TypeScript: get type of instance from Function type

Question

I want to implement getInstance into Function.prototype How can i declare type of instance

module global
{
    declare interface Function 
    {
        getInstance():new of this; /// Can i declare it?
    }
}

Function.prototype.getInstance = function(){
    if (this.instance) return this.instance;
    return this.instance = new this();
};

Function is the own object? you want add new method to this object prototype? — Hasan Fathi
– Hasan Fathi, Commented Feb 10, 2018 at 8:31
yes, I implemented getInstance into many class as static method., I realized i can make it more easier — pmingkr
– pmingkr, Commented Feb 10, 2018 at 8:42

Titian Cernicova-Dragomir · Accepted Answer · 2018-02-10 10:34:10Z

5

If you want this to be available on all classes, and return an instance of the class you can try the following declaration:

declare global {
    interface Function 
    {
        // Available on all constructor functions with no argumnets, `T` will be inferred to the class type
        getInstance<T>(this: new ()=> T): T;
    }
}


class Person {
    name: string;
}
let p = Person.getInstance(); // p is Person
let pName = p.name // works

You might want to restrict the availability of this method, right now it will be present on all classes. You could restrict that a bit, so that it is only present if a certain static member is defined (isSingleton for example):

declare global {
    interface Function 
    {
        getInstance<T>(this: { new (...args: any[]): T, isSingleton:true }): T;
    }
}
class Person {
    static isSingleton: true;
    name: string;
    public constructor(){}
}

edited Feb 10, 2018 at 10:34

answered Feb 10, 2018 at 9:02

Titian Cernicova-Dragomir

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