有人可以帮助我了解此代码的输出

Question

Hello beautiful people, I need a help with output of this program:

#include <stdio.h>

int main() {
    int x,y;
    scanf("%d %d",&x,&y);
    int t = x^y;
    int p = 0;
    while (t > 0) {
        p += t%2;
        t /= 2;
    }
    printf("%d", p);
    return 0;
}

I tried to write it down on paper and do some work by hand. So i wrote this :

lets say for x = 2 and y = 4

1. first iteration:

   p = 0 + 16mod2 which is equal to 0

   t = 8

2. second iteration:

   p = 0 + 8mod2 which is equal to 0

   t = 4

3. third iteration:

   p = 0 + 4mod2 which is equal to 0

   t = 2

4. forth iteration:

   p = 0 + 2mod2 which is equal to 0

   t = 1

And output should be 0, but somehow when I run code I get 2. Can someone help me out with this one please? And are there any other cases to consider, like what if x = 0, y = 0 or x and y are < 0 ?

Answer 1

The problem here, is that you assume that 2^4 == 16 , when in fact, it is only 6 , as the ^ operator, is actually an XOR .

You should be using

int t = pow(x, y)

Answer 2

 int t = x^y;

Is not " x raised to the power of y ". It's " x XOR y ". T starts at 6 in your example.

Answer 3

As others have mentioned x^y is XOR operation, what you had in mind is :

#include <math.h>
....
int t = pow(x,y);