linux-kernel - Re: [PATCH 0/11] LTTng-core (basic tracing infrastructure) 0.5.108

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Message-ID: <20060916173035.GA705@Krystal>
Date:	Sat, 16 Sep 2006 13:30:35 -0400
From:	Mathieu Desnoyers <mathieu.desnoyers@...ymtl.ca>
To:	Jes Sorensen <jes@....com>
Cc:	Ingo Molnar <mingo@...e.hu>, Roman Zippel <zippel@...ux-m68k.org>,
	Andrew Morton <akpm@...l.org>, tglx@...utronix.de,
	karim@...rsys.com, Paul Mundt <lethal@...ux-sh.org>,
	linux-kernel@...r.kernel.org,
	Christoph Hellwig <hch@...radead.org>,
	Ingo Molnar <mingo@...hat.com>,
	Greg Kroah-Hartman <gregkh@...e.de>,
	Tom Zanussi <zanussi@...ibm.com>, ltt-dev@...fik.org,
	Michel Dagenais <michel.dagenais@...ymtl.ca>
Subject: Re: [PATCH 0/11] LTTng-core (basic tracing infrastructure) 0.5.108

* Jes Sorensen (jes@....com) wrote:

> If you want to prove people wrong, I suggest you do some real life
> implementation and measure some real workloads with a predefined set of
> tracepoints implemented using kprobes and LTT and show us that the
> benchmark of the user application suffers in a way that can actually be
> measured. Argueing that a syscall takes an extra 50 instructions
> because it's traced using kprobes rather than LTT doesn't mean it
> actually has any real impact.
>

And about those extra cycles.. according to :
Documentation/kprobes.txt
"6. Probe Overhead

On a typical CPU in use in 2005, a kprobe hit takes 0.5 to 1.0
microseconds to process.  Specifically, a benchmark that hits the same
probepoint repeatedly, firing a simple handler each time, reports 1-2
million hits per second, depending on the architecture.  A jprobe or
return-probe hit typically takes 50-75% longer than a kprobe hit.
When you have a return probe set on a function, adding a kprobe at
the entry to that function adds essentially no overhead.

i386: Intel Pentium M, 1495 MHz, 2957.31 bogomips
k = 0.57 usec; j = 1.00; r = 0.92; kr = 0.99; jr = 1.40

x86_64: AMD Opteron 246, 1994 MHz, 3971.48 bogomips
k = 0.49 usec; j = 0.76; r = 0.80; kr = 0.82; jr = 1.07

ppc64: POWER5 (gr), 1656 MHz (SMT disabled, 1 virtual CPU per physical CPU)
k = 0.77 usec; j = 1.31; r = 1.26; kr = 1.45; jr = 1.99


So, 1 microsecond seems more like 1500-2000 cycles to me, not 50.

Mathieu




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