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 :).

Answer 1

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.