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Date: | Sun, 30 Dec 2001 13:06:04 +0200 (IST) |
From: | Eli Zaretskii <eliz AT is DOT elta DOT co DOT il> |
X-Sender: | eliz AT is |
To: | djgpp-workers AT delorie DOT com |
Subject: | Re: gcc 3.03 and libc sources |
In-Reply-To: | <200112271404.PAA16127@father.ludd.luth.se> |
Message-ID: | <Pine.SUN.3.91.1011230130545.10061B-100000@is> |
MIME-Version: | 1.0 |
Reply-To: | djgpp-workers AT delorie DOT com |
Errors-To: | nobody AT delorie DOT com |
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On Thu, 27 Dec 2001, Martin Str|mberg wrote: > According to Eli Zaretskii: > > > > On Thu, 27 Dec 2001, Martin Str|mberg wrote: > > > > > I tried to add "--std=c89" to gcc.opt and got an immense amount of > > > warnings from trying to make dosexec.o > > > > Could you post those warnings? I don't know why we should get warnings > > for the old standard. > > Ok. Here they are: > > gcc ... -c dosexec.c > dosexec.c: In function `check_talloc': > dosexec.c:140: `E2BIG' undeclared (first use in this function) > dosexec.c:140: (Each undeclared identifier is reported only once > dosexec.c:140: for each function it appears in.) > dosexec.c: In function `direct_exec_tail_1': > dosexec.c:164: `__dpmi_regs' undeclared (first use in this function) > dosexec.c:164: parse error before "r" It sounds like --std=c89 defines __STRICT_ANSI__, which masks non-standard parts of the headers. Is that true? (The GCC manual doesn't seem to say that.) If this is true, you have no hope of compiling the library with --std=c89.
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