How to convert an int value to string in Go?

Question

i := 123
s := string(i) 

s is 'E', but what I want is "123"

Please tell me how can I get "123".

And in Java, I can do in this way:

String s = "ab" + "c"  // s is "abc"

how can I concat two strings in Go?

Answer 1

Use the strconv package's Itoa function.

For example:

package main

import (
    "strconv"
    "fmt"
)

func main() {
    t := strconv.Itoa(123)
    fmt.Println(t)
}

You can concat strings simply by +'ing them, or by using the Join function of the strings package.

Answer 2

Converting int64:

n := int64(32)
str := strconv.FormatInt(n, 10)

fmt.Println(str)
// Prints "32"

Answer 3

Another option:

package main
import "fmt"

func main() {
   n := 123
   s := fmt.Sprint(n)
   fmt.Println(s == "123")
}

https://golang.org/pkg/fmt#Sprint

Answer 4

It is interesting to note that strconv.Itoa is shorthand for

func FormatInt(i int64, base int) string

with base 10

For Example:

strconv.Itoa(123)

is equivalent to

strconv.FormatInt(int64(123), 10)

Answer 5

You can use fmt.Sprintf or strconv.FormatFloat

For example

package main

import (
    "fmt"
)

func main() {
    val := 14.7
    s := fmt.Sprintf("%f", val)
    fmt.Println(s)
}

Answer 6

fmt.Sprintf("%v",value);

If you know the specific type of value use the corresponding formatter for example %d for int

More info - fmt

Answer 7

ok,most of them have shown you something good. Let'me give you this:

// ToString Change arg to string
func ToString(arg interface{}, timeFormat ...string) string {
    if len(timeFormat) > 1 {
        log.SetFlags(log.Llongfile | log.LstdFlags)
        log.Println(errors.New(fmt.Sprintf("timeFormat's length should be one")))
    }
    var tmp = reflect.Indirect(reflect.ValueOf(arg)).Interface()
    switch v := tmp.(type) {
    case int:
        return strconv.Itoa(v)
    case int8:
        return strconv.FormatInt(int64(v), 10)
    case int16:
        return strconv.FormatInt(int64(v), 10)
    case int32:
        return strconv.FormatInt(int64(v), 10)
    case int64:
        return strconv.FormatInt(v, 10)
    case string:
        return v
    case float32:
        return strconv.FormatFloat(float64(v), 'f', -1, 32)
    case float64:
        return strconv.FormatFloat(v, 'f', -1, 64)
    case time.Time:
        if len(timeFormat) == 1 {
            return v.Format(timeFormat[0])
        }
        return v.Format("2006-01-02 15:04:05")
    case jsoncrack.Time:
        if len(timeFormat) == 1 {
            return v.Time().Format(timeFormat[0])
        }
        return v.Time().Format("2006-01-02 15:04:05")
    case fmt.Stringer:
        return v.String()
    case reflect.Value:
        return ToString(v.Interface(), timeFormat...)
    default:
        return ""
    }
}

Answer 8

package main

import (
    "fmt" 
    "strconv"
)

func main(){
//First question: how to get int string?

    intValue := 123
    // keeping it in separate variable : 
    strValue := strconv.Itoa(intValue) 
    fmt.Println(strValue)

//Second question: how to concat two strings?

    firstStr := "ab"
    secondStr := "c"
    s := firstStr + secondStr
    fmt.Println(s)
}

Answer 9

In this case both strconv and fmt.Sprintf do the same job but using the strconv package's Itoa function is the best choice, because fmt.Sprintf allocate one more object during conversion.

check the benchmark here: https://gist.github.com/evalphobia/caee1602969a640a4530

see https://play.golang.org/p/hlaz_rMa0D for example.

Answer 10

fmt.Sprintf, strconv.Itoa and strconv.FormatInt will do the job. But Sprintf will use the package reflect, and it will allocate one more object, so it's not an efficient choice.