dongwei3120
dongwei3120
2016-04-19 14:02
浏览 140
已采纳

Go中的不可变字符串

is anyone able to explain me why the address of &c1.name is the same after being changed in function changeMe(). I thought strings are immutable in Go.

package main

import "fmt"

type customer struct {
    name string
    age  int
}

func main() {
    c1 := customer{"Todd", 44}
    fmt.Println(&c1.name) // 0x8201e4120

    changeMe(&c1)

    fmt.Println(c1)       // {Rocky 44}
    fmt.Println(&c1.name) // 0x8201e4120
}

func changeMe(z *customer) {
    fmt.Println(z)       // &{Todd 44}
    fmt.Println(&z.name) // 0x8201e4120
    z.name = "Rocky"
    fmt.Println(z)       // &{Rocky 44}
    fmt.Println(&z.name) // 0x8201e4120
}
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1条回答 默认 最新

  • douzhuan1467
    douzhuan1467 2016-04-19 14:31
    已采纳

    The immutability of strings is not the same as immutability of variables.

    Immutability of strings means that the characters in the string cannot be changed. This holds true for Go. Go makes use of it when slicing strings as shown in the example below.

    Variables in Go are always mutable. When a string variable is changed, the internal fields of the variable (pointer and length) are changed. The address of variable never changes.

    The example below presents the internals of Go string variable. The first integer is an address to the array of characters and the second is the length.

    See the article on internals of string in Go http://research.swtch.com/godata.

    package main
    
    import (
            "fmt"
            "reflect"
            "unsafe"
    )
    
    func main() {
        var x string = "abc"
        fmt.Println(x, &x, (*reflect.StringHeader)(unsafe.Pointer(&x)))
        x = "cde"
        fmt.Println(x, &x, (*reflect.StringHeader)(unsafe.Pointer(&x)))
        x = x[1:]
        fmt.Println(x, &x, (*reflect.StringHeader)(unsafe.Pointer(&x)))
    }
    

    Playground

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