如何在Go中将int值转换为字符串？

问题描述：

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?

  i：= 123 
s：= string（i）
  code>  pre  > 
 
  s是'E'，但我想要的是“ 123”  p> 
 
 
请告诉我如何获取“ 123”。 p> 
  
 
在Java中，我可以这样操作： p> 
 
 
 字符串s =“ ab” +“ c” // s是“ abc” 
   pre> 
 
 如何在Go中 concat  code>两个字符串？ p> 
  div>

答

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.

答

You can use fmt.Sprintf

See http://play.golang.org/p/bXb1vjYbyc for example.

答

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

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

More info - fmt

答

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)

答

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 a good choice.

答

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.

答

Converting int64:

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

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

答

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)
}

答

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 ""
    }
}

如何在Go中将int值转换为字符串？

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