On 5/17/19 1:33 PM, Arnd Bergmann wrote:
> On Fri, May 17, 2019 at 8:08 PM Alex Elder <el...@linaro.org> wrote:
>>
>> On 5/15/19 2:34 AM, Arnd Bergmann wrote:
>>>> +static void gsi_trans_tre_fill(struct gsi_tre *dest_tre, dma_addr_t addr,
>>>> + u32 len, bool last_tre, bool bei,
>>>> + enum ipa_cmd_opcode opcode)
>>>> +{
>>>> + struct gsi_tre tre;
>>>> +
>>>> + tre.addr = cpu_to_le64(addr);
>>>> + tre.len_opcode = gsi_tre_len_opcode(opcode, len);
>>>> + tre.reserved = 0;
>>>> + tre.flags = gsi_tre_flags(last_tre, bei, opcode);
>>>> +
>>>> + *dest_tre = tre; /* Write TRE as a single (16-byte) unit */
>>>> +}
>>> Have you checked that the atomic write is actually what happens here,
>>> but looking at the compiler output? You might need to add a 'volatile'
>>> qualifier to the dest_tre argument so the temporary structure doesn't
>>> get optimized away here.
>>
>> Currently, the assignment *does* become a "stp" instruction.
>> But I don't know that we can *force* the compiler to write it
>> as a pair of registers, so I'll soften the comment with
>> "Attempt to write" or something similar.
>>
>> To my knowledge, adding a volatile qualifier only prevents the
>> compiler from performing funny optimizations, but that has no
>> effect on whether the 128-bit assignment is made as a single
>> unit. Do you know otherwise?
>
> I don't think it you can force the 128-bit assignment to be
> atomic, but marking 'dest_tre' should serve to prevent a
> specific optimization that replaces the function with
>
> dest_tre->addr = ...
> dest_tre->len_opcode = ...
> dest_tre->reserved = ...
> dest_tre->flags = ...
>
> which it might find more efficient than the stp and is equivalent
> when the pointer is not marked volatile. We also have the WRITE_ONCE()
> macro that can help prevent this, but it does not work reliably beyond
> 64 bit assignments.
OK, I'll mark it volatile to avoid that potential result.
Thanks.
-Alex
>
> Arnd
>
