Referencing the constructor function

Question

Im wondering how I can get a reference to a types constructor to pass the function as a value. Basically, I would like to have a generic type registry that would allow instances to be created by calling a member function of a generic type registry instance.

For example:

class GeometryTypeInfo
{        
    constructor (public typeId: number, public typeName: string, public fnCtor: (...args: any[]) => IGeometry) {
    }
    createInstance(...args: any[]) : IGeometry { return this.fnCtor(args); }
    }
}

Later:

class Point implements IGeometry {
    constructor(public x: number, public y: number) { }

    public static type_info = new GeometryTypeInfo(1, 'POINT', Point); // <- fails
    // also fails: 
    //    new GeometryTypeInfo(1, 'POINT', new Point);
    //    new GeometryTypeInfo(1, 'POINT', Point.prototype);
    //    new GeometryTypeInfo(1, 'POINT', Point.bind(this));
}

Anyone know if it is possible to reference a classes constructor function?

standup75 · Accepted Answer · 2020-04-05 21:57:46Z

22

You can use the constructor type literal or an object type literal with a construct signature to describe the type of a constructor (see, generally, section 3.5 of the language spec). To use your example, the following should work:

interface IGeometry {
    x: number;
    y: number;
}

class GeometryTypeInfo
{        
    constructor (public typeId: number, public typeName: string, public fnCtor: new (...args: any[]) => IGeometry) {
    }
    createInstance(...args: any[]) : IGeometry { return new this.fnCtor(args); }
}

class Point implements IGeometry {
    constructor(public x: number, public y: number) { }

    public static type_info = new GeometryTypeInfo(1, 'POINT', Point);
}

Notice the constructor type literal in GeometryTypeInfo's constructor parameter list, and the new call in the implementation of createInstance.

edited Apr 5, 2020 at 21:57

standup75

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answered Oct 8, 2012 at 18:27

Brian Terlson

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1 Comment

peter Over a year ago

in // <- fails I suppose as it's the solution, it doesn't fail anymore, if that's correct maybe you can edit it to // <- doesn't fail anymore or something similar? Otherwise it's a bit misleading, thx

gaperton · Accepted Answer · 2016-06-30 19:40:11Z

10

typeof YourClass gives you constructor type which can be used in type annotations.

YourClass and this.constructor is constructor itself. So, this code compiles:

class A {}

const B : typeof A = A;

this.constructor is not recognized as value of constructor type by TypeScript (which is funny), so in situations like that you need to use some cheating casting it no any

new (<any> this.constructor)()

That's it.

answered Jun 30, 2016 at 19:40

gaperton

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6 Comments

jkjustjoshing Over a year ago

The issue with this solution is OP wants to reference the type "constructor of a type that implements an interface", and I believe you can't say typeof IGeometry when IGeometry is an interface. In the case where the passed param is a constructor for a given class or super class this is a great method.

gaperton Over a year ago

interface MyCtor { new () : MyInterface } should do the job then.

gaperton Over a year ago

But I prefer typeof Class whenever possible. It's more specific and handles inheritance chain properly.

jkjustjoshing Over a year ago

That wouldn't work in this case, since the constructor's signature is new (x: number, y: number): Point. And what happens when each class that implements it has a different constructor signature?

gaperton Over a year ago

Cause, polymorphic code you would probably write (class factory, as I understand from the context) would expect the same signature from all the subclasses' constructors.

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