javascript - 绑定(bind)一个类会失去它的静态属性

Question

给定

class someClass {
  constructor (str) {
    console.log('created', str);
  }
  static someStatic() {
  }
}

我们能做到

const someBoundClass = someClass.bind(someClass, 'xxx');
const a = new someBoundClass(); // logs 'created xxx'
console.log(someClass.someStatic) // logs someStatic () {}
console.log(someBoundClass.someStatic) // logs undefined

幕后发生了什么导致我无法访问绑定(bind)类上的静态属性？有没有办法既能达到想要的绑定(bind)效果又不丢失静态方法？

Answer 1

好吧，bind 创建了一个新的函数对象，所以我不确定为什么您会期望它具有相同的静态属性。请记住，ES6 class 主要是语法糖:

function someClass(str) {
  if (!new.target) throw "constructor must be called with new";
  console.log('created', str);
}
someClass.someStatic = function() {};

var someBoundClass = someClass.bind(null, 'xxx');
console.log(someBoundClass === someClass) // false, of course

作为解决方法，您可以使用子类化:

class someBoundClass extends someClass { constructor(...args) { super('xxx', ...args); }}
const a = new someBoundClass(); // logs 'created xxx'
console.log(someClass.someStatic) // logs someStatic () {}
console.log(someBoundClass.someStatic) // logs someStatic () {}

这里的someBoundClass继承了someClass的静态属性。