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cmd/compile: enable constant-time CFG editing

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Provide indexes along with block pointers for Preds
and Succs arrays.  This allows us to splice edges in
and out of those arrays in constant time.

Fixes worst-case O(n^2) behavior in deadcode and fuse.

benchmark                     old ns/op      new ns/op     delta
BenchmarkFuse1-8              2065           2057          -0.39%
BenchmarkFuse10-8             9408           9073          -3.56%
BenchmarkFuse100-8            105238         76277         -27.52%
BenchmarkFuse1000-8           3982562        1026750       -74.22%
BenchmarkFuse10000-8          301220329      12824005      -95.74%
BenchmarkDeadCode1-8          1588           1566          -1.39%
BenchmarkDeadCode10-8         4333           4250          -1.92%
BenchmarkDeadCode100-8        32031          32574         +1.70%
BenchmarkDeadCode1000-8       590407         468275        -20.69%
BenchmarkDeadCode10000-8      17822890       5000818       -71.94%
BenchmarkDeadCode100000-8     1388706640     78021127      -94.38%
BenchmarkDeadCode200000-8     5372518479     168598762     -96.86%

Change-Id: Iccabdbb9343fd1c921ba07bbf673330a1c36ee17
Reviewed-on: https://go-review.googlesource.com/22589
Reviewed-by: Josh Bleecher Snyder <josharian@gmail.com>
Run-TryBot: Keith Randall <khr@golang.org>
TryBot-Result: Gobot Gobot <gobot@golang.org>
This commit is contained in:
Keith Randall 2016-04-28 16:52:47 -07:00
parent bcd4b84bc5
commit 4fa050024f
33 changed files with 642 additions and 566 deletions
Download patch file Download diff file Expand all files Collapse all files

6
src/cmd/compile/internal/ssa/prove.go
View file

@ -502,7 +502,7 @@ func prove(f *Func) {
b.Kind = BlockFirst
b.SetControl(nil)
if succ == negative {
b.Succs[0], b.Succs[1] = b.Succs[1], b.Succs[0]
b.swapSuccessors()
}
}
@ -525,10 +525,10 @@ func getBranch(sdom sparseTree, p *Block, b *Block) branch {
// has one predecessor then (apart from the degenerate case),
// there is no path from entry that can reach b through p.Succs[1].
// TODO: how about p->yes->b->yes, i.e. a loop in yes.
if sdom.isAncestorEq(p.Succs[0], b) && len(p.Succs[0].Preds) == 1 {
if sdom.isAncestorEq(p.Succs[0].b, b) && len(p.Succs[0].b.Preds) == 1 {
return positive
}
if sdom.isAncestorEq(p.Succs[1], b) && len(p.Succs[1].Preds) == 1 {
if sdom.isAncestorEq(p.Succs[1].b, b) && len(p.Succs[1].b.Preds) == 1 {
return negative
}
return unknown