Re: [PATCH] alpha: spinlock: don't perform memory access in locked critical section

Will Deacon
2013-05-06T21:12:50Z

Hi Al, Matt,

On Mon, May 06, 2013 at 09:53:30PM +0100, Al Viro wrote:
> On Mon, May 06, 2013 at 01:19:51PM -0700, Matt Turner wrote:
>
> > I'm not sure of the interpretation that LDA counts as a memory access.
> >
> > The manual says it's Ra <- Rbv + SEXT(disp).
> >
> > It's not touching memory that I can see.
>
> More to the point, the same manual gives explicit list of instructions
> that shouldn't occur between LDx_L and STx_C, and LDA does not belong to any
> of those. I suspect that Will has misparsed the notations in there - LDx is
> present in the list, but it's _not_ "all instructions with mnemonics starting
> with LD", just the 4 "load integer from memory" ones. FWIW, instructions
> with that encoding (x01xxx<a:5><b:5><offs:16>) are grouped so:
> LDAx - LDA, LDAH; load address
> LDx - LDL, LDQ, LDBU, LDWU; load memory data into integer register
> LDQ_U; load unaligned
> LDx_L - LDL_L, LDQ_L; load locked
> STx_C - STL_C, STQ_C; store conditional
> STx - STL, STQ, STB, STW; store
> STQ_U; store unaligned

Your suspicions are right! I did assume that LDA fell under the LDx class,
so apologies for the false alarm. I suspect I should try and get out more,
rather than ponder over this reference manual.

The other (hopefully also wrong) worry that I had was when the manual
states that:

`If the virtual and physical addresses for a LDx_L and STx_C sequence are
not within the same naturally aligned 16-byte sections of virtual and
physical memory, that sequence may always fail, or may succeed despite
another processor’s store to the lock range; hence, no useful program
should do this'

This seems like it might have a curious interaction with CoW paging if
userspace is trying to use these instructions for a lock, since the
physical address for the conditional store might differ from the one which
was passed to the load due to CoW triggered by a different thread. Anyway,
I was still thinking about that one and haven't got as far as TLB
invalidation yet :)

> They all have the same encoding, naturally enough (operation/register/address
> representation), but that's it... See section 4.2 in reference manual for
> details; relevant note follows discussion of LDx_L and it spells the list
> out. LDx is present, LDAx isn't (and neither is LDA by itself).

Indeed, and looking at the disassembly, you can see the immediate operand to
LDA encoded into the instruction. I thought that perhaps it might behave
like ldr =<imm> on ARM, which goes and fetches the immediate value from the
literal pool.

Cheers for the explanation,

Will