Bit masks and long-long

Question

I have a masking procedure (which creates an all-ones bit mask for the bottom half for a given size):

template<class T>
T bottom_half() {
    T halfway = ((sizeof(T) * 8) / 2);
    T mask = (1 << halfway) - 1;

    return mask;
}

which works fine if I call bottom_half<int>() or long or char. But for some reason when I run it with long long, halfway is correctly set to 32, but mask is 0. Why would that be?

Answer 1

The left shift is shifting 1, which is int by default and probably 32 bits on your machine. When you shift 1<<32, the result is undefined, which means it is not predictable anymore, as it could be anything.

On some processors, 1<<32 might result in shifting the bit off the high end of the integer and resulting in 0. On other processors, the 32 shift is modulo the register size, so effective it is a zero shift, and the result is 1. In any case, it is undefined.

(See What's bad about shifting a 32-bit variable 32 bits? for a discussion on this).

Note also that sizeof returns units char or "bytes" (these are defined to be the same in C, sizeof(char) == 1 always), but C does not guarantee that a byte is 8 bits. There is standard macro CHAR_BIT to get the bit size of a char.

Try this

#include <limits.h>

template<class T>
T bottom_half() {
    T halfway = ((sizeof(T) * CHAR_BIT) / 2);
    T mask = ((T)1 << halfway) - 1;

    return mask;
}

Answer 2

The expression 1 << x has type int. Left-shifting a signed type such that the value exceeds the maximum representable value has undefined behavior. Use T(1) << x instead.