Forcing inline assembly code to be produced

[Andrew Haley]

Kristis Makris writes:
 > Hello,
 > 
 > I'm facing a problem producing inline assembly code in a C program. It
 > seems that when assembly code is added in a path that seems to be
 > unreachable (dead-code I suppose) the assembly code is not produced at
 > all.

Right.

 > I'm using gcc 3.3.5 and as 2.15.

 > Is there a way to force such assembly code to be produced even though it
 > may be unreachable ? If not should there be a way ?

If code is unreachable, there's no way to jump to it.  It's not legal
in gcc to jump into or out of an inline asm.

 > I'm attaching a simplified test case I'm using. I'm trying to generate a
 > symbol with inline assembly.
 > 
 > In a more complicated example, I'm trying to position uncommonly used
 > code in a separate memory region for better cache locality. One of the
 > inline assembly statements I'm using is not produced. This results in
 > not changing back to the ".text" section, resulting in further assembler
 > problems as described in:

Use an attribute to set the section.

         extern void foobar (void) __attribute__ ((section ("bar")));

     puts the function `foobar' in the `bar' section.

 > http://lists.gnu.org/archive/html/bug-binutils/2007-04/msg00107.html
 > 
 > Thanks for any help.
 > Kristis
 > #include <stdio.h>
 > 
 > 
 > int main(void)
 > {
 >   int a = 7;
 > 
 >   goto there;
 > 
 >  here:
 >   exit(0);
 > 
 >  there:
 >   a++;
 >   goto here;
 > 
 >   asm ("some_label_that_i_really_want_produced:\n\t"
 >        "nop\n\t"
 >        "nop\n\t"
 >        "nop\n\t"
 >        "nop\n\t" );
 > }

There's no way to guarantee that the asm will remain at the end of the
function.  gcc could convert this to:

int main(void)
{
  int a = 7;

  a++;
  goto here;

  asm ("some_label_that_i_really_want_produced:\n\t"
       "nop\n\t"
       "nop\n\t"
       "nop\n\t"
       "nop\n\t" );

 here:
  exit(0);
}


If you really need naked assembly code, do it *outside* a function:

asm ("\nfoo:\t\n"
     "\t# wibble");

int main(void)
{
  return 0;
}

Andrew.