C Program to Check if nth Bit is Set

In C, you can check if the nth bit is set by using the expression (num & (1 << n)) != 0, where num is the number and n is the bit position starting from 0.

Examples

Inputnum = 5, n = 0

OutputBit 0 is set

Inputnum = 5, n = 2

OutputBit 2 is set

Inputnum = 8, n = 1

OutputBit 1 is not set

How to Think About It

To check if the nth bit is set, think of shifting the number 1 to the left by n positions to create a mask that has only the nth bit set. Then use the bitwise AND operator & between the number and this mask. If the result is not zero, the nth bit in the number is set; otherwise, it is not.

Algorithm

1

Get the input number and the bit position n.

2

Create a mask by shifting 1 left by n positions.

3

Perform bitwise AND between the number and the mask.

4

If the result is not zero, the nth bit is set; else, it is not.

5

Print the result.

Code

c

#include <stdio.h>

int main() {
    int num, n;
    printf("Enter number and bit position: ");
    scanf("%d %d", &num, &n);

    if ((num & (1 << n)) != 0) {
        printf("Bit %d is set\n", n);
    } else {
        printf("Bit %d is not set\n", n);
    }
    return 0;
}

Output

Enter number and bit position: 5 2
Bit 2 is set

Dry Run

Let's trace the input num=5 and n=2 through the code.

1

Input values

num = 5 (binary 0101), n = 2

2

Create mask

1 << 2 = 4 (binary 0100)

3

Bitwise AND

5 & 4 = 4 (binary 0100)

4

Check result

4 != 0, so bit 2 is set

Operation	Value (decimal)	Value (binary)
num	5	0101
1 << n	4	0100
num & (1 << n)	4	0100

Why This Works

Step 1: Shift 1 to create mask

Shifting 1 left by n positions creates a number with only the nth bit set, which acts as a mask.

Step 2: Use bitwise AND

The bitwise AND between the number and mask isolates the nth bit of the number.

Step 3: Check if result is zero

If the result is not zero, it means the nth bit was set in the original number.

Alternative Approaches

Using right shift operator

c

#include <stdio.h>

int main() {
    int num, n;
    printf("Enter number and bit position: ");
    scanf("%d %d", &num, &n);

    if (((num >> n) & 1) == 1) {
        printf("Bit %d is set\n", n);
    } else {
        printf("Bit %d is not set\n", n);
    }
    return 0;
}

This method shifts the number right by n bits and checks the least significant bit; it is equally efficient and sometimes clearer.

Complexity: O(1) time, O(1) space

Time Complexity

The operation uses simple bitwise shifts and AND, which take constant time regardless of input size.

Space Complexity

No extra memory is needed beyond a few variables, so space complexity is constant.

Which Approach is Fastest?

Both left shift with AND and right shift with AND methods run in constant time and are equally fast; choice depends on readability preference.

Approach	Time	Space	Best For
Left shift and AND	O(1)	O(1)	Clear mask creation
Right shift and AND	O(1)	O(1)	Direct bit extraction