Jest's runallTicks doesn't completely exhaust the micro task queue

0
votes

I am trying to learn Jest(My jest version is 26). But, I am stuck on one problem regarding the behaviour of jest.runallTicks function.

Consider the following code (I am not directly using the console.log at the beginning of each function because I am not sure whether it is truly synchronous or not, instead I push the log to an array):

const arr: String[] = [];

const A = ()=>{
    traceFunctionStart("A");
    B().then(F);
    return C();
}
const B = () => {
    traceFunctionStart("B");
    return D().then(E);
}
const C = () => {
    traceFunctionStart("C");
    return Promise.resolve();
};

const D = () => {
    traceFunctionStart("D");
    return Promise.resolve();
}

const E = () =>{
    traceFunctionStart("E");
    return Promise.resolve();
}
const X = () => {
    traceFunctionStart("X");
    return Promise.resolve();
}
const F = () => {
    traceFunctionStart("F");
    return X();
}

const traceFunctionStart = (x: string) => {
    arr.push(x);
}


test("test", async ()=>{
    jest.useFakeTimers();
    await A();
    jest.runAllTicks();
    console.log(arr);
});

Expectation:

A gets called. Which calls B which calls D. Now, D returns a resolved promise. Hence, E gets added to the micro-queue. B returns a pending promise. Hence, F does not get added to the micro queue just yet when control goes to line with B().then(F);. C gets called and returned. A is resolved and code moves to jest.runAllTicks, which forces the micro-task queue to be exhausted. The queue only contains E as of now. So, E gets called. And, it returns a resolved promise. Which means that the promise returned by B is resolved which in turn means F gets added to the microtask queue. And, then, it should get called followed by X. As a result, the order of execution(as given by the console output) must be: A->B->D->C->E->F->X.

Observed:

I observe that the output is the following:

[ 'A', 'B', 'D', 'C', 'E' ]

Where are F and X?

If I understand the documentation properly, the jest.runAllTicks() function should exhaust the micro task queue including the tasks that get added by these tasks themselves. Hence, F and X should be there.

jestjs