Re: Fwd: Question about container_of macro.

[Laurent Pinchart <laurent.pinchart@xxxxxxxxx>]

On Thursday 17 January 2008, Alan Stern wrote:
> On Thu, 17 Jan 2008, Manish Katiyar wrote:
> > My question is instead of defining container_of as:
> >
> > #define container_of(ptr, type, member) ({ \
> >                 const typeof( ((type *)0)->member ) *__mptr = (ptr);
> >                 (type *)( (char *)__mptr - offsetof(type,member) );})
> >
> >
> > Why won't this simply work :
> >
> > #define container_of(ptr, type, member) ({ \
> >                         (type *)( (char *)(ptr) - offsetof(type,member)
> > );})
>
> Indeed, or even this:
>
> #define container_of(ptr, type, member) \
> 			((type *) ((char *)(ptr) - offsetof(type,member)))

If I'm not mistaken, that's because

const typeof( ((type *)0)->member ) *__mptr = (ptr);

will warn (or maybe even bail out ?) if ptr is not a pointer to a variable of 
member's type.

Laurent Pinchart

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