为什么我不能将 * Struct 分配给接口？

问题描述：

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

转载于:https://stackoverflow.com/questions/13511203/why-cant-i-assign-a-struct-to-an-interface

答

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

答

This is perhaps what you meant:

package main

type Interface interface{}

type Struct struct{}

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

        _, _ = pi, ps
}

Compiles OK. See also here.

为什么我不能将 * Struct 分配给接口？

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