On Mit, 2004-06-30 at 10:51, Anupam Kapoor wrote:
> On Wed, 30 Jun 2004 11:38:47 +0300 (IDT), Nir Tzachar
> <tzachar@cs.bgu.ac.il> wrote:
[...]
> > > offsetof is similar:
> > >
> > > <from include/linux/stddef.h>
> > > 12 #define offsetof(TYPE, MEMBER) ((size_t) &((TYPE *)0)->MEMBER)
> >
> > well, lets say you have a structure defined like this:
> > struct foo {
> > int a;
> > int b;
> > int c;
> > char d;
> > };
> >
> > to get the offset of member c in this struct, we need to size of all
> > members which come b4 c: offset_of_c = sizeof(a)+sizeof(b) .
> no, this is not correct. the compilers are free to add padding to meet
> alignment requirements of a given platform.
While this is true, ...
> > however, a more generic and _much_ better way:
> > lets say you had a pointer to a struct foo (foo_ptr), so you can get the
> > offset like this:
> >
> > offset_of_c = &foo_ptr->c - foo_ptr
> >
> > the macro actually saves the subtraction, by letting the compiler think
> > that foo_ptr is 0 ((TYPE *)0) (located at start of memory).
> the assumption here is that null is repersented as a bunch of zeros on
> a platform. if that is not the case, you will get strange results.
No. A (so-called) null-pointer has two different representations - one
in the C-source (this is 0, '\0', (void *)0, etc.) and one at run-time
in the memory, which is not specified in the C standard.
It is the compilers job to "convert" between these two representations
when it makes sense.
But if the run-time representation is not "all buts are zero" you will
get interesting results after a "memset(p, 0, sizeof(p)", but not for
the above (since the compiler has to - at least virtually - convert at
the right place).
And I do not know if there is any hardware on this planet where the
run-time representation is not "all bits are 0".
Bernd
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