[SOLVED] container_of:why not "const typeof(*ptr) *__mptr=(ptr);"

songcaidao · 09-21-2013, 08:59 AM

Hi,I was reading about the implementation of the kernel's linked list.And I was fascinated by the macro container_of.

Code:

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

After I figured out every word in the definition,I got another question:
why not replace const typeof( ((type *)0)->member ) *__mptr = (ptr); with "const typeof(*ptr) *__mptr=(ptr);" ?
The second can also check the incompatible initialization and seems more easily to be understood.

Mara · 09-22-2013, 02:21 PM

This is not an official answer, but that's how I see it. The de-reference for a 'member' will always point to the right field. The way you propose would probably work if you're sure 'member' is always the first field (with no padding at the beginning of the structure), and it may or may not be true depending on the compiler and platform's ABI.

johnsfine · 09-22-2013, 03:42 PM

Quote:

Originally Posted by songcaidao

why not replace const typeof( ((type *)0)->member ) *__mptr = (ptr); with "const typeof(*ptr) *__mptr=(ptr);" ?

I think the intent is to generate an error in case typeof(*(ptr)) is the wrong type.

Quote:

The second can also check the incompatible initialization and seems more easily to be understood.

Why do you think that checks for incompatible initialization?

Code:

const typeof(*ptr) *__mptr=(ptr);

That will compile even if *(ptr) is the wrong type.

Code:

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

That will compile even if __mptr is the wrong type.
An advantage of:

Code:

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

is that it won't compile if ptr is an incompatible type.

songcaidao · 09-22-2013, 10:27 PM

Quote:

Originally Posted by johnsfine

I think the intent is to generate an error in case typeof(*(ptr)) is the wrong type.

Why do you think that checks for incompatible initialization?

Code:

const typeof(*ptr) *__mptr=(ptr);

That will compile even if *(ptr) is the wrong type.

Code:

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

That will compile even if __mptr is the wrong type.
An advantage of:

Code:

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

is that it won't compile if ptr is an incompatible type.

Thank you!
first of all,i want to replace the kernel's implementation with

Code:

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

.
and then i found it can not check the initialization like this

Code:

int a;
container_of(a,type,member);

.
but the macro what i mentioned in this thread works.and it made me believe the macro can work everywhere.
thank you for your reminding!i found it could not check the wrong type in this kind of initialization:

Code:

struct dog * pdog;
container_of(&pdog,struct dog,list);

thanks again!

johnsfine · 09-23-2013, 06:24 AM

Quote:

Originally Posted by songcaidao

i found it could not check the wrong type in this kind of initialization:

Code:

struct dog * pdog;
container_of(&pdog,struct dog,list);

That isn't a wrong type. It is a wrong value. &pdog has the correct type. It just does not have a value compatible with use of that macro.

A compiler can't check that your code is logically correct.

songcaidao · 09-23-2013, 07:05 AM

Quote:

Originally Posted by johnsfine

That isn't a wrong type. It is a wrong value. &pdog has the correct type. It just does not have a value compatible with use of that macro.

A compiler can't check that your code is logically correct.

Thanks!I get it.

johnsfine · 10-26-2013, 08:35 AM

Quote:

Originally Posted by gschen123

if typeof(*ptr) isnot the typeof(member), how could you do.

Your question may need a few more words. Anyway I don't want to guess at the meaning of your question.