Is sizeof(*ptr) undefined behavior when pointing to invalid memory?

Question

We all know that dereferencing an null pointer or a pointer to unallocated memory invokes undefined behaviour.

But what is the rule when used within an expression pa

Answer 1

No. sizeof is an operator, and works on types, not the actual value (which is not evaluated).

To remind you that it's an operator, I suggest you get in the habit of omitting the brackets where practical.

int* ptr = 0;
size_t size = sizeof *ptr;
size = sizeof (int);   /* brackets still required when naming a type */