Extending Array in TypeScript

How to add a method to a base type, say Array? In the global module this will be recognized

interface Array {
   remove(o): Array;
}

but where to put the actual implementation?

You can use the prototype to extend Array:

interface Array<T> {
    remove(o: T): Array<T>;
}

Array.prototype.remove = function (o) {
    // code to remove "o"
    return this;
}

If you are within a module, you will need to make it clear that you are referring to the global Array<T>, not creating a local Array<T> interface within your module:

declare global {
    interface Array<T> {
        remove(o: T): Array<T>;
    }
}

declare global seems to be the ticket as of TypeScript 2.1. Note that Array.prototype is of type any[], so if you want to have your function implementation checked for consistency, best to add a generic type parameter yourself.

declare global {
  interface Array<T> {
    remove(elem: T): Array<T>;
  }
}

if (!Array.prototype.remove) {
  Array.prototype.remove = function<T>(this: T[], elem: T): T[] {
    return this.filter(e => e !== elem);
  }
}

Adding to Rikki Gibson's answer,

export{}
declare global {
    interface Array<T> {
        remove(elem: T): Array<T>;
    }
}

if (!Array.prototype.remove) {
  Array.prototype.remove = function<T>(elem: T): T[] {
      return this.filter(e => e !== elem);
  }
}

Without the export{} TS error "Augmentations for the global scope can only be directly nested in external modules or ambient module declarations."

From TypeScript 1.6, you can "natively" extend arbitrary expressions like inbuilt types.

What's new in TypeScript:

TypeScript 1.6 adds support for classes extending arbitrary expression that computes a constructor function. This means that built-in types can now be extended in class declarations.

The extends clause of a class previously required a type reference to be specified. It now accepts an expression optionally followed by a type argument list. The type of the expression must be a constructor function type with at least one construct signature that has the same number of type parameters as the number of type arguments specified in the extends clause. The return type of the matching construct signature(s) is the base type from which the class instance type inherits. Effectively, this allows both real classes and "class-like" expressions to be specified in the extends clause.

// Extend built-in types

class MyArray extends Array<number> { }
class MyError extends Error { }

// Extend computed base class

class ThingA {
    getGreeting() { return "Hello from A"; }
}

class ThingB {
    getGreeting() { return "Hello from B"; }
}

interface Greeter {
    getGreeting(): string;
}

interface GreeterConstructor {
    new (): Greeter;
}

function getGreeterBase(): GreeterConstructor {
    return Math.random() >= 0.5 ? ThingA : ThingB;
}

class Test extends getGreeterBase() {
    sayHello() {
        console.log(this.getGreeting());
    }
}

Extending Array in TypeScript

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