Question

据我所知

super（[arguments]）关键字调用父构造函数

有这些基本课程：

class X{
    constructor(){
       return 'x'; // doesn't matter when calling super()
    }
}
class Z extends X{
    constructor(){
        console.log('This will return the newly created object invoked by new keyword', super())
    }
}
new Z; // Z {}

为什么super()会返回全新的 Z 对象？

Answer 1

您无法从构造函数中返回基本类型。

＆＃13;

class X{
    constructor(){
       // to set correct prototype
       return Object.setPrototypeOf(new String("x"), new.target.prototype); 
       // or 
       // return new String('x') just to show you could return any object
    }
}
class Z extends X{
    constructor(){
        console.log('This will return the newly created object invoked by new keyword', super())
    }
}
console.log(new Z); // Z {}

＆＃13;

为什么super（）返回es6中新创建的对象？

1 个答案: