char类型按位操作失败到int

Question

I am try to bitwise operate in C. Here is my code:

int main() {
    char c = 127;
    c <<= 1;
    if (c == 0xfe) {
        printf("yes");
    }
    else {
        printf("No");
    }
    return 0;
}

System console print No. What I expect to receive is Yes. the value in c after bit moving is ffffffffe. It seems that system has changed 8bits to a 32bits. Can anyone help me.

Answer 1

You have correctly figured out the content of the lower byte, ie 0xfe . However, there is more to what's actually happening here: when you write

if (c == 0xfe) {
    ...
}

you are comparing c , a char , to an int constant 0xfe . In this situation C rules require promoting char c to int for comparison. Since char is signed on your platform, 0xfe gets promoted to 0xfffffffe , which is why the comparison subsequently fails.

If you want to do the comparison in char type, cast 0xfe to char , like this:

if (c == (char)0xfe) {
    ...
}

Demo 1.

Unsigned characters, on the other hand, would give you no trouble with int promotion, because they fit nicely in an int :

unsigned char c = 127;
c<<=1;
if (c == 0xfe) { // No casting is necessary
    ...
}

Demo 2.

Note: The behavior of c <<= 1 is implementation-defined due to char overflow.