TypeScript - Create an instance using the constructor property

Question

I want to create a function called createInstance that receives an instance a and creates a new instance c that is of the same type as a. Note that inside of createInstance I do not know what is the type of a, I only know it inherits from some class A. But I want c to be of type B, which is the real type of a. This is what I've got so far:

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

class B extends A {
}

const createInstance = (a: A): void => {
    const t = a.constructor
    const c = new t("c")
    console.log(c)
    console.log(c instanceof B)
}

const b = new B("b")
createInstance(b)

I've tried it in the typescript playground and it works, I get true for c instanceof B. But it shows a warning in the new t("c") line, that says: "This expression is not constructable. Type 'Function' has no construct signatures."

What is the correct way to do this? Thanks

You want to do this...why? How is it supposed to know what arguments to pass to the constructor? If you have A in scope to type the argument, why not just call new A directly? This doesn't really make any sense. — Jared Smith
– Jared Smith, Commented Apr 26, 2020 at 16:47
@JaredSmith The code I shared is a minimal example of the problem. It does not illustrate the usage. I have A in scope but I don't want c to just be of type A, I want it to be of the same type as b. In this case, B. But I can't know inside of createInstance which type of instance to create, I just know it should be a subclass of A and the same type as a. — Ella Sharakanski
– Ella Sharakanski, Commented Apr 26, 2020 at 19:03

leonardfactory · Accepted Answer · 2020-04-26 19:35:16Z

5

This actually is still a missing feature in TypeScript, since T.constructor is not of type T but just a plain function. You can force-cast it:

const t = a.constructor as { new(ref: string): A };

Edit: you can have the constructor already typed (parameters list) using ConstructorParameters:

const t = a.constructor as { new(...args: ConstructorParameters<typeof A>): A };

See a relative issue on TS repository#4536 and this similar question

Playground Link

edited Apr 26, 2020 at 19:35

answered Apr 26, 2020 at 17:06

leonardfactory

3,5311 gold badge21 silver badges27 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

jcalz Over a year ago

microsoft/TypeScript#3841 has more relevant discussion about this; the feature is missing because it's not clear how to add it in any useful or safe way. What stops someone from writing class C extends A { constructor(oops: number) { super(oops.toFixed()); } }? new C(123).constructor is incompatible with new A("").constructor, so assuming a.constructor is new (ref: string) => A is unsafe, and that's why it needs a type assertion

leonardfactory Over a year ago

Sure, thank you jcalz for the clear example. My answer assumes OP is knowing what he is doing

Ella Sharakanski Over a year ago

Thanks! Is there a way to get new(ref: string) out of A? I don't want to duplicate its parameter list

leonardfactory Over a year ago

Yes, I think you can do it with ConstructorParameters<A>. Updated the answer

Ella Sharakanski Over a year ago

@leonardfactoryt Thank you! How about const t = a.constructor as typeof A? I checked and it works. Since it's shorter, I guess it's better?

|

odinho - Velmont · Accepted Answer · 2022-11-30 20:56:32Z

0

This works:

const t = a.constructor as typeof A;

answered Nov 30, 2022 at 20:56

odinho - Velmont

21.7k6 gold badges44 silver badges34 bronze badges

Collectives™ on Stack Overflow

TypeScript - Create an instance using the constructor property

2 Answers 2

6 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

6 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related