West Midlands | 26-Jan-ITP | Fida Ali Zada | Sprint 1 | Data Groups

[alizada-dev]

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Changelist

Completed all the exercises of sprint 1 from the Data Group module.

[cjyuan]

This test should fail if the function returns the original array (instead of a copy of the original array).

The current test checks only if both the original array and the returned array contain identical elements.
In order to validate the returned array is a different array, we need an additional check.

Can you find out what this additional check is?

[alizada-dev]

By implementing return [...new Set(array)];, the test will pass correctly, because this implementation always returns a new array.

[cjyuan]

The concern here is not about your function implementation -- it was correct.

The issue is, how to prepare a test to check if the function is indeed returning a copy of the original array.

With the test you had, a function like this could also pass the test.

function  dedupe(array) {
  const set = new Set(array);
  if (set.size == array.length) return array;
  return [...set];
}  

[alizada-dev]

const input = [1, 3, 5, 7];
const result = dedupe(input);

expect(result).toEqual(input);
expect(result).not.toBe(input);

[cjyuan]

With this test, there is a chance that, even though result has incorrect elements (for example, []),
the two tests could still pass. Can you figure out why, and then fix the tests accordingly?

[alizada-dev]

const input = [1, 3, 5, 7];
const result = dedupe(input);

expect(result).toEqual([1, 3, 5, 7]);
expect(result).not.toBe([1, 3, 5, 7]);

Is it alright, this time?

[cjyuan]

Still not correct. Try looking up the difference between toBe() and toEqual().

[cjyuan]

The proposed test in dedupe.test.js is not quite correct.

All other changes look good.

[illicitonion]

Closing PR because the January ITP run has finished. Feel free to re-open if you're still working on it.