Typescript: Instance of class type

Question

I have the type of a class like here as A_Type?

class A {
  constructor(public a: string) {}
}

type A_Type = {new (a: string): A}

And Id like to get the type of the instance of the A_Type constructor, so that I could type this function explicitly

class A {
  constructor(public a: number) {}
}
//                                              ** not working **
function f<W extends { new(a: number): A }>(e: W): instanceof W {
  return new e(2)
}

let w = f(A)

I am basically searching for the reverse operation of typeof in typescript.

How do you intend to extend it? Pass other classes to f?

Lesiak
– Lesiak

2020-05-05 22:06:54 +00:00
Commented May 5, 2020 at 22:06 — Lesiak
– Lesiak, Commented May 5, 2020 at 22:06

Aron · Accepted Answer · 2020-05-06 02:04:02Z

6

You can use the built in InstanceType<T> utility to do this

class A {
    constructor(public a: number) { }
}

// declare that the return value of the class constructor returns an instance of that class
function f<W extends { new(a: number): InstanceType<W> }>(e: W) {
    return new e(2) // return type is automatically inferred to be InstanceType<W>
}

let w = f(A) // w: A

Playground link

answered May 6, 2020 at 2:04

Aron

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1 Comment

Huan Over a year ago

InstanceType<T> does not work with private constructor() because it can not infer the new () => R. Is there a way to get a reverse operation of typeof with a class withprivate constructor()?

satanTime · Accepted Answer · 2020-05-05 22:07:12Z

1

function f<W extends { new(a: number): any }, R extends W extends { new(a: number): infer U } ? U : never>(e: W): R {
  return new e(2)
}

you need to use generics to extract the return type.

But if you always expect an instance of A you can simplify it to

function f<W extends { new(a: number): A }>(e: W): A {
  return new e(2)
}

edited May 5, 2020 at 22:07

answered May 5, 2020 at 22:02

satanTime

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6 Comments

Lesiak Over a year ago

this infers w as unknown

satanTime Over a year ago

true, fixed, please verify.

Lesiak Over a year ago

Upvote, indeed it works (no idea how though). Can you explain in few words?

satanTime Over a year ago

infer extracts nested types, so it says R is CONDITION ? VALUE_TRUE: VALUE_FALSE and condition is W is a constructor, then we return U otherwise never what makes R to be U.

satanTime Over a year ago

A extends W extends {new (a: infer ARG): any} ? ARG : never - now A is number.

|

Lesiak · Accepted Answer · 2020-05-05 22:09:59Z

1

I would go with the following:

type Constructor<X> = {new (a: number): X}

function f<X>(e: Constructor<X>): X {
  return new e(2)
}

const w = f(A)

answered May 5, 2020 at 22:09

Lesiak

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