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Date:	Sun, 05 Nov 2006 12:10:41 -0800
From:	"H. Peter Anvin" <hpa@...or.com>
To:	Zachary Amsden <zach@...are.com>
CC:	Benjamin LaHaise <bcrl@...ck.org>,
	Chuck Ebbert <76306.1226@...puserve.com>,
	Linus Torvalds <torvalds@...l.org>, Andi Kleen <ak@...e.de>,
	linux-kernel <linux-kernel@...r.kernel.org>
Subject: Re: [rfc patch] i386: don't save eflags on task switch

Zachary Amsden wrote:
> Benjamin LaHaise wrote:
>> On Sat, Nov 04, 2006 at 11:09:42AM -0800, Zachary Amsden wrote:
>>  
>>> Every processor I've ever measured it on, popf is slower.  On P4, for 
>>> example, pushf is 6 cycles, and popf is 54.  On Opteron, it is 2 / 
>>> 12.  On Xeon, it is 7 / 91.
>>
>> pushf has to wait until all flag dependancies can be resolved.  On the 
>> P4 with >100 instructions in flight, that can take a long time.  Popf 
>> on the other hand has no dependancies on outstanding instructions as 
>> it resets the machine state.
> 
> Yes, but as Linus points out popf is most likely microcoded, thus much 
> slower.  Flag dependency is not unique to pushf, many much more common 
> instructions (adc, jcc, sbc, cmovcc, movs, stos, ...) have flag 
> dependencies, which can still be pipeline forwarded.  I think the raw 
> cycle counts speak for themselves, despite the fact that I only measured 
> instruction latency, not throughput.  Using a branch to eliminate a 
> pushf is thus probably not a win in most cases.
> 

The "sane" decomposition of popf into uops something like this:

- Memory read
- Mask bits that are immutable in the current mode
- Trap to microcode on changing any bit that alters the pipeline state

The "trap to microcode" can obviously be arbitrarily expensive.  So when 
timing popf, it's also important to know *which* bits change.

The simplest case, obviously, is when no flags change.  I still *fully* 
expect that to be more painful than pushf ever is.

	-hpa
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