我什么时候需要从构造函数中调用`super`？ [英] When do I need to call `super` from a constructor?

问题描述

阅读博士Axel Rauschmayer关于ES6课程的博客，我知道派生类在没有提供时会有以下默认构造函数

Reading Dr. Axel Rauschmayer's blog on ES6 classes, I understand that a derived class has the following default constructor when none is provided

constructor(...args) {
    super(...args);
}

我也明白，如果我想使用这个在构造函数中我首先需要调用 super ，否则此将不会被初始化（抛出一个ReferenceError）。

I also understand that if I want to use this within a constructor I first need to call super, otherwise this will not yet be initialized (throwing a ReferenceError).

constructor(width, height) {
    this.width = width;  // ReferenceError
    super(width, height);
    this.height = height; // no error thrown
    ...
}

以下假设然后纠正？ （如果没有，请你解释我应该明确调用 super 的条件）

Is the following assumption then correct? (and if not, could you please explain the conditions under which I should explicitly call super)

对于派生类，我只需要明确地调用 super ...

For derived classes, I only need to explicitly call super when...

1. 我需要从构造函数中访问 this


2. 超类构造函数需要不同的参数，然后是派生类构造函数

还有其他时候我应该包含对超类构造函数的调用吗？

Are there other times when I should include a call to the superclass constructor?

推荐答案

是的，这听起来很正确，虽然有点奇怪。规则应该是

Yes, that sounds correct, albeit a bit oddly formulated. The rules should be

• 在派生类中，总是 ¹ 需要调用 super（...）构造函数


• 如果你做的不是默认构造函数，你可以省略整个 constructor（）{} ，
  这反过来会使你的类代码不包含超级调用。

• In a derived class, you always¹ need to call the super(…) constructor
• If you are not doing more than the default constructor, you can omit the whole constructor(){}, which in turn will make your class code not contain a super call.

_{1：你不需要在明确的返回一个对象的可疑边缘情况下调用它，你几乎不会。}

_{1: You don't need to call it in the suspicious edge case of explicitly returning an object, which you hardly ever would.}

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