On Wed, Sep 30, 2020 at 03:59:15PM -0400, Arvind Sankar wrote:
> On Wed, Sep 30, 2020 at 12:14:03PM -0700, Nick Desaulniers wrote:
> > On Wed, Sep 30, 2020 at 10:13 AM Peter Zijlstra <pet...@infradead.org>
> > wrote:
> > >
> > > On Wed, Sep 30, 2020 at 11:10:36AM -0500, Segher Boessenkool wrote:
> > >
> > > > Since this variable is a local register asm, on entry to the asm the
> > > > compiler guarantees that the value lives in the assigned register (the
> > > > "r8" hardware register in this case). This all works completely fine.
> > > > This is the only guaranteed behaviour for local register asm (well,
> > > > together with analogous behaviour for outputs).
>
> How strict is the guarantee? This is an inline function -- could the
> compiler decide to reorder some other code in between the r8 assignment
> and the asm statement when it gets inlined?
Nope. It will be in r8 on entry to the asm. A guarantee is a
guarantee; it is not a "yeah maybe, we'll see".
> > Do we need register local storage here?
> >
> > static inline long bar(unsigned long hcall_id)
> > {
> > long result;
> > asm volatile("movl %1, %%r8d\n\t"
> > "vmcall\n\t"
> > : "=a" (result)
> > : "ir" (hcall_id)
> > : );
> > return result;
> > }
>
> This seems more robust, though you probably need an r8 clobber in there?
Oh, x86 has the operand order inverted, so this should work in fact.
Segher
