Golang中[]*Users和*[]Users的区别?

use*_*643 4 struct pointers go

虽然我不得不将一些数据指向一个结构体,但我对Golang 结构体[]*Users*[]UsersGolang 结构体之间的区别感到困惑

我有以下结构

type Users struct {
    ID int
    Name string
}
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hew*_*man 6

差别还是蛮大的:

*[]Users将是一个指向Users. 前任:

package main

import (
    "fmt"
)

type Users struct {
    ID int
    Name string
}

var (
    userList []Users
)

func main(){
    //Make the slice of Users
    userList = []Users{Users{ID: 43215, Name: "Billy"}}

    //Then pass the slice as a reference to some function
    myFunc(&userList);

    fmt.Println(userList) // Outputs: [{1337 Bobby}]
}


//Now the function gets a pointer *[]Users that when changed, will affect the global variable "userList"
func myFunc(input *[]Users){
    *input = []Users{Users{ID: 1337, Name: "Bobby"}}
}
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相反,[]*Users将是一片指向 的指针Users。前任:

package main

import (
    "fmt"
)

type Users struct {
    ID int
    Name string
}

var (
    user1 Users
    user2 Users
)

func main(){
    //Make a couple Users:
    user1 = Users{ID: 43215, Name: "Billy"}
    user2 = Users{ID: 84632, Name: "Bobby"}

    //Then make a list of pointers to those Users:
    var userList []*Users = []*Users{&user1, &user2}

    //Now you can change an individual Users in that list.
    //This changes the variable user2:
    *userList[1] = Users{ID:1337, Name: "Larry"}

    fmt.Println(user1) // Outputs: {43215 Billy}
    fmt.Println(user2) // Outputs: {1337 Larry}
}
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两者都使用指针,但方式完全不同。在Golang Playground 上为自己弄乱这两个片段并通读以获得更好的理解。