1) Do Bitwise AND with 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 0 then the all bits are not set
Thus,
IF
n & 1 ==0
Print that all bits are not same
ELSE
Right shift n by 1
n=n>>1
END IF-ELSE
2) IF n==0
Print that all bits are set
ELSE
REPEAT step 1

Example with Explanation:

Checking for 1212-> 00001100So first iteration:
n=12 //00001100
n & 1 ==0
so print that all bits are not setChecking for 77->00000111So first iteration:
n=7 //00000111
n & 1 =1
n=n>>1 (n=3) //00000011
So second iteration:
n=3 //00000011
n & 1 =1
n=n>>1 (n=1) //00000001
So third iteration:
n=1 //00000001
n & 1 =1
n=n>>1 (n=0) //00000000
So, All bits are set

C implementation

#include <stdio.h>
int main()
{
unsigned int n;
printf("enter the integer\n");
scanf("%d",&n);
while(n>0){
int temp=n&1;
if(temp == 0){ //if any bit not set
printf("all bits are not set\n");
return 0;
}
n=n>>1; //right shift operator
}
printf("all bits are set ");
printf("in its binary representation\n");
return 0;
}

Output

First run:
enter the integer
12
all bits are not set
Second run:
enter the integer
7
all bits are set in its binary representation

Pre-requisite: Input numbernAlgorithm:Example with Explanation:C implementation

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