Peter Pentchev <[EMAIL PROTECTED]> wrote:
>
> On Thu, Jun 07, 2001 at 10:20:51AM -0700, John Baldwin wrote:
> >
> > On 07-Jun-01 Peter Pentchev wrote:
> > > On Thu, Jun 07, 2001 at 07:07:22PM +0300, Peter Pentchev wrote:
> > >> Hi,
> > >>
> > >> Is free((void *) (size_t) ptr) the only way to free a const whatever *ptr
> > >> with WARNS=2? (or more specifically, with -Wcast-qual)
> > >
> > > Uhm. OK. So size_t may not be enough to hold a pointer. What is it then -
> > > caddr_t?
> >
> > uintptr_t for data pointers. In theory I think code pointers may not fit in a
> > uintptr_t.
> >
> > free((void *)(uintptr_t)ptr) should work.
> >
> > Of course, this begs the question of why you are free'ing a const. :)
>
> OK, here's a scenario:
>
> struct validation_fun {
> const char *name;
> valfun *fun;
> int dyn;
> };
>
> This is a structure for a set of validation functions, referenced by
> name from another type of object. There are some predefined functions,
> present in the code at compile-time, and hardcoded in an array, with
> names given as "strings". Thus, the 'const'.
>
> However, some of the functions may be defined at runtime, with both
> name and code sent by a server. In that case, the name is a dynamically
> allocated char *, which needs to be freed upon cleanup. So I have:
>
> [cleanup function]
> ...
> if (val->dyn)
> free(val->name);
>
> Any suggestions on how to improve the design to avoid this, if possible,
> would be greatly welcome.
>
> G'luck,
> Peter
Since some strings are non-constant (the are allocated) - I believe
the `const' qualifier in the structure declaration is incorrect.
What happens if you simply don't have it in the structure declaration?
- Dave Rivers -
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