Javascript:为什么在子类中声明属性会覆盖父类中与null相同的属性 [英] Javascript: Why does declaring a property in a subclass overwrite the same property in a super class as null

问题描述

我的问题与理解类如何为属性分配值以及如何在Javascript中实例化对象有关.我想了解更多此过程的工作原理.

My question has to do with understanding how classes assign values to properties and how objects are instantiated in Javascript. I would like to understand more how this process works.

如果我创建两个类，其中第二个类继承自第一个类

If I create two classes where the second inherits from the first

class A {
    name

    constructor(name){
        this.name = name
    }
}

class B extends A {
    name
    status

    constructor(name, status){
        super(name)
        this.status = status
    }
}

然后在我将其打印到控制台时创建类B的实例

And then create an instance of class B, when I print it to the console

x = new B('myClass', true)
console.log(x)

它显示名称变量未定义

B { name: undefined, status: true }

我很确定B中的name属性会覆盖A中的name属性，但是为什么A构造函数不将新的name变量分配为传递给它的值?

I'm pretty sure that the name property in B is over-writing the name property in A, but why doesn't the A constructor assign the new name variable to be the value passed into it?

推荐答案

目前(2020年10月)，这是正确的行为.

This is correct behaviour for the moment (October 2020).

设置时

class A {
    name
}

这将声明一个 class字段.它还不是标准，它是第3阶段的提案.它可能会改变，但变化不大.第三阶段是候选阶段，可能包括精修阶段.

This declares a class field. It's not a standard yet, it's a proposal in stage 3. It's likely to change but not by much. Stage 3 is the candidate stage and might include finishing refinements.

无论如何，根据提案的当前规范，您所看到的都是正确的.没有初始化程序的所有类字段都设置为 undefined .发生这种情况是因为您在 B 中有另一个名为 name 的类字段，而没有初始化程序.在父构造函数中发生的分配将被覆盖.

At any rate, what you are seeing is correct according to the current spec of the proposal. Any class fields without initialisers are set to undefined. This happens because you have another class field called name in B without an initialiser. The assignment that happens in the parent constructor will be overwritten.

没有初始化程序的字段设置为 undefined

无论是否存在初始化程序，公共和私有字段声明都会在实例中创建一个字段.如果没有初始化程序，则将该字段设置为 undefined .这与某些transpiler实现有所不同，后者只会完全忽略没有初始化程序的字段声明.

Fields without initializers are set to undefined

Both public and private field declarations create a field in the instance, whether or not there's an initializer present. If there's no initializer, the field is set to undefined. This differs a bit from certain transpiler implementations, which would just entirely ignore a field declaration which has no initializer.

例如，在下面的示例中， new D 将导致其 y 属性为 undefined 而不是 1的对象.

For example, in the following example, new D would result in an object whose y property is undefined, not 1.

class C {
  y = 1; 
} 
class D extends C {
  y; 
}

将没有初始化程序的字段设置为 undefined 而不是擦除字段的语义是，字段声明为确保属性存在于所创建的对象上提供了可靠的基础.这有助于程序员将对象保持在相同的一般状态，这可以使推理变得容易，有时甚至可以使实现更加优化.

The semantics of setting fields without initializers to undefined as opposed to erasing them is that field declarations give a reliable basis to ensure that properties are present on objects that are created. This helps programmers keep objects in the same general state, which can make it easy to reason about and, sometimes, more optimizable in implementations.

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