Messages in this thread | | | Date | Tue, 14 Oct 2003 00:07:13 -0500 | From | Matt Mackall <> | Subject | Re: VM code question |
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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 - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/
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