
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Multiple-Alternative Decision Form of
Nested if
• Can be used instead of nested if, when each false task
(except for the last) is followed by if-then-else statement
• The conditions are evaluated in sequence until a true
condition is reached
• Upon a true condition the statements in the else part are
skipped
• The words else and if the next condition appear on the
same line
if (x > 0)
num_pos = num_pos + 1;
else if (x < 0)
num_neg = num_neg + 1;
else /* x equal to 0 */
num_zero = num_zero + 1;
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In multiple-alternative decision, the order of conditions affects
the result. For example by placing if (salary <= 150000.00) at
the beginning of the following if statement the result for the
$40,000 salary input would be wrong:
if (salary < 0.0)
tax = -1.0;
else if (salary < 15000.00)
/* first range */
tax = 0.15 * salary;
else if (salary < 30000.00)
/* second range
*/
tax = (salary - 15000.00) * 0.18 + 2250.00;
else if (salary < 50000.00)
/* third range */
tax = (salary - 30000.00) * 0.22 + 5400.00;
else if (salary < 80000.00)
/* fourth range
*/
tax = (salary - 50000.00) * 0.27 + 11000.00;
else if (salary <= 150000.00) /* fifth range */
tax = (salary - 80000.00) * 0.33 + 21600.00;
else
tax = -1.0;
Multiple variables nested if
if (marital_status == ‘S’)
if (gender == ‘M’ )
if (age >= 18)
if (age <= 26)
printf(“ All criteria are met. \n”);
Is equal to:
if (marital_status == ‘S’ && gender == ‘M’ &&
age >= 18 && age <= 26 )
printf(“ All criteria are met. \n”);
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• Always C associates an else with the most
recent if
if (road_status == ‘S’)
if (temp > 0) {
printf(“Wet roads ahead\n”);
printf(“Stopping time doubled\n”);
}
else {
printf(“Icy roads ahead\n”);
printf(“Stopping time quadrupled\n”);
}
else
printf(“Drive carefully\n”);
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• By placing braces, else can be associated to
the further if, For example:
if (road_status == ‘S’){
if (temp > 0) {
printf(“Wet roads ahead\n”);
printf(“Stopping time doubled\n”);
}
} else
printf(“Drive carefully\n”);
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Case study: Computing Compass Bearing
/*
* Transforms a compass heading to a compass bearing using this table:
*
* HEADING
* IN DEGREES
BEARING COMPUTATION
*
* 0 - 89.999...
north (heading) east
* 90 - 179.999...
south (180.0 - heading) east
* 180 - 269.999...
south (heading - 180.0) west
* 270 - 360
north (360.0 - heading) west
*/
#include <stdio.h>
void instruct(void);
int
main(void)
{
double heading; /* Input - compass heading in degrees
*/
instruct();
/* Get compass heading.
*/
printf("Enter a compass heading> ");
scanf("%lf", &heading);
/* Display equivalent compass bearing. */
if (heading < 0.0)
printf("Error -- negative heading (%.1f)\n",
heading);
else if (heading < 90.0)
printf("The bearing is north %.1f degrees east\n",
heading);
else if (heading < 180.0)
printf("The bearing is south %.1f degrees east\n",
180.0 - heading);
else if (heading < 270.0)
printf("The bearing is south %.1f degrees west\n",
heading - 180.0);
else if (heading <= 360.0)
printf("The bearing is north %.1f degrees west\n",
360.0 - heading);
else
printf("Error--heading > 360 (%.1f)\n", heading);
return (0);
}
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CPS125
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/*
* Display program purpose and user instructions
*/
void
instruct(void)
{
printf("To convert a compass heading to a ");
printf("compass bearing,\nenter a value ");
printf("between 0.0 and 360.0 at the prompt.\n");
}
• The results of bearing computation in the printf statements can
be stored into output variable bearing. In that case it can be
used later.
• For testing the program the boundary values of 0, 90, 180,
270, 360 and also an out of range value can be used as the
program input.
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The switch Statement
• The switch statement is useful when selection is based
on the value of single value or of a simple expression
(called controlling expression).
• This value of controlling expression should be of type int
or char, but not of type double
• The controlling expression is evaluated and compared
to each of the case labels in the label sets until a match
is found
• When this match is found the statements following the
case label are executed until break statement then the
rest of switch statement is skipped.
• If no case label value matches the controlling
expression, the entire switch statement body is skipped
unless it contains a default label. In that case the
statement following the default label are executed.
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Syntax of the switch statement
switch ( integer or char expression )
{
Controlling Expression
case
const1:
statements 1
break;
Label Set1
case
const2:
statements 2;
break;
……
default:
statements d;
break;
}
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Example of a switch Statement with Type char Case Labels
switch (class) {
case 'B':
case 'b':
printf("Battleship\n");
break;
case 'C':
case 'c':
printf("Cruiser\n");
break;
case 'D':
case 'd':
printf("Destroyer\n");
break;
case 'F':
case 'f':
printf("Frigate\n");
break;
default:
printf("Unknown ship class %c\n", class);
}
Example of a switch Statement with Type int Case Labels
switch (watts) {
case 25:
life = 2500;
break;
what is the result of omission of this break ?
case 40:
case 60:
life = 1000;
break;
case 75:
case 100:
life = 750;
break;
default:
life =0;
}
what is the result of omission of this brace ?
•
What is the equivalent nested if statement for this switch statement ?
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38
Common Programming Errors
• Using 0 < x < 4 instead of (0< x && x<4) in the if
statements
• Confusing the operator = with ==
• Forgetting parentheses in the if condition
• Missing braces in compound statements of the if
statements
• Confusion when matching else with related if in
the nested if statements
• Missing break and default in the switch
statement
A. Abhari
CPS125
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