Re: [9front] Weird >>= operator behaviour in 5c/5l

[ori@eigenstate.org]

Quoth ori@eigenstate.org:
> Quoth Jonas Amoson <jonas.amoson@gmail.com>:
> > Left shifting by 32 instead of 0 does explain my
> > result, and seam indeed to be the real bug.
> > 
> > Regarding optimising out shifts where the number
> > of steps is zero: it already does that when using the
> > pattern x = x >> 0; but it doesn't for x >>= 0;
> > 
> > Thanks,
> > Jonas
> > 
> > On Mon, 18 Jan 2021 at 07:50, <ori@eigenstate.org> wrote:
> > >
> > > Quoth Noam Preil <noam@pixelhero.dev>:
> > > > It could just be architecture specific behavior.
> > > >
> > >
> > > It's a bug. Two, actually.
> > >
> > > 1) a shift of zero should be optimized out by
> > >    the compiler; cinap has a change for this in
> > >    the works already. This fix should hide the
> > >    second bug.
> > >
> > > 2) a shift by a literal zero is miscompiled by
> > >    the assembler into a shift by 32, because
> > >    of a missed special case in the instruction.
> > >    encoding. I'm working on fixing the assembler.
> 
> I think this corrects the linker.
> 
> diff -r 01125acb5565 sys/src/cmd/5l/asm.c
> --- a/sys/src/cmd/5l/asm.c	Tue Jan 19 19:56:38 2021 -0800
> +++ b/sys/src/cmd/5l/asm.c	Tue Jan 19 21:56:58 2021 -0800
> @@ -785,7 +785,7 @@
>  
>  	case 8:		/* sll $c,[R],R -> mov (R<<$c),R */
>  		aclass(&p->from);
> -		o1 = oprrr(p->as, p->scond);
> +		o1 = oprrr((instoffset == 0) ? ASLL : p->as, p->scond);
>  		r = p->reg;
>  		if(r == NREG)
>  			r = p->to.reg;
> 
> The PDF version of the arm architecture manual
> I found implied that this is the correct way
> to do it, and the official arm website is so
> bad that I can't figure out if it contradicts
> the PDF from some academic's website.
> 
> Only lightly tested, not sure if I have any
> edge cases.
> 
I like this version more.

diff -r 01125acb5565 sys/src/cmd/5l/asm.c
--- a/sys/src/cmd/5l/asm.c	Tue Jan 19 19:56:38 2021 -0800
+++ b/sys/src/cmd/5l/asm.c	Tue Jan 19 22:12:05 2021 -0800
@@ -789,6 +789,12 @@
 		r = p->reg;
 		if(r == NREG)
 			r = p->to.reg;
+		/*
+		 * R>>32 is encoded as R>>0, so flip to the
+		 * equivalent R<<0.
+		 */
+		if(instoffset == 0)
+			o1 &= ~(1<<5);
 		o1 |= r;
 		o1 |= (instoffset&31) << 7;
 		o1 |= p->to.reg << 12;