Is it possible to create an instance from a generic type?

March 31, 2023

I have a class called Entity like so:

class Entity {
    constructor(readonly someValue: string) {}
    someFunction() {}
}

Now I want to create a class which manages these entities, being able to create instances of them. Since I might potentially derive those entities, the entity class to be used should be a generic class argument like so:

class Manager1<T extends Entity> {
    create() {
        const instance = new T("theValue");  // ERROR: 'T' only refers to a type, but is being used as a value here.
    }
}

This causes the stated error. Alternatively I tried this (which doesn’t work either):

class Manager2<T extends new (someValue: string) => Entity> {
    create() {
        const instance = new T("theValue");  // ERROR: 'T' only refers to a type, but is being used as a value here.
    }
}

Is there a way to construct the Entity instance inside the class, given only the TYPE as a generic argument (somehow enforcing that the ctor of that type must match the specified signature) ?

>Solution :

You can’t create an imaginable thing
You have to provide the real(i.e. runtime) thing to it

class Manager1<T extends Entity> {
    create(entityConstructor: new(v:string)=>T) {
        const instance = new entityConstructor("theValue"); 
    }
    
    owned: new(v:string)=>T
    constructor(owned:new(v:string)=>T){this.owned = owned}
    createOwned() {
        const instance = new this.owned("theValue"); 
    }
}