Typescript: express function instance property

Question

Given a JS function

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

  // logic and stuff
  return arg1;

How can state be expressed in Typescript?

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

The property state doesn't exist on your function. It would exist on newly constructed objects using someMethod as a constructor (i.e. new someMethod(true)), but not on the function. — Silvio Mayolo
– Silvio Mayolo, Commented Mar 1, 2019 at 18:32

p.s.w.g · Accepted Answer · 2019-03-01 18:35:34Z

1

If you want someMethod to be a normal function that also has a property, you could simply declare it as having both a call signature and properties, like this:

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

However, if someMethod is actually meant to be used as a constructor function, I'd strongly recommend rewriting this as a class instead:

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

And then use it like this (note the new keyword):

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

edited Mar 1, 2019 at 18:35

answered Mar 1, 2019 at 18:28

p.s.w.g

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1 Comment

dook Over a year ago

I sort of chalked this up to a bad pattern in the first place. Good to have reaffirmation and thanks for the answer.

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