free() and implicit conversion to a function pointer
L A Walsh
cygwin@tlinx.org
Thu Mar 16 21:46:00 GMT 2017
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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