sizeof operands get evaluated?

Question

AFAIK sizeof doesn\'t evaluate its operands it C++.

E.g.

int x = 0;
sizeof(x += 1); // value of x is not changed

But what does this

Answer 1

It isn’t. In an arithmetic expression, array names decay into pointers. That says nothing about performing the calculation itself. The type of + is deducible from the types of its operands, in this case pointer and integer, yielding the same result as sizeof(int*).