Re: VM code question

[Matt Mackall <mpm@selenic.com>]

On Tue, Oct 14, 2003 at 11:52:36AM +1000, Nick Piggin wrote:
> 
> 
> William Lee Irwin III wrote:
> 
> >On Tue, Oct 14, 2003 at 11:32:27AM +1000, Darren Williams wrote:
> >
> >>I have a small question wrt some VM code.
> >>source file is include/linux/kernel.h
> >>#define container_of(ptr, type, member) ({                      \
> >>       const typeof( ((type *)0)->member ) *__mptr = (ptr);    \
> >>       (type *)( (char *)__mptr - offsetof(type,member) );})
> >>what is the use of the 0 (zero) in the typeof? I am thinking
> >>that we are casting 0 to (type *) then referencing 'member' of
> >>'type', however why do we require the 0 ?
> >>Just curious
> >>
> >
> >It's an address calculation method. We subtract the address of the
> >start of the structure from the address of the member inside the
> >structure.
> >
> 
> AFAIKS the 0 is not part of the address calculation method though. It
> is only used in the argument to the typeof operator. I think 0 is used
> simply because its as good a place as any, right?

It could be simplified to:

 ((type *)((char *)(ptr) - offsetof(type, member)))

The other bit is just there to throw errors if you cast in a pointer
of the wrong type. To do this, we've got to create a pointer of the
same type as &type.member so that assigning to it without casting will
throw a warning if ptr isn't of the right type. But we can't do
typeof(&type.member), as type is a type name and not an object. So
0 is simply the shortest, safest thing to cast to a (type *).

-- 
Matt Mackall : http://www.selenic.com : Linux development and consulting