[Python-Dev] Compiler warnings (original) (raw)

Sjoerd Mullender sjoerd at acm.org
Wed Feb 1 11:34:00 CET 2006

Thomas Wouters wrote:

On Tue, Jan 31, 2006 at 08:16:21PM -0500, Tim Peters wrote:

Is this version of gcc broken in some way relative to other gcc versions, or newer, or ... ? We certainly don't want to see warnings under gcc, since it's heavily used, but I'm not clear on why other versions of gcc aren't producing these warnings (or are they, and people have been ignoring that?). Well, I said 4.0.3, and that was wrong. It's actually a pre-release of 4.0.3 (in Debian's 'unstable' distribution.) However, 4.0.2 (the actual release) behaves the same way. The normal make process shows quite a lot of output on systems that use gcc, so I wouldn't be surprised if people did ignore it, for the most part. My main problem with fixing the warnings is that I don't see the difference between, for example, the 'ssize' variable and the 'nchannels' variable in linuxaudio's ladobuffree/ladbufsize/ladobufcount. 'ssize' gets a warning, 'nchannels' doesn't, yet how they are treated is not particularly different. The ssize output parameter gets set inside a switch, is directly followed by a break, and the switch is directly followed by a set of the nchannels output parameter. The only way through the switch is through the set of ssize. I understand the compiler doesn't "see" it this way, but who knows for how long :) I guess we ignore this until we're closer to a 2.5alpha1 ;P

I don't quite understand what's the big deal. The compiler issues a warning. We know better (and I agree, we do know better in most of these cases), but it's easy to add a "= 0" to the declaration of the variable to shut up the compiler, hopefully with a comment saying as much. That's what I've been doing in my code that generated these warnings. It's clearly a "bug" in the compiler that it isn't smart enough to figure out that variable do actually only get used after they've been set. Hence, this is Somebody Else's Problem.

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