使用构造函数在相同的方式类扩展工作?

Question

为了便于理解原型，我一直在使用类。从逻辑上讲，扩展一个类可以让您从该基类继承。我将如何使用函数构造函数做下面的代码?:

class Person {
constructor(name,age,speciality="Not Specified") {
this.name = name; 
this.age = age;
this.speciality = speciality;
}
sayName() {
return this.name;
}
}
class Athlete extends Person {
constructor(name, age, speciality) {
super(name,age)
this.speciality = speciality
}
getSport() {
return "I play " + this.speciality;
}
}
let john = new Person('john', 44)
let jim = new Athlete('jim', 54, 'basketball');

我知道我应该:

function Person(name,age,speciality="Not Specified") {
this.name = name;
this.age = age;
this.speciality = speciality;
}

我假设我将sayName()作为原型，而不是Person中的属性，如:

Person.prototype.sayName = function() {
return this.name;
}

但是，我如何拥有一个继承这些的Athlete构造函数，就像我可以扩展一个类一样?

Answer 1

有了这个类，运动员的原型链是:

instance <- Athlete.prototype <- Person.prototype

对于Functions，实例的内部原型是用来构造它的函数的.prototype属性。与类一样，new Foo()的内部原型将是Foo.prototype。因此，要手动扩展它，您需要将.prototype属性设置为继承自Person.prototype的对象:

function Person(name, age, speciality = "Not Specified") {
this.name = name;
this.age = age;
this.speciality = speciality;
}
Person.prototype.sayName = function() {
return this.name;
}
function Athlete(name, age, speciality) {
Person.call(this, name, age, speciality);
this.speciality = speciality;
}
Athlete.prototype = Object.create(Person.prototype);
Athlete.prototype.getSport = function() {
return "I play " + this.speciality;
};

let jim = new Athlete('jim', 54, 'basketball');
console.log(jim.sayName());
console.log(jim.getSport());