Bit Shifting: Shift Count >= Width Of Type

Question

The code below, when compiled, throws a warning caused by line 9:

warning: shift count >= width of type [-Wshift-count-overflow]

However, line 8 does not throw a similar warning, even though k == 32 (I believe). I'm curious why this behavior is occurring? I am using the gcc compiler system.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


int bit_shift(unsigned x, int i){
    int k = i * 8;
    unsigned n = x << k; /* line 8 */
    unsigned m = x << 32; /* line 9 */
    return 0;
} 

int main(){
    bit_shift(0x12345678, 4);
    return 0;
}

Answer 1

The value of k in bit_shift is dependent on the parameter i. And because bit_shift is not declared static it is possible that it could be called from other translation units (read: other source files).

So it can't determine at compile time that this shift will always be a problem. That is in contrast to the line unsigned m = x << 32; which always shifts by an invalid amount.