Why does decay to pointer for array argument appear not to apply to sizeof()?

Question

I read a question earlier that was closed due to being an exact duplicate of this

When a function has a specific-size array parameter, why is it replaced with a poin

Answer 1

sizeof is an operator, not a function. It's a specific one at that, too. The parentheses aren't even necessary if it's an expression:

int a;
sizeof (int); //needed because `int` is a type
sizeof a; //optional because `a` is an expression
sizeof (a); //^ also works 

As you can see, it's on this chart of precedence as well. It's also one of the non-overloadable operators.