The existing implementation allocates a new string even when the
count is 1, where we know the output is the same as the input.
While we wouldn't expect a count of 1 for hardcoded values of the
parameter, it is expected when the parameter is computed based on
a different value (e.g., the length of a input slice).
name old time/op new time/op delta
Repeat/5x0-10 2.03ns ± 0% 2.02ns ± 0% ~ (p=1.000 n=1+1)
Repeat/5x1-10 13.7ns ± 0% 2.0ns ± 0% ~ (p=1.000 n=1+1)
Repeat/5x2-10 18.2ns ± 0% 18.1ns ± 0% ~ (p=1.000 n=1+1)
Repeat/5x6-10 27.0ns ± 0% 27.0ns ± 0% ~ (p=1.000 n=1+1)
Repeat/10x0-10 2.02ns ± 0% 2.02ns ± 0% ~ (p=1.000 n=1+1)
Repeat/10x1-10 16.1ns ± 0% 2.0ns ± 0% ~ (p=1.000 n=1+1)
Repeat/10x2-10 20.8ns ± 0% 20.9ns ± 0% ~ (p=1.000 n=1+1)
Repeat/10x6-10 29.2ns ± 0% 29.4ns ± 0% ~ (p=1.000 n=1+1)
Change-Id: I48e08e08f8f6d6914d62b3d6a61d563d637bec59
GitHub-Last-Rev:
068f58e08b8f5c4105e7a210f242ca1ff3a61177
GitHub-Pull-Request: golang/go#53321
Reviewed-on: https://go-review.googlesource.com/c/go/+/411477
Reviewed-by: Ian Lance Taylor <iant@google.com>
Run-TryBot: Ian Lance Taylor <iant@google.com>
TryBot-Result: Gopher Robot <gobot@golang.org>
Auto-Submit: Ian Lance Taylor <iant@google.com>
Reviewed-by: Keith Randall <khr@google.com>
// It panics if count is negative or if
// the result of (len(s) * count) overflows.
func Repeat(s string, count int) string {
- if count == 0 {
+ switch count {
+ case 0:
return ""
+ case 1:
+ return s
}
// Since we cannot return an error on overflow,
func BenchmarkRepeat(b *testing.B) {
s := "0123456789"
for _, n := range []int{5, 10} {
- for _, c := range []int{1, 2, 6} {
+ for _, c := range []int{0, 1, 2, 6} {
b.Run(fmt.Sprintf("%dx%d", n, c), func(b *testing.B) {
for i := 0; i < b.N; i++ {
Repeat(s[:n], c)