绑定一个类会失去它的静态属性

Question

Given

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

we can do

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

What's happening behind the scenes causing me to be unable to access static properties on a bound class? Is there any way to achieve the desired bound effect without losing the static methods?

Answer 1

Well bind creates a new function object, so I'm not sure why you would expect it to have the same static properties. Remember that ES6 class is mostly syntactic sugar:

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

As a workaround, you might be able to use subclassing:

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 () {}

The someBoundClass here inherits the static properties from someClass .

Answer 2

Take a look into Javascript static vs instance, prototype keyword

There is a difference between setting a property to an object, what is done when using static and adding a method to each instance of the object, which can be done by using someClass.prototype.someMethod or in your case by just removing static .

That means undefined would also occur, if you try to call someStatic on a instance of someClass , which was instantiated directly by new someClass("xxx") .

For another approach take a look at the answer of Bergi.