Re: Why is the kfree() argument const?

[Giacomo A. Catenazzi]

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
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