锡诺间谍的功能表达

Question

is it possible to make sinon spy on function expressions? Look this code for example.

 function one() { return 1; } function two() { return 2; } function three() { return 3; } function myMethod() { var n1 = one(); var n2 = two(); var n3 = three(); return n1 + n2 + n3; } QUnit.module('My test'); QUnit.test('testing functions', (assert) => { assert.expect(3); const spyOne = sinon.spy(one); const spyTwo = sinon.spy(two); const spyThree = sinon.spy(three); myMethod(); assert.ok(spyOne.called, "called one"); assert.ok(spyTwo.called, "called two"); assert.ok(spyThree.called, "called three"); sinon.restore(); }); 

Even though I call myMethod() and I have spies on one - two - three I still get false on one.called (same for two and three )

What am I missing here?

Thanks!

Answer 1

Calling sinon.spy(fn) doesn't change fn , it merely creates a new function (the spy) that will call fn .

For you to be able to test one , two , three , you need to replace those functions (or rather, their references) with spies, and restore them afterwards:

// keep references to the original functions
var _one   = one;
var _two   = two;
var _three = three;

// replace the original functions with spies
one   = sinon.spy(one);
two   = sinon.spy(two);
three = sinon.spy(three);

// call our method
myMethod();

// test
assert.ok(one.called,   "called one");
assert.ok(two.called,   "called two");
assert.ok(three.called, "called three");

// restore the original functions
one   = _one;
two   = _two;
three = _three;

It's not ideal though, and if possible I would probably group all the functions into an object. That would also enable Sinon to restore the originals itself.