Re: WIP: Register doco

[Jim Blandy <jimb at redhat dot com>]

Andrew Cagney <ac131313@ges.redhat.com> writes:
> >> Sorry, I'm again lost.  I earlier wrote (note edits):
> >> ``No, ABI.  For instance mipsIII and o32.  The o32 ABI thinks
> >> registers have 32 bits yet the real register has 64 bits.  This gives
> >> two [cooked] views of the same [raw] register.  When o32 debug info
> >> indicates a value in two adjacent [cooked] registers, it is refering
> >> to 32 bit and not 64 bit registers.''
> >> I'm not discussing which of these should be printed since that is
> >> outside of the scope of this discussion.
> > (Sorry, the `what would this print' is a distraction.)
> > Suppose I have a program compiled to the o32 ABI which has a 64-bit
> > variable that the debug info says is in $a0.  I'm running it on a MIPS
> > III machine.  This means that half of my variable is in the low 32
> > bits of $a0, and the the other half is in the low 32 bits of $a1.
> > So, when you say that cooked registers are "ABI registers", are you
> > saying that, in the cooked register set, $a0 and $a1 would be 32-bit
> > registers, even though we're executing a 64-bit instruction set?
> > Having the register sizes disagree with the actual instructions being
> > executed is what seems like a bad idea to me.
> 
> As I pointed out in the above, there are two cooked $a0's.  One is 32
> bits and one is 64 bits.

Wow.  I read what you wrote, but I didn't get that.  So, there are
going to be two cooked register numbers for $a0, depending on whether
one is looking at it from the ABI point of view --- like debug info
does --- or from the ISA point of view.  Is that right?