Very Computer

  In studying the kernel I find many gcc inline 'asm' statements,
like so.

        asm ( assembler template
            : output operands                           (optional)
            : input operands                            (optional)
            : list of clobbered registers               (optional)
            );  

  I have read all the docs and I still can't clearly understand when it
is required to specify a clobberlist - a register or memory that will
be modified and must be preserved by gcc.

  In searching the kernel source there were very few clobbers given
and most of those were for memory.

  For example in arch/i386/lib/delay.c:

71         __asm__("mull %0"
72         :"=d" (xloops), "=&a" (d0)
73         :"1" (xloops),"" (current_cpu_data.loops_per_jiffy));
74         __delay(xloops * HZ);

  It's pretty obvious that eax gets clobbered, as I know clobber.  Why is
it not listed as such? Does the answer to this question apply in all cases?
What about memory clobbers - how do they happen?

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On Mon, 18 Mar 2002 15:27:40 -0800 (PST),

The clobber list is _extra_ clobbers, registers or memory that gcc
cannot deduce from the other parameters.  =d" (xloops), "=&a" (d0)
already tells gcc that edx and eax are used, you do not need to
explicitly specify them as clobbers.  OTOH, asm ("mov 1, %eax")
requires a clobber because the operands do not mention eax.

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In newsgroup: linux.dev.kernel

Quote:

>   I have read all the docs and I still can't clearly understand when it
> is required to specify a clobberlist - a register or memory that will
> be modified and must be preserved by gcc.

>   In searching the kernel source there were very few clobbers given
> and most of those were for memory.

>   For example in arch/i386/lib/delay.c:

> 71         __asm__("mull %0"
> 72         :"=d" (xloops), "=&a" (d0)
> 73         :"1" (xloops),"" (current_cpu_data.loops_per_jiffy));
> 74         __delay(xloops * HZ);

>   It's pretty obvious that eax gets clobbered, as I know clobber.  Why is
> it not listed as such? Does the answer to this question apply in all cases?
> What about memory clobbers - how do they happen?

It's not listed as such because it's an output (the second output),
not a clobber.  This is particularly common on x86 where registers are
so specialized that they all have their own constraint letters.

Incidentally, the above asm statement is technically incorrect, "=&a"
(d0) means eax is an "early clobber", however, the instruction doesn't
require it to be early clobbered.

        -hpa
--

"Unix gives you enough rope to shoot yourself in the foot."

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