How can I stub the following function in sinon?

Question

My test function is as follows:

const testLibrary = require("./test");

describe("Top Test", function() {

    it("should test function", function(done) {
        testLibrary.print();
        done();

    });


});

test.ts has the following two functions:

export function stubMe() {
    console.log("original function");
}

export function print() {
    stubMe();
}

When I run the test, it prints: 'original function'

I now try to stub my test as follows:

const testLibrary = require("./test");


const sinon = require("sinon");

const stubMe = sinon.stub(testLibrary, "stubMe");
stubMe.yields();

describe("Top Test", function() {

    it("should test function", function(done) {
        testLibrary.print();
        done();

    });


});

My test function still prints 'original function' indicating that the function hasn't been stubbed.

How do I stub the stubMe function?

Update:

I have modified my code based on Ankit's solution below to:

  const sinon = require("sinon");
  const testLibrary = require("./test");

  testLibrary.stubMe();

  const stubMe = new sinon.stub(testLibrary, "stubMe").callsFake(function () {
    console.log("console from stub");
  });

  testLibrary.stubMe();


  describe("Top Test", function () {

    it("should test function", function (done) {
      testLibrary.print();
      done();

    });

  });

Oddly, this prints:

original function
console from stub


  Top Test
original function

Why does the stub revert during the test?

Because in one case you're calling `testLibrary.stubMe()` and in another case you're calling `stubMe()`. This is what the answer says. — Estus Flask, Jun 29 '18 at 12:10

Ankit Agarwal · Answer 1 · 2018-06-29T11:51:13.567

0

Use callsFake() instead of yields() like:

 const testLibrary = require('./test');

  const sinon = require('sinon');

  const stubMe = new sinon.stub(testLibrary, 'stubMe');
  stubMe.callsFake(function () {
    console.log('console from stub');
  });

  describe('Top Test', function () {

    it('should test function', function (done) {
      testLibrary.print();
      done();

    });

  });

Here is the output for the above test:

edited Jun 29 '18 at 11:51

answered Jun 29 '18 at 11:39

Ankit Agarwal

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It's still printing 'original function' – Hoa Jun 29 '18 at 11:42
@Hoa it works for me in my local. What is the version of sinon? – Ankit Agarwal Jun 29 '18 at 11:50
6.0.1 - what version are you using? – Hoa Jun 29 '18 at 11:58
I am using `"sinon": "4.2.1"` @Hoa – Ankit Agarwal Jun 29 '18 at 11:59
Getting a strange result - see main edited question. Any idea why that behaviour is occurring? – Hoa Jun 29 '18 at 12:03
You won't physically be able to stub stubMe in a situation like this one. You didn't list `test` file, but it should differ from the OP. – Estus Flask Jun 29 '18 at 12:09
@Hoa you are also calling `testLibrary.stubMe();` in your test file before the test actually executes. That is why you get that first `console.log` in the console as `original function` – Ankit Agarwal Jun 29 '18 at 12:10

score 0 · Answer 2 · answered Jun 29 '18 at 11:50

There is no way how the function can be spied or mocked. stubMe is referred directly in same module.

As explained in related Jest question, a way to improve testability is to move functions to different modules:

// one module
export function stubMe() {
    console.log("original function");
}

// another module
import { stubMe } from '...';
export function print() {
    stubMe();
}

In this case testLibrary.stubMe can be spied or mocked, but only if non-native ES modules are in use, because ES module exports are read-only in native implementation.

How can I stub the following function in sinon?

2 Answers2