doushi4864 2017-08-04 15:40 采纳率: 100%
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我不能改变这个结构,我也不明白为什么

Consider the following gist linked here:

Code:

package main

import (
    "fmt"
)

type StateTransition struct {
    msg Message
}

type Message interface {
    To() *string
}

type Transaction struct {
    data txdata
}

type txdata struct {
    Recipient *string
}

func (t Transaction) To() (*string) {
    return t.data.Recipient
}


func UnMask(n **string, k string) {
    *n = &k
} 

func main() {
    toField := "Bob"
    toPtr := &toField
    txd := txdata{toPtr}
    tx := Transaction{txd}
    st := StateTransition{tx}
    n1 := st.msg.To()
    fmt.Printf("Hello, %s 
", *n1)
    UnMask(&n1, "Joe")
    fmt.Printf("Hello, %s 
", *n1)
    n2 := st.msg.To()
    fmt.Printf("Hello, %s 
", *n2)
}

Output

Hello, Bob 
Hello, Joe 
Hello, Bob 

Expected Output

Hello, Bob
Hello, Joe
Hello, Joe

The result is the sequence "Bob, Joe, Bob" is printed whereas my intuition says that it should be "Bob, Joe, Joe" (this is also what i want it to print). Can someone experienced in go please explain to me enough about combining pointers, structs, and interfaces as they relate to this problem to give me some understanding about why I'm wrong, and how to fix it?

  • 写回答

1条回答 默认 最新

  • dtvq4978 2017-08-04 15:59
    关注

    Unmask takes a pointer to a pointer, let's say pointer X to pointer Y, pointer Y points to the string value. Unmask then changes the pointer to which X is pointing to, Y is unchanged and points still to the same old string.

    You can do this:

    func UnMask(n **string, k string) {
        **n = k
    }
    

    or

    func UnMask(n *string, k string) {
        *n = k
    }
    // ....
    UnMask(n1, "Joe") // drop the '&'
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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