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Re: free() and implicit conversion to a function pointer
- From: L A Walsh <cygwin at tlinx dot org>
- To: cygwin at cygwin dot com
- Date: Thu, 16 Mar 2017 14:46:16 -0700
- Subject: Re: free() and implicit conversion to a function pointer
- Authentication-results: sourceware.org; auth=none
- References: <email@example.com> <firstname.lastname@example.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
If you aren't talking standard 'free()', then
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)
if ((void*)free != NULL)
Even setting aside that the latter should of course have been
if ((void*)free == NULL)
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)
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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