Question

我试图将10百万的int转换为Hex，然后用0＆＃39;填充它以获得一个代表十六进制数的4个字符的字符串。

到目前为止，我尝试了以下内容：

var hexNumber string
for idx := O; idx < 10000000; idx++ {
    hexNumber = fmt.Sprintf("%04x", idx)

    // Do some stuff ....
}

但fmt.Sprintf效率不高。我怎样才能以有效的方式实现这一目标？

解决方案：事实证明@peterSO strconv.AppendInt解决方案更快。

package bench

import (
    "fmt"
    "strconv"
    "strings"
    "testing"
)

var stringHex [16]string
var runesHex [16]rune

func init() {
    stringHex = [16]string{"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F"}
    runesHex = [16]rune{'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'}
}

func intToHex1(intNumber int) string {
    hexNumber := []rune("0000")
    for i, j := int(0), uint(12); i < 4; i, j = i+1, j-4 {
        hexNumber[i] = runesHex[(intNumber>>j)&0x0f]
    }
    return string(hexNumber)
}

func intToHex2(intNumber int) string {
    hexNumber := "0000"
    for i, j := int(0), uint(12); i < 4; i, j = i+1, j-4 {
        hexNumber = hexNumber[:i] + stringHex[(intNumber>>j)&0x0f] + hexNumber[i+1:]
    }
    return hexNumber
}

func BenchmarkFmtSprintf(b *testing.B) {
    b.ReportAllocs()
    for n := 0; n < b.N; n++ {
        hexNumber := fmt.Sprintf("%04x", n)
        _ = hexNumber
    }
}

func BenchmarkStrconvFormatInt(b *testing.B) {
    b.ReportAllocs()
    for n := 0; n < b.N; n++ {
        retStr := strings.Repeat("0", 4) + strconv.FormatInt(int64(n), 16)
        hexNumber := retStr[(len(retStr) - 4):]
        _ = hexNumber
    }
}

func BenchmarkAppend(b *testing.B) {
    b.ReportAllocs()
    buf := []byte{'0', '0', '0', '0', 4 + 16: 0}
    for n := 0; n < b.N; n++ {
        buf = strconv.AppendInt(buf[:4], int64(n), 16)
        hexNumber := string(buf[len(buf)-4:])
        _ = hexNumber
    }
}

func BenchmarkIntToHex1(b *testing.B) {
    b.ReportAllocs()
    for n := 0; n < b.N; n++ {
        hexNumber := intToHex1(n)
        _ = hexNumber
    }
}

func BenchmarkIntToHex2(b *testing.B) {
    b.ReportAllocs()
    for n := 0; n < b.N; n++ {
        hexNumber := intToHex2(n)
        _ = hexNumber
    }
}

所以基准：

BenchmarkFmtSprintf-2            3000000               364 ns/op              16 B/op          2 allocs/op
BenchmarkStrconvFormatInt-2      5000000               354 ns/op              15 B/op          3 allocs/op
BenchmarkAppend-2               20000000                75.6 ns/op             0 B/op          0 allocs/op
BenchmarkIntToHex1-2            10000000               162 ns/op               8 B/op          1 allocs/op
BenchmarkIntToHex2-2             3000000               536 ns/op              16 B/op          4 allocs/op

Answer 1

strconv.AppendUint似乎比fmt.Sprintf更快。例如，

hex_test.go：

package main

import (
    "fmt"
    "strconv"
    "testing"
)

func BenchmarkFmtSprintf(b *testing.B) {
    b.ReportAllocs()
    for n := 0; n < b.N; n++ {
        hexNumber := fmt.Sprintf("%04x", n&0xFFFF)
        _ = hexNumber
    }
}

func BenchmarkAppend(b *testing.B) {
    b.ReportAllocs()
    for n := 0; n < b.N; n++ {
        buf := []byte{'0', '0', '0', 3 + 4: 0}
        buf = strconv.AppendUint(buf[:3], uint64(n)&0xFFFF, 16)
        hexNumber := string(buf[len(buf)-4:])
        _ = hexNumber // Do some stuff ....
    }
}

输出：

$ go test -bench=. hex_test.go
BenchmarkSprintf-4      10000000       116 ns/op      16 B/op     1 allocs/op
BenchmarkAppend-4       100000000       19.2 ns/op     0 B/op     0 allocs/op

Answer 2

您应该能够使用strconv，然后手动填充。这可能会减少工作量。

strconv.FormatInt(idx,16)

有效的方法将int转换为十六进制字符串，然后在Golang中用0＆＃39; s填充它？

2 个答案: