Re: free() and implicit conversion to a function pointer

[Hans-Bernhard Bröker <HBBroeker at t-online dot de>]

Am 16.03.2017 um 22:46 schrieb L A Walsh:
> Going by subj and talk below, this is a bit confusing...
>
> But it looks like you are testing 'free' for a value?

Not really.  The idea is to test free for _exixtence_.  Which only makes 
sense in case of weak symbol support getting involved.  In other 
situations, there could not possibly be a need for a run-time if() test, 
because surely the code could know at build time whether free() exists 
or not.

> Isn't standard 'free' declared to take 1 arg and
> return void?

Yes.  But since the code in question doesn't actually _call_ free, 
that's both irrelevant.

> If you aren't talking standard 'free()', then
> nevermind...

We are talking standard free.  More to the point, we're discussing 
newlib, the package that actually implements free() for cygwin.

> > The only code that might actually be a slight bit better than the given
> >
> >     if (! free)
> >
> > would be
> >
> >     if (0 != free)
> >
> > The function designator `free' auto-decays into a function pointer,
> > which is compared to a null pointer constant: 0.  The ! operator does
> > that same thing implicitly, but is fully equivalent to it.
> ---
> Free autodecays to a function pointer?

In the use case at hand: yes, it does.

> In what language?

Standard C.


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