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Subject: Re: [Qemu-devel] qemu log function to print out the registers of
 the guest
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To: Steven <wangwangkang@gmail.com>
Cc: Peter Maydell <peter.maydell@linaro.org>, qemu-devel@nongnu.org, =?utf-8?B?6Zmz6Z+L5Lu7IChXZWktUmVuIENoZW4p?= <chenwj@iis.sinica.edu.tw>

> To verify what is translation time and what is the run time, I log the
> register information before disassembling each guest code. I copied
> some results from the log file, which is generated at run time of a
> guest machine.
>=20
>          EAX=3D00000000  EBX=3D00006ffc
>          IN:
>          0x00000000000f2087:  mov    $0xf5588,%eax
>=20
>          EAX=3D000f5588  EBX=3D00006ffc
>          IN:
>          0x00000000000f208B:  move 0x4(%ebx)  %eax
>=20
> The first instruction load eax with the value 0xf5588, so the eax at
> the second instruction is EAX=3D000f5588. So can I consider the memory
> address of 0x4(%ebx) as (00006ffc +  4)? I think this should be the
> run time information I need. Please correct me  if there is anything
> wrong. Thanks.

  IIRC, "-d in_asm" only give you what has been translated not executed.
Remember you said you log the register information before disassembling
each guest code? In other words, (guest) ebx might not be the value you
saw here. This is just my opinion.

Regards,
chenwj

--=20
Wei-Ren Chen (=E9=99=B3=E9=9F=8B=E4=BB=BB)
Computer Systems Lab, Institute of Information Science,
Academia Sinica, Taiwan (R.O.C.)
Tel:886-2-2788-3799 #1667
Homepage: http://people.cs.nctu.edu.tw/~chenwj