为什么不提升 ES6 类? [英] Why are ES6 classes not hoisted?

问题描述

因为 ES6 类只是 JavaScript 现有的基于原型继承的语法糖 [1](IMO)提升它的定义是有意义的:

Since ES6 classes are just a syntactical sugar over JavaScript's existing prototype-based inheritance [1] it would (IMO) make sense to hoist it's definition:

var foo = new Foo(1, 2); //this works

function Foo(x, y) {
   this.x = x;
   this.y = y;
}

但以下方法不起作用:

var foo = new Foo(1, 2); //ReferenceError

class Foo {
   constructor(x, y) {
      this.x = x;
      this.y = y;
   }
}

为什么不提升 ES6 类?

Why are ES6 classes not hoisted?

推荐答案

为什么不提升 ES6 类?

Why are ES6 classes not hoisted?

实际上它们是提升的(变量绑定在整个范围内可用)就像let 和 const 是 - 它们只是没有初始化.

Actually they are hoisted (the variable binding is available in the whole scope) just like let and const are - they only are not initialised.

提升其定义是有意义的

没有.在定义之前使用类从来都不是一个好主意.考虑例子

No. It's never a good idea to use a class before its definition. Consider the example

var foo = new Bar(); // this appears to work
console.log(foo.x)   // but doesn't

function Bar(x) {
    this.x = x || Bar.defaultX;
}
Bar.defaultX = 0;

并与它进行比较

var foo = new Bar(); // ReferenceError
console.log(foo.x);

class Bar {
    constructor (x = Bar.defaultX) {
        this.x = x;
    }
}
Bar.defaultX = 0;

正如您所期望的那样抛出错误.这是静态属性、原型混合、装饰器和所有东西的问题.这对于子类化也非常重要，当您使用具有未调整原型的类时，它在 ES5 中完全中断，但如果 extend ed 类尚未初始化，则现在会抛出错误.

which throws an error as you would expect. This is a problem for static properties, prototype mixins, decorators and everything. Also it is quite important for subclassing, which broke entirely in ES5 when you used a class with its non-adjusted prototype, but now throws an error if an extended class is not yet initialised.

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