Pre-requisite: Input number n

Algorithm:

1)  Set count=0 & store= -1
2)  Do bit wise AND between n and 1. 
    n & 1

    let n be a7a6a5a4a3a2a1a0
    1->00000001
    So doing bitwise AND (refer to published article on bitwise operators) 
    will result in all bits 0 except the LSB which will be a0. 
    If a0 is 1 then it is set, else not set.
    If a0 is 1 then the bitwise AND results in 00000001 (1) else 00000000 (0)
    
    Thus,
    IF
        n& 1 ==1 //current bit is set 
        update store to current count value.
    END IF

3)  count++;
4)  Right shift n by 1
    n = n >> 1
5)  IF n==0
        Break
    ELSE
        REPEAT step 2, 3, 4
6)  IF store==-1
        No set bit
    Else
        Print store

Example with Explanation:

Highest set bit in 12 (0-index based position, i.e., LSB is 0th bit): 12 → 00001100

Initially,
Count=0
Store=-1

So first iteration:
n=12 //00001100
n& 1 =0
store=-1
count=1
n=n>>1 (n=6) //00000110

So second iteration:
n=6 //00000110
n& 1 =0 
store=-1
count=2
n=n>>1 (n=3) //00000011

So third iteration:
n=3 //00000011
n& 1 =1 
store=2
count=3
n=n>>1 (n=1) //00000001

So fourth iteration:
n=1 //00000001
n& 1 =1 
store=3
count=4
n=n>>1 (n=0) //00000000
program terminates
Print store, 3

Highest bit set: 3

C implementation

#include <stdio.h>

int main()
{
	unsigned int n;
	printf("enter the integer\n");
	scanf("%d",&n);
	int count=0,store=-1;
	while(n!=0){
		if(n & 1 == 1) //if current bit is set
			store=count; //update store
		n=n>>1; //right shift
		count++; //increase count
	}

	if(store==-1){//if store not updated
		printf("No bit is set\n"); //no set bit present at all
		return 0;
	}
	printf("Highest bit set ");
	//printing highest set bit
	printf("in its binary representation: %d \n",store); 

	return 0;
}

Output

enter the integer
12
Highest bit set in its binary representation: 3