Alternate way of computing size of a type using pointer arithmetic

Question

Is the following code 100% portable?

int a=10;
size_t size_of_int = (char *)(&a+1)-(char*)(&a); // No problem here?

std::cout<

        

Answer 1

It's not 100% portable for the following reasons:

1. Edit: You'd best use int a[1]; and then a+1 becomes definitively valid.
2. &a invokes undefined behaviour on objects of register storage class.
3. In case of alignment restrictions that are larger or equal than the size of int type, size_of_int will not contain the correct answer.

Disclaimer:

I am uncertain if the above hold for C++.