类继承 - 添加参数而不重写父构造函数参数

Question

我想扩展以下类：

class Person {
    constructor(name, age) {
        this._name = name;
        this._age = age;
    }
  
    get name() {
        return this._name;
    }
  
    get age() {
        return this._age;
    }
}

...通过添加 sport 及其 getter：

class Athlete extends Person {
    constructor(name, age, sport) {
        super();
        this._name = name;
        this._age = age;
        this._sport = sport;
    }
  
    get sport() {
        return this._sport;
    }
}

虽然上述方法有效，但我想避免重复基父构造函数的参数。以下方法不起作用：

class Athlete extends Person {
    constructor(sport) {
        super();
        this._sport = sport;
    }
  
    get sport() {
        return this._sport;
    }
}

let athlete = new Athlete('Peter', 29, 'cricket');
console.log(athlete.name, athlete.age, athlete.sport); // name and age are not inherited, is returned 'undefined'

那么，如何在不重写基类的情况下向子类添加字段？

Answer 1

可以实现您想要的（使用destructured parameters (...args) 语法来表示任意数量的参数），但是，不建议这样做（至少我这样做的方式），并且它出现在某些情况下成本，其中一些是可读性、易于调试、不易出错的代码等等（因为这会使代码更加晦涩难懂，使得参数的顺序不那么清晰，并且可以传递错误数量的论据）：

class Person {
    constructor(name, age) {
        this._name = name;
        this._age = age;
    }
  
    get name() {
        return this._name;
    }
  
    get age() {
        return this._age;
    }
}

class Athlete extends Person {
    constructor(...args) {
        let sport = args.splice(-1)[0];
        super(...args);
        this._sport = sport;
    }
    
    get sport() {
        return this._sport;
    }
}

let athlete = new Athlete('Peter', 29, 'cricket');
console.log(athlete.name, athlete.age, athlete.sport);

同样可以使用arguments对象实现：

class Person {
    constructor(name, age) {
        this._name = name;
        this._age = age;
    }
  
    get name() {
        return this._name;
    }
  
    get age() {
        return this._age;
    }
}

class Athlete extends Person {
    constructor() {
        let args = Array.from(arguments);
        let sport = args.splice(-1)[0];
        super(...args);
        this._sport = sport;
    }
    
    get sport() {
        return this._sport;
    }
}

let athlete = new Athlete('Peter', 29, 'cricket');
console.log(athlete.name, athlete.age, athlete.sport);

Answer 2

您需要传递参数，但您可以通过将它们传递给 super() 来避免分配所有参数：

class Athlete extends Person {
    constructor(name, age, sport) {
        super(name, age); // pass arguments to super()
        this._sport = sport;
    }
  
    get sport() {
        return this._sport;
    }
}

let athlete = new Athlete('Peter', 29, 'cricket');
console.log(athlete.name, athlete.age, athlete.sport);