Davide Angelocola <[EMAIL PROTECTED]> wrote:
> hash.c with USE_OBSTACK fails to compile with gcc 3.3.4:
> hash.c:38:16: #if with no expression
Define it to `1' if you want to enable that:
i.e., -DUSE_OBSTACK=1 on the command line
or
#define USE_OBSTACK 1
___
Davide Angelocola <[EMAIL PROTECTED]> writes:
> hash.c:38:16: #if with no expression
You're supposed to define USE_OBSTACK to 1, not to no expression.
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According to Davide Angelocola on 2/17/2006 11:24 AM:
> Hi,
> hash.c with USE_OBSTACK fails to compile with gcc 3.3.4:
> hash.c:38:16: #if with no expression
Are you correctly defining USE_OBSTACK with a non-zero value, or did you
just define it as
Hi,
hash.c with USE_OBSTACK fails to compile with gcc 3.3.4:
hash.c:38:16: #if with no expression
hash.c:78:16: #if with no expression
hash.c:590:16: #if with no expression
hash.c:669:16: #if with no expression
hash.c:716:16: #if with no expression
hash.c:832:16: #if with no expression
hash.c:89