原型继承：如何获取instanceof？

Question

看了这个video后，我学到了一种用Javascript实现继承的好方法：原​​型继承。此方法使用Object.create / Object.assign基于另一个实例创建新对象。这似乎非常有趣，因为它很容易理解发生了什么。例如：

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const human = {
  species: 'human',
  create: function(values) {
    const instance = Object.create(this);
    Object.assign(instance, values);
    return instance;
  },
  sayName: function() {
    console.log(`Hi, my name is ${this.name}`);
  },
  saySpecies: function() {
    console.log(`Hi, I'm a ${this.species}`);
  }
}

const musician = human.create({
  species: 'musician',
  playInstrument: function() {
    console.log(`I'm a musician and I play ${this.instrument}`);
  }
});

const aHuman = human.create({ name: 'Paul' });
aHuman.sayName();
aHuman.saySpecies();

const aMusician = musician.create({ name: 'David', instrument: 'Guitar' });
aMusician.sayName();
aMusician.saySpecies();
aMusician.playInstrument();

// how to check if musician is an instance of human?
console.log(musician instanceof human);

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使用经典继承，我可以使用instanceof来了解给定对象的链中是否有给定的 proto （例如musician instanceof Human）。

这就是 proto 链的样子，看起来与使用new运算符创建的非常相似。

问题 如何通过原型继承实现同样的目标？

Answer 1

您的代码不使用继承。创建新对象的唯一位置是这一行：

    return Object.assign({}, this, values);
//                       ^^ here

{}创建一个继承自Object.prototype的新对象。这就像做Object.create(Object.prototype)或new Object()。

您的代码中的任何内容都不会继承自human或musician。

Answer 2

您的尝试有两个主要问题：

• instanceof的RHS必须是一个函数，因此您的类型必须建立在函数之上。
• 您不区分静态/类型方法和实例方法。 create是一种类型方法。 sayName是一种实例方法。

以下内容应该按照您的预期运作。

function Human(){}

Object.assign(Human, {
  create: function(values) {
    let newObj = Object.create(this.prototype);
    return Object.assign(newObj, values);
  }
});

Object.assign(Human.prototype, {
  species: 'human',
  sayName: function() {
    console.log(`Hi, my name is ${this.name}`);
  },
  saySpecies: function() {
    console.log(`Hi, I'm a ${this.species}`);
  }
});

function Musician(){}

Object.assign(Musician, Human);

Musician.prototype = Human.create({
  species: 'musician',
  playInstrument: function() {
    console.log(`I'm a musician and I play ${this.instrument}`);
  }
});

const aHuman = Human.create({ name: 'Paul' });
aHuman.sayName();
aHuman.saySpecies();

const aMusician = Musician.create({ name: 'David', instrument: 'Guitar' });
aMusician.sayName();
aMusician.saySpecies();
aMusician.playInstrument();

// how to check if musician is an instance of human?
console.log(aMusician instanceof Human);

Answer 3

@ JLRishe的答案有效。但是如果你想避免使用函数并继续使用对象，你可以实现自己的instanceof版本。此代码段也有效：

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const human = {
  species: 'human',
  create: function(values) {
    let newObj = Object.create(this);
    return Object.assign(newObj, values);
  },
  instanceof: function(obj) {
    if (!Object.getPrototypeOf(this)) {
      return false;
    }
    let isPrototypeOf = (
      this === obj ||
      Object.getPrototypeOf(this) === obj ||
      this === Object.getPrototypeOf(obj) ||
      Object.getPrototypeOf(this) === obj.prototype
    );
    if (!isPrototypeOf) {
      isPrototypeOf = this.instanceof.call(Object.getPrototypeOf(this), obj);
    }
    return isPrototypeOf;
  },
  sayName: function() {
    console.log(`Hi, my name is ${this.name}`);
  },
  saySpecies: function() {
    console.log(`Hi, I'm a ${this.species}`);
  }
}

const musician = human.create({
  species: 'musician',
  playInstrument: function() {
    console.log(`I'm a musician and I play ${this.instrument}`);
  }
});

const aHuman = human.create({ name: 'Paul' });
aHuman.sayName();
aHuman.saySpecies();

const aMusician = musician.create({ name: 'David', instrument: 'Guitar' });
aMusician.sayName();
aMusician.saySpecies();
aMusician.playInstrument();

var test = {};

console.log(human.instanceof(human));
console.log(musician.instanceof(human));
console.log(musician.instanceof(Object));
console.log(aHuman.instanceof(human));
console.log(aMusician.instanceof(human));
console.log(aHuman.instanceof(musician));
console.log(aHuman.instanceof(Object));
console.log(aMusician.instanceof(musician));
console.log(musician.instanceof(test));
console.log(human.instanceof(test));
console.log(aMusician.instanceof(test));
console.log(aHuman.instanceof(test));

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