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

[L A Walsh <cygwin@tlinx.org>]

Going by subj and talk below, this is a bit confusing...

But it looks like you are testing 'free' for a value?

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

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


Hans-Bernhard Bröker wrote:
> [Sorry, forgot to reply-all...]
>
> Am 15.03.2017 um 23:48 schrieb Jeffrey Walton:
>
>> Since Coverity is
>> complaining about an implicit conversion, maybe the following will
>> help to avoid the implicit part (and sidestep the finding):
>>
>>     if (free != NULL)
>>         break;
>>
>> Or perhaps:
>>
>>     if ((void*)free != NULL)
>>         break;
>
> Even setting aside that the latter should of course have been
>
>      if ((void*)free == NULL)
>          break;
>
> those are both worse than the original code.  (void *) is _not_ 
> suitable for use with function pointers.  Neither is NULL in the 
> general case, because it may very well be ((void *)0).
>
> The reason this is wrong is that C by design treats data and functions 
> as living in separate realms, i.e. its virtual machine has a Harvard 
> architecture.  One of the consequences of this is that pointers to 
> functions and pointers to data are incommensurable, i.e. any and all 
> conversions or comparisons across this divide are wrong.  (void *) are 
> compatible to all data pointers, but not to function pointers.
>
> 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 what language?

It's not a C-function nor a C function pointer.



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