Infer constructor type in TypeScript

Question

Is it possible to infer the constructor type of a class in TypeScript? I tried this but it seems not to work:

type Constructor<K> = K extends { new: infer T } ? T : any;

Gawky

What about: type Constructor<K> = K extends { new: () => infer T } ? T : any; – Gilberte

class Foo {}, type CtorOfFoo = typeof Foo – Eurhythmic

Can you edit this to show a minimal reproducible example of what you're trying to accomplish? – Egin

Answer 1

There is already a predefined conditional type that allows you to extract the instance type from a class type, called InstanceType

class A { private x: any}

type AInstance = InstanceType<typeof A> // same as A

The definition of this type is:

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

Flatt

Looks interesting but unfortunately, I need it the other way around. – Gawky

@MarkusMauch not clear what you mean, you need to get the class type based on the instance type ? Don't think that information is accessible from the instance type – Flatt

Yeah, it's a known issue that a class instance doesn't have a strongly-typed constructor property. So unless you just want to get Function, you're probably stuck with new (...args: any) => T unless you manually strengthen the constructor property on all the classes you care about. – Egin

Answer 2

Instead of trying to infer it, can you refer to class types by their constructor functions like this?

type Constructor<K> = { new(): K };

const x: Constructor<String> = String; 

const s = new x();

Perfumer

Answer 3

In most cases, doing typeof ClassA is what you need. But you can also create a util type to manually infer the constructor type

type Constructor<T extends new (...args: any) => any> = T extends new (...args: infer A) => infer R ? new (...args: A) => R : neverÏ

Colp