Re: [PATCH 09/18] soc: qcom: ipa: GSI transactions

[Arnd Bergmann]

On Mon, May 20, 2019 at 2:50 PM Alex Elder <elder@xxxxxxxxxx> wrote:
>
> On 5/20/19 4:25 AM, Arnd Bergmann wrote:
> > On Sun, May 19, 2019 at 7:11 PM Alex Elder <elder@xxxxxxxxxx> wrote:
> >> On 5/17/19 1:44 PM, Alex Elder wrote:
> >>> On 5/17/19 1:33 PM, Arnd Bergmann wrote:
> >>>> On Fri, May 17, 2019 at 8:08 PM Alex Elder <elder@xxxxxxxxxx>
> >>
> >> So it seems that I must *not* apply a volatile qualifier,
> >> because doing so restricts the compiler from making the
> >> single instruction optimization.
> >
> > Right, I guess that makes sense.
> >
> >> If I've missed something and you have another suggestion for
> >> me to try let me know and I'll try it.
> >
> > A memcpy() might do the right thing as well. Another idea would
>
> I find memcpy() does the right thing.
>
> > be a cast to __int128 like
>
> I find that my environment supports 128 bit integers.  But...
>
> > #ifdef CONFIG_ARCH_SUPPORTS_INT128
> > typedef __int128 tre128_t;
> > #else
> > typedef struct { __u64 a; __u64 b; } tre128_t;
> > #else
> >
> > static inline void set_tre(struct gsi_tre *dest_tre, struct gs_tre *src_tre)
> > {
> >      *(volatile tre128_t *)dest_tre = *(tre128_t *)src_tre;
> > }
> ...this produces two 8-bit assignments.  Could it be because
> it's implemented as two 64-bit values?  I think so.  Dropping
> the volatile qualifier produces a single "stp" instruction.

I have no idea how two 8-bit assignments could do that,
it sounds like a serious gcc bug, unless you mean two
8-byte assignments, which would be within the range
of expected behavior. If it's actually 8-bit stores, please
open a bug against gcc with a minimized test case.

> The only other thing I thought I could do to encourage
> the compiler to do the right thing is define the type (or
> variables) to have 128-bit alignment.  And doing that for
> the original simple assignment didn't change the (desirable)
> outcome, but I don't think it's really necessary in this
> case, considering the single instruction uses two 64-bit
> registers.
>
> I'm going to leave it as it was originally; it's the simplest:
>         *dest_tre = tre;
>
> I added a comment about structuring the code this way with
> the intention of getting the single instruction.  If a different
> compiler produces different result.

Ok, that's probably the best we can do then.

      Arnd