Convert int32 to string in Golang

Question

I need to convert int32to stringin Golang. Is it possible to convert int32to stringin Golang without first converting to intor int64?

Itoarequired int. FormatIntrequired int64.

Answer 1

One line response fmt.Sprint(i).

In any case, there are many conversions, even inside a standard library function, such as fmt.Sprint(i), so you have some options (try Go Playground ):

---

1- You can 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- fmt.Sprint(i) ()
:

// 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 strconv.Itoa(int(i)) ()
:

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

---

4- strconv.FormatInt(int64(i), 10) ()
:

// 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
}

---

( 50000000 ):

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

:

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:])
        }
    }
}

Answer 2

Sprint .

package main

import (
     "fmt"
)

func main() {

      var sampleInt int32 = 1

      sampleString := fmt.Sprint(sampleInt)
      fmt.Printf("%+V %+V\n", sampleInt, sampleString)
}

// %!V(int32=+1) %!V(string=1)

. .

Answer 3

conversion strconv.FormatInt int32 . .

s := strconv.FormatInt(int64(n), 10)

, , , strconv.Itoa:

func formatInt32(n int32) string {
    return strconv.FormatInt(int64(n), 10)
}

All code for integer integer value code in the standard library works with values int64. Any answer to this question using the formatting code in the standard library (including the fmt package) requires conversion to int64somewhere. The only way to avoid the conversion is to write a formatting function from scratch, but there is nothing to do.