[PATCH] D21663: [MBP] Enhance cost based branch prob threshold computation to handle general control flows

[Xinliang David Li via llvm-commits]

The answer is, in this case, it is better not to choose BB->Succ as the
fall through.

Let's take a look at an example: suppose BB and Pred2 in example C has a
predecessor 'S'. With your choice, the layout is

S, BB, Succ, Pred2, Pred1 with cost F2 + F3 + F2 + F3

The optimal layout is in fact

S, BB, P1, P2, Succ with cost F1 + F2 + F3

David


On Thu, Jun 23, 2016 at 9:51 PM, Dehao Chen <danielcdh at gmail.com> wrote:

> For case C, if F1=50, F2=40, F3=20, it seems it's most beneficial to
> choose BB->Succ as fall-through than Pred1->Succ or Pred2->Succ, or am I
> missing something?
>
> On Thu, Jun 23, 2016 at 8:38 PM, David Li <davidxl at google.com> wrote:
>
>> davidxl added inline comments.
>>
>> ================
>> Comment at: lib/CodeGen/MachineBlockPlacement.cpp:610
>> @@ +609,3 @@
>> +    //         (C)                 (D)
>> +    //  (D) is a degenerated case of (C) where Pred1 does not exist. In
>> (D),
>> +    //  Let
>> ----------------
>> danielcdh wrote:
>> > You mean in (C), right?
>> yes.
>>
>> ================
>> Comment at: lib/CodeGen/MachineBlockPlacement.cpp:624
>> @@ +623,3 @@
>> +    //  order for BB->Succ to be selected, we must have F1 > F2 + F3. So
>> the
>> +    //  backward probability threshold is (backward checking igores other
>> +    //  predecessors other than Pred2):
>> ----------------
>> danielcdh wrote:
>> > This is a little confusing, I think the condition should be:
>> >
>> > F1 > max(2*F3, F2)
>> >
>> > So when checking Pred1, the backward threhold should be 0.66, when
>> checking Pred2, the backward threshold should be 0.5?
>> >
>> > But for the current implementation, I tried all testcase, and some
>> other testcases with different triangle-diamond combination, and it always
>> gives me optimal solution.
>> It is F1 > F2 + F3.
>>
>> Basically, we are comparing selecting BB->Succ as fall through vs
>> selecting BB->Pred1 and Pred2->Succ as fall throughs. In order for BB->Succ
>> to succeed, the minimal min(F1) = F2 + F3.
>>
>> When computing backward probability, only BB->Succ and Pred2->Succ edges
>> are considered, so the probability threshold
>>   T = min(F1) /(min(F1) + F2) = (F2 + F3 )/(2*F2 + F3).
>>
>> I will update the comments.
>>
>>
>> http://reviews.llvm.org/D21663
>>
>>
>>
>>
>
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