Simon Kagstrom writes:
> At Tue, 22 Aug 2006 15:27:14 +0100,
> Andrew Haley wrote:
> > I see this:
> >
> > static inline int
> > puts (const char *string)
> > {
> > register unsigned long __v0 asm ("$2");
> > __asm__ volatile (".set push\n.set noreorder\n");
> > if (__builtin_constant_p (string))
> > {
> > __asm__ volatile (".long 0xfffd0000\n" ".long %0\n"::"i,!r" (string));
> > }
> > else
> > {
> > __asm__ volatile (".short 0xfffe\n" ".short %0\n"::"r" (string));
> > };
> > __asm__ volatile (".short 0xffff\t#" "puts" "\n\t"
> > ".short %1\n":"=&r" (__v0):"i" (1));
> > __asm__ volatile (".set\tpop\n");
> > return (int) __v0;
> > };
> >
> > and it's pretty clear that string is not a constant in the context of
> > this function: it's a parameter.
>
> I don't see why that would be a problem:
I know. :-)
> the function is inlined after all. Also, I've tried the same with
> exit (defined in the same way, but takes an integer), that works as
> expected.
It's a metter of _when_ the function gets inlined. If
__builtin_constant_p is folded before the function is inlined then
it'll return zero. If the constant doesn't happen to get propagated
it'll return zero. C'est la vie.
> The funny thing is that GCC realizes that it can put the constant
> address in the inline assembly with the "i,!r" constraint, but that
> __builtin_constant_p still returns 0. This is what I think looks
> inconsistent.
Does it work if you make the whole thing a macro?
Andrew.
