Re: gcc using clobbered register as input with inline assembly

[Bernd Schmidt <bernds at redhat dot com>]

On Mon, 26 Mar 2001, Zack Weinberg wrote:

> On Mon, Mar 26, 2001 at 01:08:32PM -0600, Timur Tabi wrote:
> > ** Reply to message from "Zack Weinberg" <zackw@stanford.edu> on Mon, 26 Mar
> > 2001 10:20:21 -0800
> >
> >
> > > > According to the documentation, because I specified eax and edx as
> > > > clobbered registers, gcc should not be using them as inputs or outputs.
> > > > But that's exactly what's happening in my case.
> > >
> > > Clobbers happen *after* the asm completes.  You have to do the moves
> > > separately, or let the compiler do them for you, e.g:
> >
> > I don't understand this.  What good is it to specify clobbers if they happen
> > after the assembly?  I have no control over what happens outside the assembly
> > code!  I thought the point behind clobbers was to inform the compiler of
> > specific registers that my assembly code modifies.
>
> Mm, let me rephrase.  You wrote
>
> >     __asm__ __volatile__ (
> >         "rdtsc \n\t"
> >         "mov %%eax, (%1) \n\t"
> >         "mov %%edx, 4(%1) \n\t"
> >         : "=m" (*pTime)
> >         : "r" (pTime)
> >         : "eax", "edx");
>
> As far as GCC is concerned, that is *one* machine instruction which:
> 	- takes one input, the value of pTime, in a register
> 	- places a value into memory at *pTime
> 	- destroys the values in eax and edx
> in that order.  It has no way of knowing that there's three machine
> instructions embedded in that text string, and that the destruction of
> eax and edx happens before the input is used.

That's not correct.  Clobbered registers are unavailable for inputs.
All gcc versions before 2.95 (and that includes egcs-1.1 which Timur was
using) failed to get this right on x86, however.


Bernd