Go 字符串连接+=与strings.Join性能对比

Posted 2021-01-26 promenader

Go字符串连接

对于字符串的连接大致有两种方式：
1、通过+号连接

func StrPlus1(a []string) string {
    var s, sep string
    for i := 0; i < len(a); i++ {
        s += sep + a[i]
        sep = " "
    }
    return s
}

2、通过strings.Join连接

func StrPlus2(a []string) string {
    return strings.Join(a, " ")
}

对比两种方式的效率，通过压力测试进行对比

import "testing"

func BenchmarkStrPlus1(b *testing.B) {
    for i := 0; i < b.N; i++ {
        StrPlus1([]string{"xxx", "bbb", "aaa"})
    }
}

func BenchmarkStrPlus2(b *testing.B) {
    for i := 0; i < b.N; i++ {
        StrPlus2([]string{"xxx", "bbb", "aaa"})
    }
}

运行压力测试go test -test.bench=".*"

goos: darwin
goarch: amd64
BenchmarkStrPlus1-4     10000000               127 ns/op
BenchmarkStrPlus2-4     20000000                78.7 ns/op

从本机来看通过+号连接字符串每个操作消耗127ns时间，strings.Join消耗78.7ns。效率上strings.Join更高

来看下strings包中Join的实现

// Join concatenates the elements of a to create a single string. The separator string
// sep is placed between elements in the resulting string.
func Join(a []string, sep string) string {
    switch len(a) {
    case 0:
        return ""
    case 1:
        return a[0]
    case 2:
        // Special case for common small values.
        // Remove if golang.org/issue/6714 is fixed
        return a[0] + sep + a[1]
    case 3:
        // Special case for common small values.
        // Remove if golang.org/issue/6714 is fixed
        return a[0] + sep + a[1] + sep + a[2]
    }
    n := len(sep) * (len(a) - 1)
    for i := 0; i < len(a); i++ {
        n += len(a[i])
    }

    b := make([]byte, n)
    bp := copy(b, a[0])
    for _, s := range a[1:] {
        bp += copy(b[bp:], sep)
        bp += copy(b[bp:], s)
    }
    return string(b)
}

可以看出当连接字符串数量较大时，是先通过make分配足够的空间，然后把每个字符串copy到空间里面，而不是每次通过+号来多次分配内存。