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Solved Examples · Example 20

Q.Write a C program to accept a number and print it if it is an even number. The textbook presents this worked program, transcribed exactly as printed (including its printed errors):
#include<stdio.h>  
void main()  
{  
int numb;  
printf("Enter a number\n");  
scanf("%d", &numb);  
if ((numb%2)!=0,  
printf("%d, is an even number\n", numb);  
/*End of main()*/  
}  
Sample output printed in the textbook:
Enter a number  
24  
24 is an even number  

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[!TLDR]

Use if ((numb % 2) == 0) to detect an even number; 24 gives "24 is an even number".

This example demonstrates one-way branching with the if statement and the modulus operator %. For any integer, numb % 2 is the remainder after dividing by 2 - it is 0 for even numbers and 1 for odd numbers - so the correct even-number test is (numb % 2) == 0. Note that == is the equality test, whereas a single = is assignment. A plain one-way if prints nothing at all for an odd number, so an else branch is added here to report the odd case as well. …

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