javascript - javascript mixins 中的构造函数和类属性

Question

我正在尝试理解 javascript 中的 mixin，以及到目前为止我读过的所有示例和文章都讨论了添加方法而不是属性。

我找到了Alex Jover Morales' article真的很有用，我稍微修改了他的示例，在 mixin 中包含了一个额外的 mixin 和具有新属性的构造函数 here 。

我在下面所做的事情是反模式吗？ mixin 中的构造函数和属性有问题吗？ 在每个 mixin 的构造函数中调用 super() 是否存在问题？

const PlayMixin = superclass => class extends superclass {

  constructor(args) {
    let { favouriteGame } = args        
    super(args);
    this.favouriteGame=favouriteGame;
  }

  play() {
    console.log(`${this.name} is playing ${this.favouriteGame}`);
  }
};


const FoodMixin = superclass => class extends superclass {

  constructor(args) {
    let { genericFood } = args        
    super(args);
    this.genericFood=genericFood;
  }

  eat() {
    console.log(`${this.name} is eating ${this.genericFood}`);
  }

  poop() {
    console.log("Going to 💩");
  }
};


class Animal {
  constructor(args) {
    let {name} = args
    this.name = name
  }
}

class Dog extends PlayMixin(FoodMixin(Animal)) {
  constructor(...args) {    
    super(...args)
}

  bark() {
    console.log("Woff woff!")
  }

  haveLunch() {
    this.eat();
    this.poop();
  }
}

const jack = new Dog({name:"Jack", genericFood:"lobster", 
favouriteGame:"chess"});
jack.haveLunch();
jack.play();

.as-console-wrapper { max-height: 100%!important; top: 0; }

Answer 1

Is what i have done below an anti-pattern?

不，不是。

Is there a problem with having a constructor and properties within a mixin?

不，只要您以适用于所有混合类的方式调用 super(...) 即可。

Is there a problem with calling super() within each mixin's contructor?

不，super始终指向扩展类，调用该构造函数没有问题。