Re: Why is the kfree() argument const?

["Giacomo A. Catenazzi" <cate@cateee.net>]

Jakob Oestergaard wrote:
> On Thu, Jan 17, 2008 at 01:25:39PM -0800, Linus Torvalds wrote:
> ...
>> Why do you make that mistake, when it is PROVABLY NOT TRUE!
>>
>> Try this trivial program:
>>
>> 	int main(int argc, char **argv)
>> 	{
>> 	        int i;
>> 	        const int *c;
>> 	
>> 	        i = 5;
>> 	        c = &i;
>> 	        i = 10;
>> 	        return *c;
>> 	}
>>
>> and realize that according to the C rules, if it returns anything but 10, 
>> the compiler is *buggy*.
> 
> That's not how this works (as we obviously agree).
> 
> Please consider a rewrite of your example, demonstrating the usefulness and
> proper application of const pointers:
> 
> extern foo(const int *);
> 
> int main(int argc, char **argv)
> {
>  int i;
> 
>  i = 5;
>  foo(&i);
>  return i;
> }
> 
> Now, if the program returns anything else than 5, it means someone cast away
> const, which is generally considered a bad idea in most other software
> projects, for this very reason.
> 
> *That* is the purpose of const pointers.

"restrict" exists for this reason. const is only about lvalue.

You should draw a line, not to make C more complex!

Changing the name of variables in your example:

extern print_int(const int *);

int main(int argc, char **argv)
{
   extern int errno;

   errno = 0;
   print_int(&i);
   return errno;
}

print_int() doesn't know that errno is also the argument.
and this compilation unit doesn't know that print_int() will
modify errno.

Ok, I changed int to extern int, but you see the point?
Do you want complex rules about const, depending on
context (extern, volatile,...) ?

ciao
	cate