Following are the decision making statements −
- simple – if statement
- if – else statement
- nested – if else statement
- else – if ladder
- switch statement
Simple – if statement
The ‘if’ keyword is used to execute a set of statements when the logical condition is true.
Syntax
if (condition){
Statement (s)
}
Example
The following example checks whether a number is greater than 50.
#include<stdio.h>
main (){
int a;
printf (“enter any number:\n”);
scanf (“%d”, &a);
if (a>50)
printf (“%d is greater than 50”, a);
}Output
1) enter any number: 60 60 is greater than 50 . 2) enter any number 20 no output
The if else statement
The if –else statement takes either True or False condition.
Syntax
if (condition){
True block statement(s)
}
else{
False block statement(s)
}Flowchart
Example
Following is the program to check for even or odd number −
#include<stdio.h>
main (){
int n;
printf (“enter any number:\n”);
scanf (“%d”, &n);
if (n%2 ==0)
printf (“%d is even number”, n);
else
printf( “%d is odd number”, n);
}Output
1) enter any number: 10 10 is even number
Nested if - else statement
Here ‘if’ is placed inside another if (or) else −
Syntax
if (condition1){
if (condition2)
stmt1;
else
stmt2;
}
else{
if (condition3)
stmt3;
else
stmt4;
}Flowchart
Example
Following example is to print the largest of 3 numbers from the given numbers −
#include<stdio.h>
main (){
int a,b,c;
printf (“enter 3 numbers”);
scanf (“%d%d%d”, &a, &b, &c);
if (a>b){
if (a>c)
printf (“%d is largest”, a);
else
printf (“%d is largest”, c);
} else {
if (b>c)
printf (“%d is largest”, b);
else
printf (“%d is largest”, c);
}
}Output
enter 3 numbers = 10 20 30 30 is largest
Else – if ladder
It is a multi-way decision condition.
Syntax
if (condition1) stmt1; else if (condition2) stmt2; - - - - - - - - - - else if (condition n) stmt n; else stmt x;
Flowchart
Example
Following example finds roots of quadratic equation −
#include <math.h>
main (){
int a,b,c,d;
float r1, r2
printf ("enter the values a b c");
scanf (“%d%d%d”, &a, &b, &c);
d= b*b – 4*a*c ;
if (d>0){
r1 = (-b+sqrt(d)) / (2*a);
r2 = (-b-sqrt(d)) / (2*a);
printf (“root1 ,root2 =%f%f”, r1, r2);
}
else if (d== 0){
r1 = -b / (2*a);
r2 = -b/ (2*a);
printf (“root1, root2 = %f%f”, r1, r2);
}
else
printf ("roots are imaginary”);
}Output
1) enter the values of a b c : 1 4 3 Root 1 = -1 Root 2 = -3
Switch statement
It is helpful in selecting one among multiple decisions.
Syntax
switch (expression){
case value1 : stmt1;
break;
case value2 : stmt2;
break;
- - - - - -
default : stmt – x;
}Syntax
Example
#include<stdio.h>
main (){
int n;
printf (“enter a number”);
scanf (“%d”, &n);
switch (n){
case 0 : printf (“zero”)
break;
case 1 : printf (‘one”);
break;
default : printf (‘wrong choice”);
}
}Output
enter a number 1 One