访问超类上的类字段

有什么方法可以访问超类的类字段?解决方案 为什么会这样?我猜想super不能用于访问类字段是的.类字段是实例属性，但是super尝试访问超类的原型对象上的属性.您的Animal类根本没有doSomething方法-相反，每个Animal对象都有一个包含绑定函数的属性. 但是我该怎么办?如果我将其更改为传统方法，则可以使用是的，您应该做到这一点.方法和super就是这样.避免箭头功能在不需要它们时，尤其是当它们不起作用时.也可以看看类属性中的箭头函数可能不如我们想象的那么大. 有什么方法可以访问超类的类字段?是-它是一个实例属性，您可以在覆盖它之前在其构造函数中对其进行访问:class Animal { constructor() { this.doSomething = () => { return 'Hi.'; }; }}class Dog extends Animal { constructor() { super(); const superDoSomething = this.doSomething; this.doSomething = () => { return superDoSomething() + ' Woof!'; }; }}或者，使用类字段proposal而不使用显式构造函数:class Animal { doSomething = () => { return 'Hi.'; }}class Dog extends Animal { doSomething = (superDoSomething => () => { return superDoSomething() + ' Woof!'; })(this.doSomething)}I have a file with the following code:class Animal { doSomething = () => { return 'Hi.'; };}class Dog extends Animal { doSomething = () => { return super.doSomething() + ' Woof!'; };}console.log(new Dog().doSomething());Note: trying to run the snippet above probably won't work because I can't figure out how to make it use my Babal settings.Anyway, when I compile it using Babel and run it in Node, I get the following error:/Users/elias/src/classFieldTest/build/classFieldTest.js:15 return super.doSomething() + ' Woof!'; ^TypeError: (intermediate value).doSomething is not a function at Dog.doSomething (/Users/elias/src/classFieldTest/build/classFieldTest.js:15:26) at Object.<anonymous> (/Users/elias/src/classFieldTest/build/classFieldTest.js:21:23) at Module._compile (module.js:652:30) at Object.Module._extensions..js (module.js:663:10) at Module.load (module.js:565:32) at tryModuleLoad (module.js:505:12) at Function.Module._load (module.js:497:3) at Function.Module.runMain (module.js:693:10) at startup (bootstrap_node.js:188:16) at bootstrap_node.js:609:3I am using Babel 6.26.0 with the stage-2 preset, and Node 8.11.1. I can show the commands I'm using if anyone cares.Why is this happening? I am guessing that super can't be used to access a class field, but what am I supposed to do about this? If I change the doSomething method of Animal to a traditional method (doSomething() { return 'Hi.'; }), it works, but I would rather avoid traditional methods, with the way they redefine this and all the confusion it causes.Is there any way to access a class field of a superclass? 解决方案 Why is this happening? I am guessing that super can't be used to access a class fieldYes. Class fields are instance properties, but super tries to access properties on the superclass' prototype object. Your Animal class simply doesn't have a doSomething method - instead, every Animal object has a property that contains a bound function. but what am I supposed to do about this? If I change it to a traditional method, it worksYes, you are supposed to do exactly that. This is how methods and super work.Avoid arrow functions when you don't need them, and especially when they don't work. Also have a look at Arrow Functions in Class Properties Might Not Be As Great As We Think. Is there any way to access a class field of a superclass?Yes - it is an instance property, and you can access it in your constructor before overwriting it:class Animal { constructor() { this.doSomething = () => { return 'Hi.'; }; }}class Dog extends Animal { constructor() { super(); const superDoSomething = this.doSomething; this.doSomething = () => { return superDoSomething() + ' Woof!'; }; }}Alternatively, with the class fields proposal and no explicit constructor:class Animal { doSomething = () => { return 'Hi.'; }}class Dog extends Animal { doSomething = (superDoSomething => () => { return superDoSomething() + ' Woof!'; })(this.doSomething)} 这篇关于访问超类上的类字段的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！