douchenchepan6465 2016-09-12 01:09
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在Golang中将int32转换为字符串

I need to convert an int32 to string in Golang. Is it possible to convert int32 to string in Golang without converting to int or int64 first?

Itoa needs an int. FormatInt needs an int64.

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  • ds2128629 2016-09-12 05:46
    关注

    One line answer is fmt.Sprint(i).

    Anyway there are many conversions, even inside standard library function like fmt.Sprint(i), so you have some options (try The Go Playground):


    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
    }
    

    Comparison & Benchmark (with 50000000 iterations):

    s = String(i)                       takes:  5.5923198s
    s = String2(i)                      takes:  5.5923199s
    s = strconv.FormatInt(int64(i), 10) takes:  5.9133382s
    s = strconv.Itoa(int(i))            takes:  5.9763418s
    s = fmt.Sprint(i)                   takes: 13.5697761s
    

    Code:

    package main
    
    import (
        "fmt"
        //"strconv"
        "time"
    )
    
    func main() {
        var s string
        i := int32(-2147483648)
        t := time.Now()
        for j := 0; j < 50000000; j++ {
            s = String(i) //5.5923198s
            //s = String2(i) //5.5923199s
            //s = strconv.FormatInt(int64(i), 10) // 5.9133382s
            //s = strconv.Itoa(int(i)) //5.9763418s
            //s = fmt.Sprint(i) // 13.5697761s
        }
        fmt.Println(time.Since(t))
        fmt.Println(s)
    }
    
    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:])
            }
        }
    }
    
    func String2(n int32) string {
        buf := [11]byte{}
        pos := len(buf)
        i, q := int64(n), int64(0)
        signed := i < 0
        if signed {
            i = -i
        }
        for {
            pos--
            q = i / 10
            buf[pos], i = '0'+byte(i-10*q), q
            if i == 0 {
                if signed {
                    pos--
                    buf[pos] = '-'
                }
                return string(buf[pos:])
            }
        }
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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