linux-kernel - Re: [RFC PATCH 1/3] Unified trace buffer

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Date:	Thu, 25 Sep 2008 16:53:28 +0200
From:	Peter Zijlstra <peterz@...radead.org>
To:	Martin Bligh <mbligh@...igh.org>
Cc:	Martin Bligh <mbligh@...gle.com>,
	Steven Rostedt <rostedt@...dmis.org>,
	Linus Torvalds <torvalds@...ux-foundation.org>,
	linux-kernel@...r.kernel.org, Ingo Molnar <mingo@...e.hu>,
	Thomas Gleixner <tglx@...utronix.de>,
	Andrew Morton <akpm@...ux-foundation.org>,
	prasad@...ux.vnet.ibm.com,
	Mathieu Desnoyers <compudj@...stal.dyndns.org>,
	"Frank Ch. Eigler" <fche@...hat.com>,
	David Wilder <dwilder@...ibm.com>, hch@....de,
	Tom Zanussi <zanussi@...cast.net>,
	Steven Rostedt <srostedt@...hat.com>
Subject: Re: [RFC PATCH 1/3] Unified trace buffer

On Thu, 2008-09-25 at 07:33 -0700, Martin Bligh wrote:
> > I rather like this idea, as it gives small entries (the common case) the
> > least overhead but does allow for larger ones.
> >
> > By also putting the time in there you can do the merge sort iterator,
> > Linus was right that everybody wants this anyway.
> >
> > As for delta encoding the time, we could make the tick log the absolute
> > time packet, that's at least 100Hz and it already has to compute the
> > full gtod thing anyway.
> >
> > I don't much like Linus' idea of bringing type information back into the
> > primitive header (sorry Linus ;-)). I'd much rather keep that
> > abstraction in the next layer.
> 
> There is part of the type stuff that belongs in the lower layer, it seems -
> the padding events for the up-to-end-of-page buffering, and the timestamp
> extensions. It seems wrong to split those across two layers.

Hmm, you've got a point there, then it would be 3 package types:
 
 - regular
 - full time
 - nop

Which can be encoded using 2 bits

> But perhaps we can keep a couple of bits for this, and three of the bits
> to represent the length of the data payload (maybe in 4 byte multiples
> rather than bytes?) That'd let up to 28 bytes as a payload in a short event.

Right - if you use raw tsc you're dependent on clock speed, if we'd
normalize that on ns instead you'd need at least:

l(10000000)/l(2)
23.25349666421153643532

bits to handle HZ=100, leaving us with 32-2-24 = 6 bits for size.

Sounds doable (unless I mis-counted on the 0's).

Also, I agree on the 4byte alignment, rather than the 8byte Steve seems
to favour.

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