On Wed, 18 Nov 1998, R Shapiro wrote:
> Linus Torvalds writes:
> > it (ie instead of using
> >
> > num = va_arg(args, short);
> >
> > it might be proper to use
> >
> > num = (short) va_args(args, int);
However, I also suspect that the first version really _should_ work, and
that it really is a problem in gcc or the varargs header files that makes
it not work.
Alternatively, I'd really like to know whether maybe the C standard really
says that you _have_ to use the second form. It's possible it does.
Reading section 7.8.1.2 of the ANSI C standard (/word/ corresponds to
italics):
#include <stdarg.h>
/type/ va_arg (va_list ap, /type/)
...if /type/ is not compatible with the type of the actual next
argument (as promoted according to the default argument
promotions), the behavior is undefined.
Section 6.1.2.6 defines compatible types:
Two types have /compatible types/ if their types are the same...
There are additional rules for compatibility but none apply in this
case AFAICS.
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