Is storing an invalid pointer automatically undefined behavior?

Question

Obviously, dereferencing an invalid pointer causes undefined behavior. But what about simply storing an invalid memory address in a pointer variable?

Consi

Answer 1

Your code is undefined behavior for a different reason:

the expression begin - 1 does not yield an invalid pointer. It is undefined behavior. You are not allowed to perform pointer arithmetics beyond the bounds of the array you're working on. So it is the subtraction itself that is invalid, and not the act of storing the resulting pointer.