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strings: smarter growth of temporal buffer and avoid copying on return
The implementation for single strings had two optimization opportunities:
1. Grow the temporary buffer by known size before appending.
2. Avoid a full copy of the result since the underlying buffer won't be mutated afterward.
Both things were leveraged by using a Builder instead of a byte slice.
Relevant benchmark results:
name old time/op new time/op delta
SingleMatch-8 32.0µs ± 3% 26.1µs ± 3% -18.41% (p=0.000 n=9+10)
name old speed new speed delta
SingleMatch-8 469MB/s ± 3% 574MB/s ± 3% +22.56% (p=0.000 n=9+10)
name old alloc/op new alloc/op delta
SingleMatch-8 81.3kB ± 0% 49.0kB ± 0% -39.67% (p=0.000 n=10+10)
name old allocs/op new allocs/op delta
SingleMatch-8 19.0 ± 0% 11.0 ± 0% -42.11% (p=0.000 n=10+10)
Change-Id: I23af56a15875206c0ff4ce29a51bec95fd48bb11
GitHub-Last-Rev: 403cfc3c27
GitHub-Pull-Request: golang/go#47766
Reviewed-on: https://go-review.googlesource.com/c/go/+/343089
Run-TryBot: Ian Lance Taylor <iant@golang.org>
TryBot-Result: Go Bot <gobot@golang.org>
Reviewed-by: Ian Lance Taylor <iant@golang.org>
Trust: Robert Griesemer <gri@golang.org>
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@ -387,7 +387,7 @@ func makeSingleStringReplacer(pattern string, value string) *singleStringReplace
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}
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func (r *singleStringReplacer) Replace(s string) string {
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var buf []byte
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var buf Builder
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i, matched := 0, false
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for {
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match := r.finder.next(s[i:])
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@ -395,15 +395,16 @@ func (r *singleStringReplacer) Replace(s string) string {
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break
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}
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matched = true
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buf = append(buf, s[i:i+match]...)
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buf = append(buf, r.value...)
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buf.Grow(match + len(r.value))
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buf.WriteString(s[i : i+match])
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buf.WriteString(r.value)
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i += match + len(r.finder.pattern)
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}
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if !matched {
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return s
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}
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buf = append(buf, s[i:]...)
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return string(buf)
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buf.WriteString(s[i:])
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return buf.String()
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}
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func (r *singleStringReplacer) WriteString(w io.Writer, s string) (n int, err error) {
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