Question about container

Post by Manish Katiyar
#define container_of(ptr, type, member) ({ \
const typeof( ((type *)0)->member ) *__mptr = (ptr);
(type *)( (char *)__mptr - offsetof(type,member) );})
#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)))

Alan Stern

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Laurent Pinchart

2008-01-18 00:01:54 UTC

#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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Manish Katiyar

2008-01-18 04:15:27 UTC

Hi Laurent,

#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

Post by Laurent Pinchart
Laurent Pinchart

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Manish Katiyar

2008-01-18 04:29:50 UTC

Post by Manish Katiyar
Hi Laurent,

#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:19: warning: assignment from incompatible pointer type
/home/mkatiyar> ./a.out
a = 1

Sorry, I pasted the wrong command output........the program doesn't
give any warnings....

Post by Manish Katiyar
Thanks

Post by Laurent Pinchart
Laurent Pinchart

Laurent Pinchart

2008-01-18 09:14:44 UTC

Post by Manish Katiyar
Hi Laurent,

Post by Manish Katiyar
#define container_of(ptr, type, member) ({ \
const typeof( ((type *)0)->member ) *__mptr =
(ptr); (type *)( (char *)__mptr - offsetof(type,member) );})
#define container_of(ptr, type, member) ({ \
(type *)( (char *)(ptr) -
offsetof(type,member) );})

#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.

That's why you need

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

in addition to

(type *)( (char *)__mptr - offsetof(type,member) );

otherwise you'll get no warning.

Laurent Pinchart

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Manish Katiyar

2008-01-18 09:48:49 UTC

Post by Manish Katiyar
Hi Laurent,

Post by Manish Katiyar
#define container_of(ptr, type, member) ({ \
const typeof( ((type *)0)->member ) *__mptr =
(ptr); (type *)( (char *)__mptr - offsetof(type,member) );})
#define container_of(ptr, type, member) ({ \
(type *)( (char *)(ptr) -
offsetof(type,member) );})

#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.

That's why you need
const typeof( ((type *)0)->member ) *__mptr = (ptr);
in addition to
(type *)( (char *)__mptr - offsetof(type,member) );
otherwise you'll get no warning.

Got it ..!!! .... thanks a lot laurent... have been scratching my head
over this for past 2 days :-) ..

Post by Laurent Pinchart
Laurent Pinchart

Manish Katiyar

2008-01-18 16:37:46 UTC

Post by Manish Katiyar
Hi Laurent,

Post by Manish Katiyar
#define container_of(ptr, type, member) ({ \
const typeof( ((type *)0)->member ) *__mptr =
(ptr); (type *)( (char *)__mptr - offsetof(type,member) );})
#define container_of(ptr, type, member) ({ \
(type *)( (char *)(ptr) -
offsetof(type,member) );})

#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.

That's why you need
const typeof( ((type *)0)->member ) *__mptr = (ptr);
in addition to
(type *)( (char *)__mptr - offsetof(type,member) );
otherwise you'll get no warning.

But again, keeping in mind that this macro is one of frequently used
macros in kernel code, i am not sure how much it is justified to
allocate a local variable every time and then removing it without any
significant purpose. If we are keeping it just to ensure some rare
warnings, I would say it is an overkill and probably we can get rid of
it.

Comments/ suggetions ??

Thanks

Post by Laurent Pinchart
Laurent Pinchart

Laurent Pinchart

2008-01-18 22:19:53 UTC

Post by Manish Katiyar

Post by Manish Katiyar
Hi Laurent,
On Jan 18, 2008 5:31 AM, Laurent Pinchart