c - _Bool type and strict aliasing

Question

I was trying to write some macros for type safe use of _Bool and then stress test my code. For evil testing purposes, I came up with this dirty hack:

_Bool b=0;
*(unsigned char*)&b = 42;

Given that _Bool is 1 byte on the implementation sizeof(_Bool)==1), I don't see how this hack violates the C standard. It shouldn't be a strict aliasing violation.

Yet when running this program through various compilers, I get problems:

#include <stdio.h>

int main(void)
{
  _Static_assert(sizeof(_Bool)==1, "_Bool is not 1 byte");

  _Bool b=0;
  *(unsigned char*)&b = 42;
  printf("%d ", b);
  printf("%d", b!=0 );

  return 0;
}

(The code relies on printf implicit default argument promotion to int)

Some versions of gcc and clang give output 42 42, others give 0 0. Even with optimizations disabled. I would have expected 42 1.

It would seem that the compilers assume that _Bool can only be 1 or 0, yet at the same time it happily prints 42 in the first case.

Q1: Why is this? Does the above code contain undefined behavior?

Q2: How reliable is sizeof(_Bool)? C17 6.5.3.4 does not mention _Bool at all.

Answer 1

Q1: Why is this? Does the above code contain undefined behavior?

Yes, it does. The store is valid, but subsequently reading that as a _Bool is not.

6.2.6 Representations of types

6.2.6.1 General

5 Certain object representations need not represent a value of the object type. If the stored value of an object has such a representation and is read by an lvalue expression that does not have character type, the behavior is undefined. [...]

Q2: How reliable is sizeof(_Bool)? C17 6.5.3.4 does not mention _Bool at all.

It will reliably tell you the number of bytes that are needed to store one _Bool. 6.5.3.4 also doesn't mention int, but you're not asking whether sizeof(int) is reliable, are you?