在Go中调用嵌入式类型的重载方法的正确方法 [英] Proper way to call overloaded method of embedded type in Go

问题描述

我有一个界面:

 package pkg
 type BaseInterface interface {
     func Nifty() bool
     func Other1() 
     func Other2()
     ...
     func Other34123()
 }

和实现它的结构:

 package pkg
 type Impl struct {}
 func (Impl) Nifty() bool { ... }

然后是另一个结构，该结构要嵌入第一个结构并自己做Nifty():

Then along comes another struct which wants to embed the first and do it's own Nifty():

 package myOtherPackage
 import "pkg"
 type ImplToo struct {
     *pkg.Impl
 }
 func (it ImplToo) Nifty() bool { ... something else ... }

这有点像类继承，在OOP语言中有方法重写.我想知道如何做相当于implToo.super().Nifty()的方法-也就是说，从ImplToo Nifty()实现中，调用pkg.Impl Nifty()实现.

This is sort of like class inheretance with method override in an OOP language. I want to know how to do the equivalent of implToo.super().Nifty() - that is, from the ImplToo Nifty() implementation, call the pkg.Impl Nifty() implementation.

要在 it 上使用的正确转换是什么，以便我可以完成此转换?我尝试的所有操作都会对ImplToo的Nifty()产生无限制的递归，或者会产生一些编译器错误，例如:

What is the proper conversion to use on it so that I can accomplish this? Everything I try yields either unbounded recursion on ImplToo's Nifty(), or some compiler error such as:

无效的类型断言:(& it).(BaseInterface)(非接口类型，*它在左侧)

...或对此有很多变化.

... or many variations on that.

推荐答案

您在寻找；

 type ImplToo struct {
     pkg.Impl
 }

func (it ImplToo) Nifty() bool { return it.Impl.Nifty() }

您使用的指针不一致，可能是问题的一部分(不是肯定的).如果要使嵌入式类型成为指针，则使您的方法也使接收类型成为指针，以免发生此问题.

Your use of pointers isn't consistent which is probably (not positive) part of your problem. If you want to make the embedded type a pointer then make your methods receiving type a pointer as well to avoid this problem.

如果要在嵌入式类型中显式使用方法，则可以使用通常具有属性名称的类型来引用该方法.

If you want to explicitly use a method in the embedded type you reference it using the type where you would normally have a property name.

这篇关于在Go中调用嵌入式类型的重载方法的正确方法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！