On Wed, Apr 23, 2014 at 11:25 PM, Kenneth Graunke <kenn...@whitecape.org> wrote:
> On 04/18/2014 11:56 AM, Matt Turner wrote:
>> ---
>> src/mesa/drivers/dri/i965/brw_fs_cse.cpp | 19 +++++++++++++++----
>> 1 file changed, 15 insertions(+), 4 deletions(-)
>
> I originally thought this needed to go earlier in the patch series,
> since by this point you're emitting opcodes with more than 3 sources.
> However, CSE ignores SHADER_OPCODE_LOAD_PAYLOAD, so this code will never
> run for opcodes that could break. So, it's probably fine.
>
>> diff --git a/src/mesa/drivers/dri/i965/brw_fs_cse.cpp
>> b/src/mesa/drivers/dri/i965/brw_fs_cse.cpp
>> index 94f657d..e40567f 100644
>> --- a/src/mesa/drivers/dri/i965/brw_fs_cse.cpp
>> +++ b/src/mesa/drivers/dri/i965/brw_fs_cse.cpp
>> @@ -94,10 +94,20 @@ is_expression_commutative(enum opcode op)
>> }
>>
>> static bool
>> -operands_match(enum opcode op, fs_reg *xs, fs_reg *ys)
>> +operands_match(fs_inst *a, fs_inst *b)
>> {
>> - if (!is_expression_commutative(op)) {
>> - return xs[0].equals(ys[0]) && xs[1].equals(ys[1]) &&
>> xs[2].equals(ys[2]);
>> + fs_reg *xs = a->src;
>> + fs_reg *ys = b->src;
>> +
>> + if (!is_expression_commutative(a->opcode)) {
>> + bool match = true;
>> + for (int i = 0; i < a->sources; i++) {
>> + if (!xs[i].equals(ys[i])) {
>> + match = false;
>> + break;
>> + }
>> + }
>> + return match;
>> } else {
>> return (xs[0].equals(ys[0]) && xs[1].equals(ys[1])) ||
>> (xs[1].equals(ys[0]) && xs[0].equals(ys[1]));
>
> It strikes me as a bit asymmetric to have the first block handle an
> arbitrary number of sources, and the second only check 0 and 1. It
> makes sense, since our commutative opcodes are all binops, but...
>
> How about adding
>
> /* All commutative opcodes are binary operations. */
> assert(a->sources == 2 && b->sources == 2);
>
> here in the commutative case?
What is an example of a commutative non-binary operator?
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