golang 进制转换_java string转integer

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发布于 2022-09-28 11:15:55

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1- You may write your conversion function (Fastest):

func String(n int32) string {
    buf := [11]byte{}
    pos := len(buf)
    i := int64(n)
    signed := i < 0
    if signed {
        i = -i
    }
    for {
        pos--
        buf[pos], i = '0'+byte(i%10), i/10
        if i == 0 {
            if signed {
                pos--
                buf[pos] = '-'
            }
            return string(buf[pos:])
        }
    }
}

2- You may use fmt.Sprint(i) (Slow) See inside:

// Sprint formats using the default formats for its operands and returns the resulting string.
// Spaces are added between operands when neither is a string.
func Sprint(a ...interface{}) string {
    p := newPrinter()
    p.doPrint(a)
    s := string(p.buf)
    p.free()
    return s
}

3- You may use strconv.Itoa(int(i)) (Fast) See inside:

// Itoa is shorthand for FormatInt(int64(i), 10).
func Itoa(i int) string {
    return FormatInt(int64(i), 10)
}

4- You may use strconv.FormatInt(int64(i), 10) (Faster) See inside:

// FormatInt returns the string representation of i in the given base,
// for 2 <= base <= 36. The result uses the lower-case letters 'a' to 'z'
// for digit values >= 10.
func FormatInt(i int64, base int) string {
    _, s := formatBits(nil, uint64(i), base, i < 0, false)
    return s
}

以上内容出自： https://stackoverflow.com/questions/39442167/convert-int32-to-string-in-golang.

本人自用代码：

func Test_conver(t *testing.T) {
	var stairName string
	s := "test886400"
	id := int32(886400)
	iDStr := strconv.FormatInt(int64(id), 10)
	if strings.Contains(s, iDStr) {
		stairName = strings.TrimSuffix(s, iDStr)
	}

	fmt.Println("stairName:", stairName)
}

效果如下：