Flip least significant one-bit by using negation of a size_t

Question

I've recently encountered a piece of code that supposedly works fine, but I don't quite understand why.

size_t a = 19;
std::cout<<std::bitset<8>(a)<<std::endl;
a ^= a & -a;
std::cout<<std::bitset<8>(a)<<std::endl;

This piece of code will invert the least significant bit of a given unsigned integer. I would prefer to just write a ^= 1;, but I'm puzzled by why the piece of code above actually works. I would think that making an unsigned int negative will result in undefined behavior?

Answer 1

a & -a gives you the least significant 1-bit set in a. For an odd number it is indeed 1, but that's not the case in general of course.

Making an unsigned negative is a well-defined and occasionally useful notation: -a for positive a is -a + 2^N where N is the number of bits in the type. An alternative to writing size_t a = std::numeric_limits<size_t>::max(); is to write size_t a = -1; for example.

So a ^= a & -a; flips the least significant 1-bit to 0.

Rather clever really.