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Message-ID: <20090729183756.7c1314ce@skybase>
Date: Wed, 29 Jul 2009 18:37:56 +0200
From: Martin Schwidefsky <schwidefsky@...ibm.com>
To: Daniel Walker <dwalker@...o99.com>
Cc: linux-kernel@...r.kernel.org, Ingo Molnar <mingo@...e.hu>,
Thomas Gleixner <tglx@...utronix.de>,
john stultz <johnstul@...ibm.com>
Subject: Re: [RFC][patch 02/12] remove clocksource inline functions
On Wed, 29 Jul 2009 08:52:50 -0700
Daniel Walker <dwalker@...o99.com> wrote:
> On Wed, 2009-07-29 at 17:32 +0200, Martin Schwidefsky wrote:
> > Hmm, you have an object of type struct clocksource and you do
> > cs->read(cs). If that is not clear enough then I don't know what is.
>
> It's not as clear as it could be .. In the case above you have to look
> in at least two places to know what's going on.. First to see the
> cs->read() , and second to see if "cs" is actually a clocksource or
> something else.. "cs" could be declared anyplace with any name.
Well you have something like that in the code:
struct clocksource *clock;
clock = timekeeper.clock;
cycle_now = clock->read(clock);
If I read the function top to bottom I immediately see that clock is a
clocksource and that the code does a read on it. That is not the case
if I need to lookup the clocksource_read wrapper.
> If you see clocksource_read(cs) , you might need to once check what
> clocksource_read() is actually doing, but only once.. After that when
> you see that function you know that variable is a clocksource, and it's
> "read()" is getting called. So you only need to review one line in the
> simplest case.
After you learned (once) that timekeeper.clock is a clock source you
have no trouble to understand the 6 occurrences of clock->read(clock)
there are in the code.
Anyway this seems to be a matter of personal preference, in the end I
don't care too much about the inline functions.
--
blue skies,
Martin.
"Reality continues to ruin my life." - Calvin.
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