javascript - ES6 - 使用基类的参数调用回调

Question

我试图将通用功能封装在基类中，并计划从各自的实例调用派生类特定的函数。不知何故，我没有在派生类中获取参数。不确定这里出了什么问题 -

class A {
  constructor(callback){
    this.callback = callback;
  }
  
  write(){
    console.log('Writing A');
    this.callback(10);
  }
  
}

class B extends A {
  
  constructor(){
    super(()=> this.read())
  }
  
  read(n){
    console.log('Read B: ' + n);
  }  
}

const b = new B();
b.write();

输出

"Writing A"
"Read B: undefined"

然而，我希望 -

"Writing A"
"Read B: 10"

Answer 1

你有

super(()=> this.read())

回调不带任何参数，也不向 this.read 传递任何内容 - 因此 10 会丢失。

确保回调获取所需参数并传递它。

class A {
  constructor(callback){
    this.callback = callback;
  }
  
  write(){
    console.log('Writing A');
    this.callback(10);
  }
  
}

class B extends A {
  
  constructor(){
    super((n)=> this.read(n))
  }
  
  read(n){
    console.log('Read B: ' + n);
  }  
}

const b = new B();
b.write();

更一般地说，随着传播:

class A {
  constructor(callback){
    this.callback = callback;
  }
  
  write(){
    console.log('Writing A');
    this.callback(10);
  }
  
}

class B extends A {
  
  constructor(){
    super((...args)=> this.read(...args))
  }
  
  read(n){
    console.log('Read B: ' + n);
  }  
}

const b = new B();
b.write();