[FFmpeg-devel] compile issues under mingw-cross-env

[John Calcote]

On 7/10/2010 3:10 PM, Eli Friedman wrote:
> On Sat, Jul 10, 2010 at 1:57 PM, Ramiro Polla <ramiro.polla at gmail.com> wrote:
>   
>> On Sat, Jul 10, 2010 at 5:53 PM, John Calcote <john.calcote at gmail.com> wrote:
>>     
>>> On 7/10/2010 1:25 PM, M?ns Rullg?rd wrote:
>>>       
>>>> John Calcote <john.calcote at gmail.com> writes:
>>>>         
>>>>> Anyone ever tried compiling ffmpeg under mingw-cross-env
>>>>> (http://www.nongnu.org/mingw-cross-env/)? I've been playing with it and
>>>>> I've found the following issues with CPPFLAGS in the generated config.mak:
>>>>>
>>>>> 1) I had to change -std=c99 to -std=gnu99 because _STRICT_ANSI is
>>>>> enabled if -std=c99 is used which means functions like strcasecmp aren't
>>>>> available (at least the prototypes aren't).
>>>>>
>>>>>           
>>>> Wrong.  strcasecmp() is part of C99 and on any conforming system is
>>>> declared when using those flags.
>>>>
>>>>         
>>> I'm not seeing strcasecmp as being part of C99. I'm seeing it as being
>>> part of POSIX 2001... Are you sure you're correct on this one? When I
>>> google search strcasecmp and c99, I get very little information on the
>>> pair. However, most of what I do get appear to be comments about how
>>> ffmpeg doesn't compile under mingw.
>>>       
>> 'man standards' says:
>> "POSIX.1-2001 is aligned with C99, so that all of the library
>> functions standardised in C99 are also standardised in POSIX.1-1001."
>>     
> strcasecmp is defined in strings.h, which is not part of C99.  So
> -std=c99 vs. -std=gnu99 shouldn't have any affect.
>   

I don't think this is an accurate assessment of the way these flags
work. In the first place, gnu99 is technically less restrictive than
c99. However, the real issue, in my experience, with compiler flags that
ask for compliance with a standard is that they tend to be
compound-restrictive, not compound-additive. That is to say these flags
are designed to ensure that only the subset of library routines that
comply with the requested standard are available, and that uses of all
other routines are flagged as errors.

To put it another way, they're not designed to provide access to a
specific set of functions, but to disallow anything outside the desired
set. The reason for this approach is that the flags are designed to help
developers write portable code. If you know that three target platform
tool sets all conform to C99, and you use the --std=c99 flag on the gcc
command line, then you're telling gcc that you don't want to allow
anything _except_ C99-compliant functions, not that you want to ensure
that you have access to all C99 functions. That way, you can be sure
your code will compile on your other two platforms.

Thus, if you supply flags that say you want C99 _and_ POSIX 2001, you're
likely (on some platforms, at least) to get the intersection of the two
sets, not the union of them. Often this situation goes unnoticed because
the these two sets overlap to a high degree (in fact C99 is completely
contained within POSIX 2001).

John