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

There was a debate on a similar question.

See the comments on my answer to that question for some pointers at why this is not only non-portable, but also is undefined behaviour by the standard.