How can I assert a function has `await`ed an async function using should.js

Question

I have an async function f that calls another async function g. To test if f calls g, I'm stubbing g using sinon and assert it's been called using should.js.

'use strict';

require('should-sinon');
const sinon = require('sinon');

class X {
    async f(n) {
        await this.g(n);
        // this.g(n); // I forget to insert `await`!
    }
    async g(n) {
        // Do something asynchronously
    }
}

describe('f', () => {
    it('should call g', async () => {
        const x = new X();
        sinon.stub(x, 'g').resolves();
        await x.f(10);
        x.g.should.be.calledWith(10);
    });
});

But this test passes even when I forget to use await when calling g in f.

One of the ways to catch this error is to make the stub return a dummy promise and check if its then is called.

it('should call g', async () => {
    const x = new X();
    const dummyPromise = {
        then: sinon.stub().yields()
    };
    sinon.stub(x, 'g').returns(dummyPromise);
    await x.f(10);
    x.g.should.be.calledWith(10);
    dummyPromise.then.should.be.called();
});

But this is a bit bothersome. Are there any convenient ways to do this?

Does `should-sinon` pull in `sinon`? Presumably you'd want to `require('sinon')` *before* `require('should-sinon')`? — James, Nov 28 '18 at 23:32

score 0 · Answer 1 · answered Nov 28 '18 at 23:52

0

Your example for f shows flawed code design which becomes more obvious if you write the same function without async/await syntax:

f(n) { return g(n).then(()=>{}); }

This achieves the same behavior - whether g resolved becomes hard to tell (assuming you don't know if f returned g's promise, which is the same as not knowing whether f awaited g). If f is not interested in the result of g it should just simply return it, not hide it. Then you can simply test for the result.

If your point is that f might have to trigger several async calls sequentially awaiting several g_1, g_2,... to resolve, then you can build a test chain by asserting in the stub of g_n+1 that the dummy-promise of g_n has been resolved. In general your approach to test a dummy-promise for its status is fine.

answered Nov 28 '18 at 23:52

B M

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I don't necessarily agree that this is flawed code design. It could be that he doesn't want the promise returned by `f` to expose the value with which `g` resolves. Your example with `then` syntax is exactly how one might do this without `async/await`. – sripberger Dec 03 '18 at 16:08
I see your point, and yes there might be cases when you don't want to return what `g` returns. You might want to call several `g_1`, `g_2`,... as I mentioned, or just one but might want to prevent access to a reference - however even in those cases I still think it's cleaner to return some specific mapping, an instance ID for example. Completly hiding the result of a proxied function while wanting testable guarantees that the proxied function ran seems a bit paradoxical. – B M Dec 03 '18 at 16:51
That's reasonable. Really it all comes down to what `g` actually does, and what of that operation is acceptable to expose through `f`. Certainly the test would be cleaner without having to track the completion status of `g` manually, as I do in my answer. – sripberger Dec 03 '18 at 17:19

sripberger · Answer 2 · 2018-12-03T17:53:44.310

Instead of stubbing then, you're best off stubbing g in such a way that it sets some boolean on the next event loop iteration. Then, you can check this boolean after calling f to make sure f waited for it:

it('should call g', async () => {
    const x = new X();
    let gFinished = false;
    sinon.stub(x, 'g').callsFake(() => {
        return new Promise((resolve) => {
            setImmediate(() => {
                gFinished = true;
                resolve();
            });
        });
    });
    await x.f(10);
    x.g.should.be.calledWith(10);
    gFinished.should.be.true();
});

Edit: Of course, this isn't a perfect guarantee because you could have f wait on any promise that waits at least as long as it takes for g to resolve. Like so:

async f(n) {
    this.g(n);
    await new Promise((resolve) => {
        setImmediate(() => {
            resolve();
        });
    });
}

This would cause the test I wrote to pass, even though it's still incorrect. So really it comes down to how strict you're trying to be with your tests. Do you want it to be literally impossible to have a false positive? Or is it ok if some obvious trickery can potentially throw it off?

In most cases I find that the latter is ok, but really that's up to you and/or your team.

How can I assert a function has `await`ed an async function using should.js

2 Answers2