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

Answer 1

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);
 });
});