Home » C programs » C bitwise operators programs

C program to Check if nth Bit in a 32-bit Integer is set or not

In this article, we are going to see how to find nth bit of a 32-bit integer is set or not using bitwise operator?
Submitted by Radib Kar, on January 06, 2019

Problem statement

Write a C program to check if nth bit is set or not in a 32 bit integer.

Pre-requisite: input no(32 bit longer), nth bit

Algorithm

1. Right shift by n times to get the nth bit at LSB
2. Do a bitwise and with 1(only LSB is set of 1, other bits 0).
3. IF result is 1, then nth bit is set
   Else
   Bit not set

Example with explanation

Let input no be 67600 and n is 11

This is the binary representation of 67600

After right shifting 11 times,

So bitwise ANDing this shifted no with 1

Results in 00000...0001(31 0's & one 1) which is 1... Thus n th bit is set...

C implementation to Check if nth Bit in a 32-bit Integer is set or not

#include <stdio.h>

int main() {
  int n, k;

  printf("enter a 32 bit number\n");
  scanf("%d", & k);

  printf("enter the bit no to check...\n");
  printf("bit-no 0-indexed & 0 starts from LSB...\n");
  scanf("%d", & n);

  if (n > 32) {
    printf("enter between 0-31\n");
    return -1;
  }

  k = k >> n; //right shift the no to get nth bit at LSB
  if (k & 1 == 1) //check whether nth bit set or not
    printf("%d th bit is set\n", n);
  else
    printf("%d th bit not set\n", n);

  return 0;
}

Output

First run:
enter a 32 bit number
67600
enter the bit no to check...
bitno 0-indexed & 0 starts from LSB...
11
11 th bit is set


Second run:
enter a 32 bit number
67600
enter the bit no to check...
bit-no 0-indexed & 0 starts from LSB...
10
10 th bit not set   

C Bitwise Operators Programs »