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From: | Martin Str|mberg <ams AT ludd DOT luth DOT se> |
Message-Id: | <200112270247.DAA14839@father.ludd.luth.se> |
Subject: | Re: gcc 3.03 and libc sources |
In-Reply-To: | <000401c18e3d$5c0b2cb0$cef8e0d5@zastaixp> from Tim Van Holder at "Dec 26, 2001 07:44:32 pm" |
To: | djgpp-workers AT delorie DOT com |
Date: | Thu, 27 Dec 2001 03:47:48 +0100 (MET) |
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According to Tim Van Holder: > > But what GCC flags in the cases you mentioned is something very > > different: there's no comparison between signed and unsigned > > values in those cases. What we have is an expression that > > sometimes yields signed values and sometimes unsigned values. > > What's wrong with that? > > I think there's a rule in either ANSI C++ or C99 that says that > both values in a conditional expression (?:) need to be the exact > same type (including signedness). > gcc3 enforces this rule by default; so I don't think a gcc bug > report is in order (unless it's mistakenly identifying the two > types as different, of course). Does this warning happen if you > use --std=c89 (or whatever thay option is)? I tried to add "--std=c89" to gcc.opt and got an immense amount of warnings from trying to make dosexec.o so I'm not sure what to try now. But it's strange that -Wsign-compare affects this case. As Eli says there's no comparision with signed and unsigned going on. Right, MartinS
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