Take the following code snippet:
#include <stdio.h>
#include <stdint.h>
void decodeStatus(uint8_t d)
{
uint64_t a = d << 31;
uint64_t b = d << 32;
}
When shifting an uint8_t eight or more places to the left, it becomes 0, which results in a warning. The compiler is obligated to make sure that an uint8_t is exactly 8 bits (C11 spec 7.20.1.1 sections 1 & 2), so that both shifts should result in a warning. But apparently, only the second shift creates a warning:
gcc -c main.c -o a.out
main.c: In function ‘decodeStatus’:
main.c:7:20: warning: left shift count >= width of type [-Wshift-count-overflow]
uint64_t b = d << 4*8;
Is this a compiler bug, or is there a logical explanation? I verified this on the following compilers:
$> gcc --version
gcc (Ubuntu 7.5.0-3ubuntu1~18.04) 7.5.0
$> arm-linux-gnueabihf-gcc --version
arm-linux-gnueabihf-gcc (GCC) 9.2.0