How can I specify that a type is an instance of a given class type in TypeScript?

Question

I want to make a Constructor type that takes a class as a type parameter and returns the signature of its constructor.

This would be useful in defining a static property make method on a class that can create an instance of that class rather than creating it with a new ClassInstance(...) statement.

This is what I've got so far:

type ClassConstructorArgs<T> = T extends { new (...args: infer A): any }
  ? A
  : never;

type Constructor<T> = (...args: ClassConstructorArgs<T>) => T;

class Person {
  constructor(public name: string, public age: number) {}

  static make: Constructor<typeof Person> = (...args) => new Person(...args);
  //                                                     ^^^^^^^^^^^^^^^^^^^
  // Type 'Person' is missing the following properties from type 'typeof Person': prototype, make
}

The problem is that the static make has an error: Type 'Person' is missing the following properties from type 'typeof Person': prototype, make

I understand that this is because my Constructor type is wrong, constructors of class T don't return the class T itself but an instance of T.

But then I don't know how I can express that Constructor returns an instance of T rather than the class T itself. Is that even possible in TypeScript?

Aleksey L. · Accepted Answer · 2020-04-19 07:41:19Z

2

Typescript has builtin utilities both for ConstructorParameters and InstanceType:

type Constructor<T extends new (...args: any) => any> = 
  (...args: ConstructorParameters<T>) => InstanceType<T>;

class Person {
  constructor(public name: string, public age: number) {}

  static make: Constructor<typeof Person> = (...args) => new Person(...args);
}

const p = Person.make('some name', 1); // p is of type Person

Playground

If you wonder how these utilities are defined and what's wrong with your attempt, here you go:

type ConstructorParameters<T extends new (...args: any) => any> = 
  T extends new (...args: infer P) => any ? P : never;

type InstanceType<T extends new (...args: any) => any> = 
  T extends new (...args: any) => infer R ? R : any;

Source

edited Apr 19, 2020 at 7:41

answered Apr 19, 2020 at 7:30

Aleksey L.

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2 Comments

Aron Over a year ago

Interesting, I had no idea! Clever how the InstanceType is defined. Do you know why in both types the type constraints are duplicated in the parameter declaration and inside the type body? It seems redundant

Aleksey L. Over a year ago

I don't know why it is duplicated, but one could define something like type Newable = new (...args: any) => any; and reuse it

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