How to stub a Typescript-Interface / Type-definition?

Question

I work with Typescript on an AngularJS 1.X project. I use different Javascript libraries for different purposes. To unit-test my source I would like to stub some dependencies using the Typings (= interfaces). I don't want to use the ANY-type and neither to write an empty method for each interface method.

Im looking for a way to do something like that:

let dependency = stub(IDependency);
stub(dependency.b(), () => {console.log("Hello World")});
dependency.a(); // --> Compile, do nothing, no exception
dependency.b(); // --> Compile, print "Hello World", no exception

The pain I have right now, is that I either use any and implement all methods which get called in my test case or I implement the interface and implement the full interface. That's too much useless code :(.

How can I generate an object that has an empty implementation for each method and is typed? I use Sinon for mocking purposes, but im open to use other libraries too.

PS: I know that Typescript erases the interfaces...but I still would like to solve that :).

Solution 1

I think the short answer is that this is not possible in Typescript, as the language offers no compile-time or run-time "reflection". It's not possible for a mock library to iterate the members of an interface.

See thread: https://github.com/Microsoft/TypeScript/issues/1549

This is unfortunate for TDD developers, in which mocking a dependency is a central part of the development workflow.

There are a number of techniques for quickly stubbing the methods, however, as described by the other answers. These options might do the job, with a little mental adjustment.

Edit: The Typescript Abstract Syntax Tree, AST, is a compile-time "introspection" - which could probably be used to generate mocks. However, I don't know if anyone has made a practical library.

Solution 2

I have been writing Typescript tests using qUnit and Sinon, and I have experienced exactly the same pain you are describing.

Let's assume you have a dependency on an interface like:

interface IDependency {
    a(): void;
    b(): boolean;
}

I have managed to avoid the need of additional tools/libraries by using a couple of approaches based on sinon stubs/spies and casting.

• Use an empty object literal, then directly assign sinon stubs to the functions used in the code:

  //Create empty literal as your IDependency (usually in the common "setup" method of the test file)
  let anotherDependencyStub = <IDependency>{};
  
  //Set stubs for every method used in your code 
  anotherDependencyStub.a = sandbox.stub(); //If not used, you won't need to define it here
  anotherDependencyStub.b = sandbox.stub().returns(true); //Specific behavior for the test
  
  //Exercise code and verify expectations
  dependencyStub.a();
  ok(anotherDependencyStub.b());
  sinon.assert.calledOnce(<SinonStub>anotherDependencyStub.b);
  

• Use object literal with empty implementations of the methods needed by your code, then wrap methods in sinon spies/stubs as required

  //Create dummy interface implementation with only the methods used in your code (usually in the common "setup" method of the test file)
  let dependencyStub = <IDependency>{
      a: () => { }, //If not used, you won't need to define it here
      b: () => { return false; }
  };
  
  //Set spies/stubs
  let bStub = sandbox.stub(dependencyStub, "b").returns(true);
  
  //Exercise code and verify expectations
  dependencyStub.a();
  ok(dependencyStub.b());
  sinon.assert.calledOnce(bStub);
  

They work quite nice when you combine them with sinon sandboxes and common setup/teardown like the one provided by qUnit modules.

• In the common setup you create a new sandbox and the mock object literals for your dependencies.
• In the test you just specify the spies/stubs.

Something like this (using the first option, but would work the same way if you were using the second option):

QUnit["module"]("fooModule", {
    setup: () => {
        sandbox = sinon.sandbox.create();
        dependencyMock = <IDependency>{};
    },
    teardown: () => {
        sandbox.restore();
    }
});

test("My foo test", () => {
    dependencyMock.b = sandbox.stub().returns(true);

    var myCodeUnderTest = new Bar(dependencyMock);
    var result = myCodeUnderTest.doSomething();

    equal(result, 42, "Bar.doSomething returns 42 when IDependency.b returns true");
});

I would agree this is still not the ideal solution but it works reasonably well, doesn't require extra libraries and keeps the amount of extra code needed to a low manageable level.

Solution 3

Latest TypeMoq (ver 1.0.2) supports mocking TypeScript interfaces, as long as the runtime (nodejs/browser) supports the Proxy global object introduced by ES6.

So, assuming IDependency looks like this:

interface IDependency {
    a(): number;
    b(): string;
}

then mocking it with TypeMoq would be as simple as this:

import * as TypeMoq from "typemoq";
...
let mock = TypeMoq.Mock.ofType<IDependency>();

mock.setup(x => x.b()).returns(() => "Hello World");

expect(mock.object.a()).to.eq(undefined);
expect(mock.object.b()).to.eq("Hello World");

Solution 4

From npmjs:

• https://www.npmjs.com/package/ts-mockito#mocking-interfaces

Mocking interfaces
You can mock interfaces too, just instead of passing type to mock function, set mock function generic type Mocking interfaces requires Proxy implementation

let mockedFoo:Foo = mock<FooInterface>(); // instead of mock(FooInterface)
const foo: SampleGeneric<FooInterface> = instance(mockedFoo);

ts-mockito supports mocking interfaces since version 2.4.0:

• https://github.com/NagRock/ts-mockito/releases/tag/v2.4.0

Solution 5

There are few libraries that allows to do that TypeMoq, TeddyMocks and Typescript-mockify are probably one of the more popular ones.

Check the github repositories and pick the one you like better : links:

• TeddyMocks: https://github.com/mbraude/TeddyMocks
• TypeMoq: https://github.com/florinn/typemoq
• TypeScriptMockify: https://github.com/brechtbilliet/typescript-mockify

You can also use more popular libs like Sinon, but first you have to use an <any> type and then narrow it to <IDependency> type (How do I use Sinon with Typescript?)

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Comments

[user1879408]

They all need a class to create a mock, an interface is not enough. I guess the type erasure makes it impossible, without hacking Typescript itself --> stackoverflow.com/questions/13142635/…

[PolishDeveloper]

What about creating an empty object that implements your interface ? and pass it as a object to your mock ?

[user1879408]

That does not create the methods --> Type erasure ;)

[PolishDeveloper]

Right, then only solution is to create a tool that does that :/

[Maus]

Suggesting your personal fork of typescript doesn't actually answer the question-- it's usually assumed that when people mention a language in a question, they mean an official release of that language. Hence my downvote.

[pcan]

@Maus It's your opinion. The question asks "How to stub a Typescript-Interface / Type-definition?". The answer provides a way to do this. If you read the official Typescript issues on github A LOT of people is trying to do this, but the team doesn't care at all, and does not give any means to do this in a clean way. I proved that this kind of thing is feasible: if many people ask for this feature, maybe the Typescript core team will listen to users requests.

[Maus]

I think it's impressive and important work, but i still don't think it's a great answer for this question

[James McMahon]

This is wrong, there is are few libraries that accomplish type safety, see other answers for a few examples.

[ostrumvulpes]

I've recently found @salesforce/ts-sinon to be useful for this, as it includes a stubInterface method (as well as other methods such as fromStub) which makes using Sinon in TypeScript much nicer.