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cmd/syncthing: Speed up the CPU benchmark

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Not the measured performance, just the wall clock time it takes to
complete. The random generation was dominating.
This commit is contained in:
Jakob Borg 2017-02-06 13:42:39 +01:00
parent 67acef1794
commit e14741a58c
1 changed files with 8 additions and 8 deletions
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16
cmd/syncthing/usage_report.go
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@ -288,9 +288,13 @@ func (s *usageReportingService) Stop() {
// cpuBench returns CPU performance as a measure of single threaded SHA-256 MiB/s // cpuBench returns CPU performance as a measure of single threaded SHA-256 MiB/s
func cpuBench(iterations int, duration time.Duration, useWeakHash bool) float64 { func cpuBench(iterations int, duration time.Duration, useWeakHash bool) float64 {
dataSize := 16 * protocol.BlockSize
bs := make([]byte, dataSize)
rand.Reader.Read(bs)
var perf float64 var perf float64
for i := 0; i < iterations; i++ { for i := 0; i < iterations; i++ {
if v := cpuBenchOnce(duration, useWeakHash); v > perf { if v := cpuBenchOnce(duration, useWeakHash, bs); v > perf {
perf = v perf = v
} }
} }
@ -300,17 +304,13 @@ func cpuBench(iterations int, duration time.Duration, useWeakHash bool) float64
var blocksResult []protocol.BlockInfo // so the result is not optimized away var blocksResult []protocol.BlockInfo // so the result is not optimized away
func cpuBenchOnce(duration time.Duration, useWeakHash bool) float64 { func cpuBenchOnce(duration time.Duration, useWeakHash bool, bs []byte) float64 {
dataSize := 16 * protocol.BlockSize
bs := make([]byte, dataSize)
rand.Reader.Read(bs)
t0 := time.Now() t0 := time.Now()
b := 0 b := 0
for time.Since(t0) < duration { for time.Since(t0) < duration {
r := bytes.NewReader(bs) r := bytes.NewReader(bs)
blocksResult, _ = scanner.Blocks(r, protocol.BlockSize, int64(dataSize), nil, useWeakHash) blocksResult, _ = scanner.Blocks(r, protocol.BlockSize, int64(len(bs)), nil, useWeakHash)
b += dataSize b += len(bs)
} }
d := time.Since(t0) d := time.Since(t0)
return float64(int(float64(b)/d.Seconds()/(1<<20)*100)) / 100 return float64(int(float64(b)/d.Seconds()/(1<<20)*100)) / 100