Re: [RFC Patch]: Implement remainder() as built-in function [PR fortran/24518]

["H. J. Lu" <hjl at lucon dot org>]

On Wed, Oct 25, 2006 at 01:49:15PM -0600, Roger Sayle wrote:
> 
> Hi Mike,
> 
> On Wed, 25 Oct 2006, Mike Stump wrote:
> > On Oct 25, 2006, at 6:42 AM, Kaveh R. GHAZI wrote:
> > > I agree with Richard that transforming printf("hello") into
> > > fputs("hello",__get_stdout()) may not be a good option.
> >
> > For -Os, that would be bad.  For -O2, I'd let the numbers speak for
> > themselves, though, I'd also tend to think it might not be worth it.
> 
> 
> My first thought was to argue that a __get_stdout function should be
> a very fast subroutine call.  It doesn't need to set up a stack frame,
> doesn't have any arguments, does it's decode magic and places the
> result in a register.  Often, it something as simple as
> 
> __get_stdout:	mov %eax, __stdout
> 		ret
> 
> or similar.  This is compounded by the fact that we can mark the
> function as "const", and CSE and re-use this pointer in the caller.
> With call prediction and target buffers, etc.. the cost of a lightweight
> libcall is likely to be cheaper than the saved cycles between printf
> and fputs or fwrite (for example).
> 
> 
> However, it then occured to me that we can be even cleverer still:
> 
> extern FILE *__libgcc_stdout = stdout;
> 
> 
> We can completely avoid the function call.  Our issue is that libc's
> stdout isn't guaranteed to be mapped to a single simple symbol, and
> is often #defined to something cryptic.  libgcc can avoid this by
> declaring the required stable symbol itself.  Hence, in generated
> code we can simply refer to "__libgcc_stdout".
> 

Will

#include <stdio.h>

FILE *__libgcc_stdout;

static
init_stdout (void)
{
  __libgcc_stdout = stdout;
}

static void (*const init_array []) (void)
  __attribute__ ((section (".init_array"), aligned (sizeof (void *))))
=
{
  &init_stdout
};

work?