Why does (1 >> 0x80000000) == 1?

Question

The number 1, right shifted by anything greater than 0, should be 0, correct? Yet I can type in this very simple program which prints 1.

#include <stdio.h>

int main()
{
        int b = 0x80000000;
        int a = 1 >> b;
        printf("%d\n", a);
}

Tested with gcc on linux.

Answer 1

6.5.7 Bitwise shift operators:

If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand, the behavior is undefined.

The compiler is at license to do anything, obviously, but the most common behaviors are to optimize the expression (and anything that depends on it) away entirely, or simply let the underlying hardware do whatever it does for out-of-range shifts. Many hardware platforms (including x86 and ARM) mask some number of low-order bits to use as a shift-amount. The actual hardware instruction will give the result you are observing on either of those platforms, because the shift amount is masked to zero. So in your case the compiler might have optimized away the shift, or it might be simply letting the hardware do whatever it does. Inspect the assembly if you want to know which.