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

Pre-requisite: Input numbernAlgorithm:Example with Explanation:Highest set bit in 12 (0-index based position, i.e., LSB is 0th bit): 12 → 00001100C implementation

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