Re: Can anyone explain "movl %eax %eax"?

[Jidong Xiao]

Thank you David and Peter.

And I found more information in the following website, other people
who are not very clear can have a look at this document:

http://www.x86-64.org/documentation/assembly.html

======================
Implicit zero extend

Results of 32-bit operations are implicitly zero extended to 64-bit
values. This differs from 16 and 8 bit operations, that don't affect
the upper part of registers. This can be used for code size
optimisations in some cases, such as:

  movl $1, %eax                 # one byte shorter movq $1, %rax
  xorq %rax, %rax		# three byte equivalent of mov $0,%rax
  andl $5, %eax			# equivalent for andq $5, %eax
======================

Regards
Jidong

On Wed, Feb 9, 2011 at 6:05 PM, David Miller <davem@xxxxxxxxxxxxx> wrote:
> From: Jidong Xiao <jidong.xiao@xxxxxxxxx>
> Date: Wed, 9 Feb 2011 18:02:09 -0500
>
>> Oh, I see. Thank you. So similarly, the operation "xorl %eax,%eax" is
>> used for the same reason, right? I see that appears in more files.
>
> The xorl clears the entire register.
>
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