From:	Mario Deschenes <mdeschen AT btg DOT bombardier DOT com>
Newsgroups:	comp.os.msdos.djgpp
Subject:	cmpl takes 14 clk cycles on a Pentium ???
Date:	Thu, 12 Feb 1998 16:46:29 -0500
Organization:	Bell Network Solutions
Lines:	59
Message-ID:	<34E36DB5.71744432@btg.bombardier.com>
NNTP-Posting-Host:	207.61.235.2
Mime-Version:	1.0
To:	djgpp AT delorie DOT com
DJ-Gateway:	from newsgroup comp.os.msdos.djgpp

Hi everyone,

  I'm using RDTSC to profile a routine and I got something strange.  My
routine looks like:

void rts( void )
{
   static int i=0;

   __asm__("
           cli
           movl   _TSC,%eax
           cdq
           cdq
           rdtsc
           movl   %eax,_TSC
           movl   %edx,_TSC+4

           /* Some code */

           je     RTS_end             /* <-- This test ALWAYS jumps to
RTS_end (always true) */

          /* More code NEVER reached */

           cmpl   %eax,_ROM

           .align 4,0x90
RTS_end:
           rdtsc
           subl   _TSC,%eax
           sbbl   _TSC+4,%edx
           addl   %eax,_TOTAL
           adcl   %edx,_TOTAL+4
           sti");

   i++;
   printf("%10ld", TOTAL/i);
}

   What I do is boot the computer under DOS (not a shell or restart in
MS_DOS mode) and I average the time this routine
takes and print it.  This routine as you see it takes 84 clk cycles.
But if I remove the line "cmpl   %eax,_ROM" my program
takes 70 clk cycles.  What an improvement!!! The strange thing is that
this part of code is NEVER executed.  If I replace the "cmpl" with
6 nop (to take the same number of bytes), I still got 84 clk cycles.
Looking at the assembly listing, my code always start at
0x810 and the je RTS_end is at 0x868.  But the RTS_end is at 0x8C0
without the "cmpl" and at 0x8D0 with the "cmpl" opcode.

This is not an easy one and I would really appreciate if someone got a
hint.  Thanks

Mario Deschenes
Software Engineer
Bombardier transport group
mdeschen AT btg DOT bombardier DOT com

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