On Mon, Nov 9, 2020 at 1:16 PM Luc Van Oostenryck
<luc.vanoostenryck@xxxxxxxxx> wrote:
>
> There is no phi-source here and the CBR at .L3 could be merged
> with the one at .L0, removing the false imbalance but it's not.
> I thought that sparse was doing this sort of branch simplification
> but it doesn't, or at least it doesn't in this (simple) situation.
The branch rewriting sparse does is very simplistic, afaik. It's also
confusingly written. I blame myself.
Honestly, in that example you quote:
cbr %arg1, .L1, .L2
.L1: context 1
br .L3
.L2: call use, %r1
br .L3
.L3: cbr %arg1, .L5, .L4 ;; false imbalance here
we never even try to combine the two cbr's, because they don't jump
directly to each other.
I think it would be easy to simplify if we just added some logic to
change "unconditional branch to conditional branch", and moved the cbr
up into L1 and L2, but I don't think we have any such logic (and it's
a simplification that might end up being the opposite of a
simplification - it would duplicate the conditional).
So we could try to remove L3 entirely, by moving it into both parents:
cbr %arg1, .L1, .L2
.L1: context 1
cbr %arg1, .L5, .L4
.L2: call use, %r1
cbr %arg1, .L5, .L4
and at that point, the branch simplification we *do* have woudl see
that both of those duplicated conditional branches are now dominated
by the first one, and we'd end up with
cbr %arg1, .L1, .L2
.L1: context 1
br .L5
.L2: call use, %r1
br .L4
and then we'd join L2 and L4 together and the end result would look nice.
But as it stands now, L3 isn't dominated by one side of the original
conditional branch (because we reach L3 from both sides), and the cbr
in L3 isn't something we can simplify without duplicating it and
moving it up into the parents.
Is duplicating the conditional branch worth it? In this case, clearly
yes. But in general? Maybe we could do it in cases like this, when the
*only* thing in a basic block is that conditional branch.
Linus
