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Message-ID: <Pine.LNX.4.58.0801161934500.17781@gandalf.stny.rr.com>
Date: Wed, 16 Jan 2008 19:36:30 -0500 (EST)
From: Steven Rostedt <rostedt@...dmis.org>
To: Mathieu Desnoyers <mathieu.desnoyers@...ymtl.ca>
cc: john stultz <johnstul@...ibm.com>,
LKML <linux-kernel@...r.kernel.org>, Ingo Molnar <mingo@...e.hu>,
Linus Torvalds <torvalds@...ux-foundation.org>,
Andrew Morton <akpm@...ux-foundation.org>,
Peter Zijlstra <a.p.zijlstra@...llo.nl>,
Christoph Hellwig <hch@...radead.org>,
Gregory Haskins <ghaskins@...ell.com>,
Arnaldo Carvalho de Melo <acme@...stprotocols.net>,
Thomas Gleixner <tglx@...utronix.de>,
Tim Bird <tim.bird@...sony.com>,
Sam Ravnborg <sam@...nborg.org>,
"Frank Ch. Eigler" <fche@...hat.com>,
Steven Rostedt <srostedt@...hat.com>,
Paul Mackerras <paulus@...ba.org>,
Daniel Walker <dwalker@...sta.com>
Subject: Re: [RFC PATCH 16/22 -v2] add get_monotonic_cycles
On Wed, 16 Jan 2008, Steven Rostedt wrote:
> On Wed, 16 Jan 2008, Mathieu Desnoyers wrote:
> >
> > !0 is not necessarily 1. This is why I use cpu_synth->index ? 0 : 1 in
>
> How about simply "cpu_synth->index ^ 1"? Seems the best choice if you ask
> me, if all you are doing is changing it from 1 to zero and back to 1.
>
FYI:
rostedt@...bo:~/c$ cat flipme.c
int flip1 (int x)
{
return !x;
}
int flip2 (int x)
{
return x ? 0 : 1;
}
int flip3(int x)
{
return x ^ 1;
}
rostedt@...bo:~/c$ gcc -O2 -c flipme.c
rostedt@...bo:~/c$ objdump -d flipme.o
flipme.o: file format elf32-i386
Disassembly of section .text:
00000000 <flip1>:
0: 55 push %ebp
1: 31 c0 xor %eax,%eax
3: 89 e5 mov %esp,%ebp
5: 83 7d 08 00 cmpl $0x0,0x8(%ebp)
9: 5d pop %ebp
a: 0f 94 c0 sete %al
d: c3 ret
e: 66 90 xchg %ax,%ax
00000010 <flip2>:
10: 55 push %ebp
11: 31 c0 xor %eax,%eax
13: 89 e5 mov %esp,%ebp
15: 83 7d 08 00 cmpl $0x0,0x8(%ebp)
19: 5d pop %ebp
1a: 0f 94 c0 sete %al
1d: c3 ret
1e: 66 90 xchg %ax,%ax
00000020 <flip3>:
20: 55 push %ebp
21: 89 e5 mov %esp,%ebp
23: 8b 45 08 mov 0x8(%ebp),%eax
26: 5d pop %ebp
27: 83 f0 01 xor $0x1,%eax
2a: c3 ret
So, if you know for sure that x is only 1 or 0, then using x ^ 1 to invert
it, seems the most efficient.
-- Steve
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