Why std::bitset::reference::operator~?

Question

I was reading the documentation of std::bitset and I was wondering why std::bitset::reference explicitly define operator~ because I don't see any performance or design reasons. Without it, I think it would work equally well:

bool b = ~mybitset[i];

because the reference would be converted to a bool, on which the ~ operator would be applied.

Any explanation for this design decision?

Answer 1

bool b = true;
b = ~b;

The value of b after this operation is true!

This is because ~ promotes the bool to int of value 1, then performs the bitwise-not on the result, which resolves to -2, and then casts that back to bool which is true.

So it has to provide an operator so that the result is how you would expect it.

Answer 2

Thanks to the integral promotions, ~true is ~1, which is most assuredly nonzero, and hence not false, when converted back to bool.