javascript - Typescript:表达函数实例属性

Question

给定一个 JS 函数

function someMethod (arg1: boolean) {
  this.state = { };

  // logic and stuff
  return arg1;

如何在 Typescript 中表达状态？

someMethod(true);
someMethod.state; // Property 'state' does not exist on type '(arg1: boolean) => boolean'

Answer 1

如果您希望 someMethod 成为一个也具有属性的普通函数，您可以简单地将其声明为同时具有调用签名和属性，如下所示:

declare const someMethod: {
  (arg1: boolean): boolean;
  state: any; // or whatever type you intend someMethod.state to have
}

但是，如果 someMethod 实际上要用作构造函数，我强烈建议将其重写为 class相反:

declare class someClass {
  constructor(arg1: boolean);
  state: any;
}

然后像这样使用它(注意new关键字):

const instance = new someClass(true);
instance.state = {};