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runtime: fix MapCycle test
It wasn't actually testing what it says it was testing. A random permutation isn't cyclic. It only probably hits a few elements before entering a cycle. Use an algorithm that generates a random cyclic permutation instead. Fixing the test makes the previous CL look less good. But it still helps. (Theory: Fixing the test makes it less cache friendly, so there are more misses all around. That makes the benchmark slower, suppressing the differences seen. Also fixing the benchmark makes the loop iteration count less predictable, which hurts the raw loop implementation somewhat.) (baseline = tip, experiment = tip+previous CL, noswiss = GOEXPERIMENT=noswissmap) goos: darwin goarch: arm64 pkg: runtime cpu: Apple M2 Ultra │ baseline │ experiment │ │ sec/op │ sec/op vs base │ MapCycle-24 20.59n ± 4% 18.99n ± 3% -7.77% (p=0.000 n=10) khr@Mac-Studio src % benchstat noswiss experiment goos: darwin goarch: arm64 pkg: runtime cpu: Apple M2 Ultra │ noswiss │ experiment │ │ sec/op │ sec/op vs base │ MapCycle-24 16.12n ± 1% 18.99n ± 3% +17.83% (p=0.000 n=10) Change-Id: I3a4edb814ba97fec020a6698c535ce3a87a9fc67 Reviewed-on: https://go-review.googlesource.com/c/go/+/625900 Reviewed-by: Michael Pratt <mpratt@google.com> LUCI-TryBot-Result: Go LUCI <golang-scoped@luci-project-accounts.iam.gserviceaccount.com> Reviewed-by: Keith Randall <khr@google.com>
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@ -258,12 +258,41 @@ func BenchmarkMapLast(b *testing.B) {
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}
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}
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func cyclicPermutation(n int) []int {
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// From https://crypto.stackexchange.com/questions/51787/creating-single-cycle-permutations
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p := rand.New(rand.NewSource(1)).Perm(n)
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inc := make([]int, n)
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pInv := make([]int, n)
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for i := 0; i < n; i++ {
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inc[i] = (i + 1) % n
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pInv[p[i]] = i
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}
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res := make([]int, n)
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for i := 0; i < n; i++ {
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res[i] = pInv[inc[p[i]]]
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}
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// Test result.
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j := 0
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for i := 0; i < n-1; i++ {
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j = res[j]
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if j == 0 {
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panic("got back to 0 too early")
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}
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}
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j = res[j]
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if j != 0 {
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panic("didn't get back to 0")
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}
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return res
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}
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func BenchmarkMapCycle(b *testing.B) {
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// Arrange map entries to be a permutation, so that
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// we hit all entries, and one lookup is data dependent
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// on the previous lookup.
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const N = 3127
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p := rand.New(rand.NewSource(1)).Perm(N)
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p := cyclicPermutation(N)
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m := map[int]int{}
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for i := 0; i < N; i++ {
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m[i] = p[i]
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