duanli9591 2017-02-05 02:15
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在struct方法中更改struct的指针值

I am trying to wrap my head around pointer in go. I have this code right here

package main

import (
    "fmt"
)

// LinkedList type
type LinkedList struct {
    data int
    next *LinkedList
}

// InsertList will insert a item into the list
func (node *LinkedList) InsertList(data int) {
    newHead := LinkedList{data, node}
    node = &newHead
}

func main() {
    node := &LinkedList{}
    node.InsertList(4)
    fmt.Printf("node = %+v
", node)
}

and The output is

node = &{data:0 next:<nil>}

I would like to understand that why is node = &newHead my InsertList method did not reference the node pointer to a different struct at all

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3条回答 默认 最新

  • douxunwei8259 2017-02-05 02:45
    关注

    The receiver node is passed by value just like other parameters, so any changes you make in the function are not seen by the caller. If you want a function to modify something that exists outside the function, the function needs to be dealing with a pointer to that object. In your case, node is a pointer, but what you really want is a pointer to something that represents the list itself. For example:

    package main
    
    import (
        "fmt"
    )
    
    type LinkedListNode struct {
        data int
        next *LinkedListNode
    }
    
    type LinkedList struct {
        head *LinkedListNode
    }
    
    // InsertList will insert a item into the list
    func (list *LinkedList) InsertList(data int) {
        newHead := &LinkedListNode{data, list.head}
        list.head = newHead
    }
    
    func main() {
        var list LinkedList
        list.InsertList(4)
        fmt.Printf("node = %+v
    ", list.head)
        list.InsertList(7)
        fmt.Printf("node = %+v
    ", list.head)
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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