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.
Besides, for most debugging-enabled free() implementations, free() does indeed
touch the memory pointed to by its argument, which makes giving it a const
pointer completely bogus except for a single potential optimized special-case
where it might actually not touch the memory.
--
/ jakob
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