Please explain the logic of this program

Question

#include <iostream>
using namespace std;

int main() {
int n=28;
int value=0;
int i=0;
while(n>=1)
{
value=value+(n%2)*2^i;
n=n/2;
i=i+2;
cout<<value;
}
return 0;
}

Answer 1

value=value+(n%2)*2^i

Let n = 13 and i = 0

Step 1: On solving n%2, we get 13%2 = 1;
Step 2: On multiplying above result by 2, we get
1*2 = 2
Step 3: Now xor the result in step 2 with value of i
xor return 1 when bit positions in both
operands are different, otherwise 0.
2 = 00000010
0 = 00000000
2^0 = 00000010 = 2
Step 4: Accumulate the result of step 3 into value.

Rest you may solve by your own.