soywiz
soywiz

Reputation: 438

Typescript 0.9: module functions

It seems that you can't do this anymore in TypeScript 0.9:

declare module "mymodule" {
    export function(): any;
}

Is there a way of creating typings that allow to call the exported module in 0.9?

Notice that the exported function doesn't have a name. That's the way to declare callable modules in previous versions of typescript, because there are a lot of node modules exporting just a function.

Upvotes: 5

Views: 8316

Answers (4)

Dale Anderson
Dale Anderson

Reputation: 1711

@Weston was close, I found you also need to add an internal module of the same name as the function:

declare module "mymodule" {
    function placeholder(): any;
    module placeholder {}
    export = placeholder;
}

Upvotes: 4

Weston
Weston

Reputation: 2002

It seems that you can create a my-module.d.ts file like this:

declare module "my-module" {
    function placeholder(arg1: any): void;
    export = placeholder;
}

This will let you consume the module in your index.ts file:

/// <reference path="./my-module.d.ts" />
import myModule = require("my-module");
myModule("Test Arg");

Very non-intuitive IMO.

Edit: another pitfall that confused me was the Ambient External Modules section, which makes it sound like the declare module "my-module" wrapper could be omitted in this case. Does anyone know if that's possible?

Upvotes: 3

danatcofo
danatcofo

Reputation: 733

I have an example of how I'm doing it.

https://gist.github.com/danatcofo/9116918

/*
 * This is an example of how to turn your typescript modules into functions. 
 */
function Example(command: string, ...params: any[]): void;
function Example(command: string, ...params: any[]): any {
  var isConstructor = false;
  if (this instanceof Example && !this.__previouslyConstructedByExample) {
    isConstructor = true;
    this.__previouslyConstructedByExample = true;
  }
  switch (typeof (Example[command])) {
    case "function":
      if (isConstructor) {
        return (function(cls, args){
          function F(): void {
            return cls.apply(this, args);
          }
          F.prototype = cls.prototype;
          return new F();
        })(Example[command], params);           
      } 
      return Example[command].apply(Example, params);
    case "undefined": throw "unknown command call";
    default: 
      if (isConstructor) throw "unknown command call";
      return Example[command];
    }
}

module Example {
  export function Func0(parm1:string): string {
    var ret = "Func0 was called: parm1 = " + parm1;
    console.debug(ret);
    return ret;
  }
  export function Func1(parm1:string, parm2: string): string {
    var ret = "Func1 was called: parm1 = " + parm1 + ", parm2 = " + parm2;
    console.debug(ret);
    return ret;
  }

  export class Test {
    public ret: string;
    constructor(parm1: string, parm2: string){
      this.ret = Func1(parm1, parm2);
    }
  }
}

var func0 = Example.Func0("hello world");
var func0_fn = <any>Example("Func0", "hello world");
console.assert(func0 == func0_fn, "single param example")


var func1 = Example.Func1("hello", "world");
var func1_fn = <any>Example("Func1", "hello", "world");
console.assert(func1 == func1_fn, "multi param example")

var test = new Example.Test("hello", "world");
var test_fn = new Example("Test", "hello", "world");

console.assert(test instanceof Example.Test, "class example");
console.assert(test_fn instanceof Example.Test, "function class example");

Upvotes: -1

Jude Fisher
Jude Fisher

Reputation: 11284

The change appears deliberate, and there is no longer a way to do that:

The ‘module’ keyword no longer creates a type

Description: In 0.9.0, a clearer distinction is made between roles of namespaces, types, and values. Modules now contribute only to namespaces and values and will no longer contribute a type. 
Reason: This simplification of the role of modules allow modules to now extend classes and functions more easily.

From: http://blogs.msdn.com/b/typescript/archive/2013/04/22/announcing-0-9-early-previews.aspx

Upvotes: 6

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