Stub a function inside a function without passing it as an argument

Question

I have the following code:

const {compose} = require('./compose');


const complicatedFunction = async function (argA, argB) {
    const result = await getValue(argA)
    const value = await getValue2(argB)
    const composition = compose(result, value)
    validator(composition)
    return composition

I am calling the complicatedFunction to test the "validator" function. To test the validator function, I must stub the compose function to make it return whatever I want.

It would be easy to stub it and pass it as an argument, but it is not an option. How can stub compose to make it return any value? I know that proxyquire allows to mock dependencies but I don't understand how could I insert it in that situation

score 1 · Accepted Answer · answered Mar 20 '23 at 03:41

Try:

complicatedFunction.js:

const { compose } = require('./compose');

const complicatedFunction = (argA, argB) => {
  const result = 1;
  const value = 2;
  const composition = compose(result, value);
  return composition;
};

module.exports = complicatedFunction;

compose.js:

module.exports = {
  compose() {},
};

complicatedFunction.test.js:

const proxyquire = require('proxyquire');
const sinon = require('sinon');

describe('75761657', () => {
  it('should pass', () => {
    const composeStub = sinon.stub().withArgs(1, 2).returns(100);
    const complicatedFunction = proxyquire('./complicatedFunction', {
      './compose': {
        compose: composeStub,
      },
    });
    const actual = complicatedFunction();
    sinon.assert.match(actual, 100);
  });
});

Stub a function inside a function without passing it as an argument

1 Answers1