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Re: [Qemu-devel] qemu log function to print out the registers of the gue


From: Wei-Ren Chen
Subject: Re: [Qemu-devel] qemu log function to print out the registers of the guest
Date: Sun, 19 Aug 2012 16:33:13 +0800
User-agent: Mutt/1.5.21 (2010-09-15)

On Fri, Aug 17, 2012 at 03:57:55PM +0400, Max Filippov wrote:
> On Fri, Aug 17, 2012 at 3:14 PM, 陳韋任 (Wei-Ren Chen)
> <address@hidden> wrote:
> >> > On Thu, Aug 16, 2012 at 7:49 PM, Steven <address@hidden> wrote:
> >> > [...]
> >> >> I want to get the guest memory address in the instruction mov
> >> >> 0x4(%ebx)  %eax, whic is 0x4(%ebx).
> >> >> Since %ebx is not resolved until the execution time, the code in
> >> >> softmmu_header.h does not generate any hit or miss information.
> >> >> Do you know any place that I could resolve the memory access address? 
> >> >> Thanks.
> >> >
> >> > You'll have to generate code.  Look at how helpers work.
> >> Hi, Laurent,
> >> do you mean the target-i386/op_helper.c/helper.c or the tcg helper? Thanks.
> >
> >   What do you mean by "resolve the memory access address"? Do you want
> > to get guest virtual address for each guest memory access, right? As Max
> > mentioned before (you can also read [1]), there are fast and slow path
> > in QEMU softmmu, tlb hit and tlb miss respectively. Max provided patch
> > for slow path. As for fast path, take a look on tcg_out_tlb_load (tcg
> > /i386/tcg-target.c). tcg_out_tlb_load will generate native code in the
> > code cache to do tlb lookup, I think you cannot use the trick Max used
> > since tcg_out_tlb_load will not be called when the fast path executed,
> 
> That's why I've posted the following hunk that should have made all
> accesses go via slow path:

  Ya, I know. :) Just try to explain what Laurent want to say.

Regards,
chenwj

-- 
Wei-Ren Chen (陳韋任)
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



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