有符号与无符号比较

Question

#include <iostream>

int main()
{
    signed int a = 5;
    unsigned char b = -5;
    unsigned int c = a > b;

    std::cout << c << std::endl;
}

This code prints 0 .

Can anyone please explain what is happening here? I am guessing that compiler converts a and b to same type( unsigend int maybe) and compares them.

Answer 1

Let's see how the computer stores the value b:
5 is 00000101 , so -5 will be 11111011 , so, when you convert it to unsigned char , it will became some positive number with value 11111011 in binary, which is larger than 00000101 .
So, that's why a = 00000101 is smaller than b (0 means false).

Answer 2

It is printing 0 because a < b and 0 means false. The type of b is unsigned so it cannot hold negative numbers. Because of that -5 becomes 251 which is grater than 5 .

Answer 3

Lets go to the third line in main c take the value of 0 because a is not greater than b . This is because in C zero is considered false and everything then else is true.

With regards to b . Most platforms store negative integers using 2s complement format. So when we negate an number we flip all the bits and add 1. So -5 unsigned become 0xfa which is greater than 5.