为什么我不能将* Struct分配给*接口？ [英] Why can't I assign a *Struct to an *Interface?

问题描述

我只是通过去参观，我对指针和接口感到困惑。为什么这个Go代码不能编译？

  package main 
 
类型接口接口{} 
 $ b类型struct struct {} 
 
 func main（）{
 var ps * struct 
 var pi *接口
 pi = ps 
 
 _，_ = pi，ps 
} 
  

即如果 Struct 是一个接口，为什么不会有一个 * Struct 是一个 *接口？

我得到的错误信息是：

  prog.go：10：不能使用ps（类型* Struct）作为类型*赋值中的接口：
 *接口是指向接口的指针，而不是接口
  

解决方案

当你有一个实现接口的结构时，该结构也会自动实现该接口。这就是为什么在函数原型中永远不会有 * SomeInterface ，因为这不会在 SomeInterface 中添加任何内容，并且你在变量声明中不需要这样的类型（参见）。

一个接口值不是具体结构的值（因为它有一个可变的大小，这是不可能的），但它是一种指针（更精确的指向结构的指针和指向类型的指针）。 Russ Cox完全描述这里：

接口值表示为一个双字对，给出一个指针
给存储在接口中的类型信息和一个指向
关联数据的指针。

这就是为什么界面，而不是 *界面是正确的键入一个指向实现接口的结构体的指针。

所以你必须简单地使用

  var pi界面
  

I'm just working through the Go tour, and I'm confused about pointers and interfaces. Why doesn't this Go code compile?

package main

type Interface interface {}

type Struct struct {}

func main() {
    var ps *Struct
    var pi *Interface
    pi = ps

    _, _ = pi, ps
}

i.e. if Struct is an Interface, why wouldn't a *Struct be a *Interface?

The error message I get is:

prog.go:10: cannot use ps (type *Struct) as type *Interface in assignment:
        *Interface is pointer to interface, not interface

解决方案

When you have a struct implementing an interface, a pointer to that struct implements automatically that interface too. That's why you never have *SomeInterface in the prototype of functions, as this wouldn't add anything to SomeInterface, and you don't need such a type in variable declaration (see this related question).

An interface value isn't the value of the concrete struct (as it has a variable size, this wouldn't be possible), but it's a kind of pointer (to be more precise a pointer to the struct and a pointer to the type). Russ Cox describes it exactly here :

Interface values are represented as a two-word pair giving a pointer to information about the type stored in the interface and a pointer to the associated data.

This is why Interface, and not *Interface is the correct type to hold a pointer to a struct implementing Interface.

So you must simply use

var pi Interface

这篇关于为什么我不能将* Struct分配给*接口？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！