Extending an interface in TypeScript

Question

In JavaScript, it is straight-forwardd to add functions and members to the prototype of any type. I'm trying to achieve the same in TypeScript as follows:

interface Date
{
    Minimum: Date;
}

Date.prototype.Minimum = function () { return (new Date()); }

This produces the following error:

Type '() => Date' is not assignable to type 'Date'.
Property 'toDateString' is missing in type '() => Date'.

Considering TS is strongly-types, how could we achieve this?

Since I'm writing a custom utility library in TS, I'd rather not resort to JS.

Solution 1

You can have it like this:

interface Date
{
    Minimum(): Date;
}

(<any>Date.prototype).Minimum = function () { return (new Date()); }

let d = new Date();
console.log(d.Minimum());

Hope this helps.

Solution 2

Interfaces don't get transpiled to JS, they're just there for defining types.

You could create a new interface that would inherit from the first:

interface IExtendedDate extends Date {
    Minimum: () => Date;
}

But for the actual implementations you will need to define a class. For example:

class ExtendedDate implements IExtendedDate {
    public Minimum(): Date {
        return (new ExtendedDate());
    }
}

However note that you can do all this without an interface.

Comments

[David Sherret]

Note that the assertion to any is not necessary.

[Aaron Beall]

Upvoted. In my view this is better than messing with the prototype of a buit-in object, even though JS let's you do it.