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<<size_of_int;// or printf("%zu",size_of_int);

P.S: The question is only for learning purpose. So please don’t give answers like Use sizeof() etc

Answer 1

From ANSI-ISO-IEC 14882-2003, p.87 (c++03):

“75) Another way to approach pointer
arithmetic is first to convert the
pointer(s) to character pointer(s): In
this scheme the integral value of the
expression added to or subtracted from
the converted pointer is first
multiplied by the size of the object
originally pointed to, and the
resulting pointer is converted back to
the original type. For pointer
subtraction, the result of the
difference between the character
pointers is similarly divided by the
size of the object originally pointed
to.”

This seems to suggest that the pointer difference equals to the object size.

If we remove the UB’ness from incrementing a pointer to a scalar a and turn a into an array:

int a[1];
size_t size_of_int = (char*)(a+1) - (char*)(a);

std::cout<<size_of_int;// or printf("%zu",size_of_int);

Then this looks OK. The clauses about alignment requirements are consistent with the footnote, if alignment requirements are always divisible by the size of the object.

UPDATE: Interesting. As most of you probably know, GCC allows to specify an explicit alignment to types as an extension. But I can’t break OP’s “sizeof” method with it because GCC refuses to compile it:

#include <stdio.h>

typedef int a8_int __attribute__((aligned(8)));

int main()
{
 a8_int v[2];

 printf("=>%d\n",((char*)&v[1]-(char*)&v[0]));
}

The message is error: alignment of array elements is greater than element size.