Question

I can usually understand the reason behind a compiler warning, but this one seems just plain wrong.

#include <stdint.h>    
uint8_t myfunc(uint8_t x,uint8_t y)
{
    x |= y;
    return x;
}

The intel compiler with -Wall complains:

conversion from "int" to "uint8_t={unsigned char}" may lose significant bits
  x |= y;
    ^

Is this right? Is the above code non-portable and non-standard somehow?

Answer 1

That’s integer promotions at work.

in

x |= y;

both operands of the | operator are promoted to int

x = (int)x | (int)y;

then the result is converted back to uint8_t losing precision.