Write a C program to input any number from user and count number of trailing zeros in the given number using bitwise operator. How to find total number of trailing zeros in any given number using bitwise operator in C programming.

Input any number: 22

Output trailing zeros: 1

Happy coding ;)

**Example:**Input any number: 22

Output trailing zeros: 1

### Required knowledge:

Basic C programming, Bitwise operator, If else, Loop### Logic:

Counting trailing zeros in a binary number is an easy task if you know how to find first set bit in a binary number. As total number of trailing zeros is equal to the order of the first set bit.### Program:

/** * C program to count trailing zeros in a binary number using bitwise operator */ #include <stdio.h> #define INT_SIZE sizeof(int) * 8 int main() { int num, count=0, i; //Reads a number from user printf("Enter any number: "); scanf("%d", &num); //Iterate over each bit of the number for(i=0; i<INT_SIZE; i++) { //If a set bit is found if((num >> i ) & 1) { //No need to run further break; } count++; } printf("Total number of trailing zeros in %d is %d.", num, count); return 0; }

### Program: Another approach

/** * C program to count trailing zeros in a binary number using bitwise operator */ #include <stdio.h> int main() { int num, count=0; //Reads a number from user printf("Enter any number: "); scanf("%d", &num); while(!(num & 1)) { count++; num >>= 1; } printf("Total number of trailing zeros = %d.", count); return 0; }

Output

Enter any number: 48

Total number of trailing zeros in 48 is 4.

Total number of trailing zeros in 48 is 4.

Happy coding ;)

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