CODEWITHSUNDEEP
javascripttypescriptconstructorinstanceinstanceof
Ella Sharakanski
Ella Sharakanski

Reputation: 2773

TypeScript - Create an instance using the constructor property

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

Upvotes: 1

Views: 880

Answers (2)

odinho - Velmont
odinho - Velmont

Reputation: 21506

This works:

const t = a.constructor as typeof A;

Upvotes: 0

leonardfactory
leonardfactory

Reputation: 3501

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

Upvotes: 5

Related Questions