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

The code above will portably compute sizeof(int) on a target platform but the latter is implementation defined - you will get different results on different platforms.