Re: Fwd: Question about container_of macro.

["Manish Katiyar" <mkatiyar@xxxxxxxxx>]

Hi Laurent,


On Jan 18, 2008 5:31 AM, Laurent Pinchart <laurent.pinchart@xxxxxxxxx> wrote:
>
> 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.
>

But even if ptr is not a variable of member's type, in the second line
we are force typecasting it to char * and then subtracting. I guess it
will not warn. I tried in a simple program and it doesn't shout.
Below is the program

#include <stdio.h>
#include <stddef.h>

struct t_s{
        int a;
        int b;
};

#define container_of(ptr, type, member) \
                               ((type *) ((char *)(ptr) -
offsetof(type,member)))

int main()
{
        struct t_s init;
        char *c;
        init.a = 1;
        init.b = 2;

        c = (char *)&(init.b);

        printf("a = %d\n",(container_of(c,struct t_s,b))->a);
        return 0;
}

/home/mkatiyar> gcc -Wall b.c
b.c: In function 'main':
b.c:19: warning: assignment from incompatible pointer type
/home/mkatiyar> ./a.out
a = 1


Thanks

> Laurent Pinchart
>



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********************************************
Manish Katiyar  ( http://mkatiyar.googlepages.com )
3rd Floor, Fair Winds Block
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