将方法或函数的指针设置为nil

Question

Why both these destroy functions do not change pointer to nil and how can I create such function?

package main

import (
    "fmt"
)

type position struct {
    x int
    y int
}

func (p *position) destroy() {
    p = nil
}

func destroy(p *position) {
    p = nil
}

func main() {
    p1 := &position{1,1}
    p2 := &position{2,2}
    p1.destroy()
    destroy(p2)

    if p1 == nil {
        fmt.Println("p1 == nil")
    } else {
        fmt.Println(p1)
    }

    if p2 == nil {
        fmt.Println("p2 == nil")
    } else {
        fmt.Println(p2)
    }

}

Outputs:

&{1 1}
&{2 2}

https://play.golang.org/p/BmZjX1Hw24u

Answer 1

You need a pointer to pointer to change a pointer's value.

Here's your code sample, modified to do this (playground):

package main

import (
    "fmt"
)

type position struct {
    x int
    y int
}

func destroy(p **position) {
    *p = nil
}

func main() {
    p1 := &position{1, 1}
    destroy(&p1)

    if p1 == nil {
        fmt.Println("p1 == nil")
    } else {
        fmt.Println(p1)
    }
}

---

In your current code

func destroy(p *position) {
    p = nil
}

Inside destroy, p is a value the holds the address of a position struct. By assigning something to p itself, you're simply making it hold the address of some other position struct (or nil). You're not modifying the original pointer passed in.

This isn't different from a function trying to modify its argument by assigning to it:

// This will not actually modify the argument passed in by the caller
func setto2(value int) {
  value = 2
}

The go spec says, in the section about calls and call parameters:

After they are evaluated, the parameters of the call are passed by value to the function and the called function begins execution. The return parameters of the function are passed by value back to the calling function when the function returns.