如何在Go中实现抽象类? [英] How to implement an abstract class in Go?

问题描述

如何在Go中实现抽象类?由于Go不允许我们在接口中包含字段，因此这将是一个无状态的对象.因此，换句话说，Go中的方法是否可以具有某种默认实现?

How to implement an abstract class in Go? As Go doesn't allow us to have fields in interfaces, that would be a stateless object. So, in other words, is it possible to have some kind of default implementation for a method in Go?

考虑一个例子:

type Daemon interface {
    start(time.Duration)
    doWork()
}

func (daemon *Daemon) start(duration time.Duration) {
    ticker := time.NewTicker(duration)

    // this will call daemon.doWork() periodically  
    go func() {
        for {
            <- ticker.C
            daemon.doWork()
        }
    }()
}

type ConcreteDaemonA struct { foo int }
type ConcreteDaemonB struct { bar int }

func (daemon *ConcreteDaemonA) doWork() {
    daemon.foo++
    fmt.Println("A: ", daemon.foo)
}

func (daemon *ConcreteDaemonB) doWork() {
    daemon.bar--
    fmt.Println("B: ", daemon.bar)
}

func main() {
    dA := new(ConcreteDaemonA)
    dB := new(ConcreteDaemonB)

    start(dA, 1 * time.Second)
    start(dB, 5 * time.Second)

    time.Sleep(100 * time.Second)
}

由于无法将接口用作接收器，因此无法编译.

This won't compile as it's not possible to use interface as a receiver.

事实上，我已经回答了我的问题(请参见下面的答案).但是，这是实现这种逻辑的惯用方式吗?除了语言的简单性之外，还有什么理由不使用默认实现?

In fact, I have already answered my question (see the answer below). However, is it an idiomatic way to implement such logic? Are there any reasons not to have a default implementation besides language's simplicity?

推荐答案

一个简单的解决方案是将daemon *Daemon移至参数列表(从而从界面中删除start(...)):

An easy solution is to move daemon *Daemon to the argument list (thus removing start(...) from the interface):

type Daemon interface {
    // start(time.Duration)
    doWork()
}

func start(daemon Daemon, duration time.Duration) { ... }

func main() {
    ...
    start(dA, 1 * time.Second)
    start(dB, 5 * time.Second)
    ...
}

这篇关于如何在Go中实现抽象类?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！