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Message-ID: <389ACC87.134F06F1@inti.gov.ar>
Date: Fri, 04 Feb 2000 09:56:39 -0300
From: salvador <salvador AT inti DOT gov DOT ar>
Organization: INTI
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To: djgpp AT delorie DOT com
Subject: Re: Clobbering registers in asm and GCC 2.952
References: <EHMPDBMAIKAABAAA AT shared1-mail DOT whowhere DOT com>
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Reply-To: djgpp AT delorie DOT com

Nimrod Alonzo Abing wrote:

> Hello!
>
> Just wondering, if there is a more down-to-
> earth explanation of that register clobbering
> stuff in asm using GCC 2.952.

[snip]

>
> My reaction to this reasonably obfuscated
> description of the problem and its solution was "Huh?!?" And the examples given were too
> short to demonstrate the principles behind it.
>
> Q: Where can I find a more subtle explanation of
> what's going on, aside from the cut-and-dry
> explanations from the GCC FAQ?
>
> Help will be greatly appreciated. Thanks...

:-))), isn't that obfuscated, but you aren't the first to ask for clarifications.
Basically: you can't list an input register in the clobber list. So if you will pass a parameter to your
assembler in ECX and you know it will be clobbered (example: is the counter of a loop) you can't put it in
the clobber list. Forget about why it was possible in the past and why isn't now, the explanation is quite
weak for x86 architecture. So now you are wondering how to indicate gcc that this register will be clobbered,
because you know it will be clobbered. Simple: use it as an output register, but then you'll ask: and where
do I put the content of ECX? and isn't it a waste of variables? To the first: use a "dummy" local variable.
To the second: if you use this variable only to hold the output value and nothing more gcc will optimize it
and remove the load and allocation of the local variable. So gcc will know the ECX carries new information
(different to the one he put in ECX) and will also know you don't care about this information.

SET

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