Question

我对JS还是很陌生，似乎所有事情都有这么多种方法使您感到困惑。无论如何，这就是我要尝试的方法：

//SuperClass
function Animal(name, origin, type) {
  this.name = name;
  this.origin = origin;
  this.type = type;

  //Some function that does something
  this.makeSound = function() {
    return "foo";
  }
}

function Mamal(name, origin, type, weight, height) {
  Animal.call(this, name, origin, type);
  this.weight = weight;
  this.height = height;

  this.makeSound = function() {
    return "I am a Mamal";
  }
}

所以问题是我如何在之后的Animal类中添加一个函数已经声明它，以便Mamal类（或任何子类）也继承并可以使用它。

Answer 1

如果我正确理解了您的问题，那么您实际上想要实现的是Mamal类继承了Animal类的方法？正如您在您所提的问题中将它们称为“类”一样，我提供了一个带有类而非函数的示例。

如果是这样，您可以这样声明Mamal：

class Mamal extends Animal {
    constructor(weight, height) {
        super();
        this.weight = weight;
        this.height = height;
    }

    // if you want to execute the Animal implementation of makeSound
    makeSound() {
        return super.makeSound();
    }

    // if you want to overwrite the Animal implementation of makeSound
    makeSound() {
        return "I am a mammal";
    }
}

extends关键字用于类声明或类表达式中创建一个班级作为另一个班级的子级。

Sub classing with extends Super class calls with super

更新

在此处找到原型继承的替代方法：

function Animal(name, origin, type) {
    this.name = name;
    this.origin = origin;
    this.type = type;
}

Animal.prototype.makeSound = function () {
    return "foo";
};

function Mammal(weight, height) {
    this.weight = weight;
    this.height = height;
    Animal.call(this); // call super constructor.
}

Mammal.prototype = Object.create(Animal.prototype); //inherit the Animal object through the prototype chain

Mammal.prototype.makeSound = function () {
    return "I am a mammal";
}; //Overwrite the makeSound method inherited from Animal

const cheetah = new Animal('cheetah', 'Africa', 'mammal');
cheetah.makeSound();
//foo

const mammal = new Mammal('100kg', '1m');
mammal.makeSound();
//I am a mammal

向超类添加将由子类使用的函数

1 个答案: