How to add prototype functions in "TypeScript style" to external modules?

Question

I want to modify the prototype of the nodejs module module.

@types/node has an interface defined for this (NodeModule) but don't has anything when I import the module, so I have to require it.

import * as Module from "module"; // Error
var Module = require("module"); // Ok

The problem is that Module don't has any definition. Is returned as any.

I have this

export interface CustomNodeModule extends NodeModule {
    __thingy:()=>void;
}

Module.prototype.__thingy = function() {
    // things!
}

I have to create my own CustomNodeModule interface, but, how can I modify the Module prototype and also those functions know that they are part of a CustomNodeModule instance?

Daniel Rosenwasser · Accepted Answer · 2016-11-14 06:07:00Z

2

In some global file (one with no imports/exports), add the following:

declare module "module" {
    export = Module;

    var Module: ModuleConstructor;

    interface ModuleConstructor {
        new (id: any, parent: any): NodeModule;
        readonly prototype: NodeModule;

        Module: typeof Module;
        globalPaths: string[];
        /** @deprecated */
        requireRepl(...args: string[]): any;
        runMain(): void;
        wrap(script: any): any;
        wrapper: string[];
    }
}

Note: sending a pull request to DefinitelyTyped with this would probably be accepted.

Then, in the module where you actually tack a new property on, you can write:

import m = require("module");

// This is a *global augmentation*.
// It adds to declarations in the global scope.
declare global {
    interface NodeModule {
        __thingy: () => void;
    }
}

m.prototype.__thingy = function() {
    // ...
};

answered Nov 14, 2016 at 6:07

Daniel Rosenwasser

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4 Comments

Jorge Fuentes González Over a year ago

So I need to create my own module. Thank you! +1 for the This is a *global augmentation*.. Was important here.

Jorge Fuentes González Over a year ago

Just noticed that the problem still persists. __thingy() function still don't knows that this is a NodeModule.

Daniel Rosenwasser Over a year ago

You can add a this type annotation in the type of __thingy (i.e. __thingy: (this: NodeModule) => void) TypeScript should be able to infer the type of this in the function body itself afterwards. Read more here, and turn on --noImplicitThis to be safer.

Daniel Rosenwasser Over a year ago

You're welcome! For the record, you shouldn't even need to add the annotation on the function expression, but it's definitely a style preference thing.

toskv · Accepted Answer · 2016-11-12 11:41:42Z

2

Since TypeScript interfaces are open ended you shouldn't have to need to create another interface, you can just extend the existing one with your method.

export interface NodeModule {
    __thingy:()=>void;
}

answered Nov 12, 2016 at 11:41

toskv

31.8k7 gold badges77 silver badges77 bronze badges

6 Comments

Jorge Fuentes González Over a year ago

Oh, that's true. Now I remember that I've read about that some months ago when I started with TypeScript :-/ +1 for that

Jorge Fuentes González Over a year ago

Now the important question, how can I overwrite Module prototype so they know that the this instance is a NodeModule?

Jorge Fuentes González Over a year ago

Oh yeah, now I remember why I've done this. When I don't do extend the interface, NodeModule is not mixed (maybe because comes from a @types, I don't know), so I created a custom interface. Was not a big deal. Check i.imgur.com/VtQHOyb.png and i.imgur.com/ojd2APj.png

toskv Over a year ago

@JorgeFuentesGonzález if you have a better solution for your particular case please post it and mark it as the correct answer, so people will know. :)

Jorge Fuentes González Over a year ago

I still don't have a solution. I was just telling why I created an interface extending NodeModules instead of directly adding parameters to it. Going to ask in GitHub as the question here don't has much visits. Thank you for your time :-)

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