Two arrays with same elements but different orders should be considered as equal

[ghost]

equal([1,2,3], [1,2,3]) -> true

equal([1,2,3], [3,2,1]) -> false

How to get the true from the second statement?

[xgbuils]

Hi @kitz17

I don't understand why two arrays with same elements but different orders should be considered as equal.

You can get true if uses:

equal([1,2,3].sort(), [3,2,1].sort())

[MrBenJ]

I agree with @xgbuils - in regular old JS, the order of elements inside an Array matters. @ghost - If you're trying to evaluate if 2 Arrays have the same stuff, but don't care for the order, maybe using a different data type like Map, WeakMap, Set, or WeakSet would be better for you.

[jmakeig]

If you don’t care about order or repeats, you should turn your Array into a Set. However, then you run into #46.

[wilfredjonathanjames]

Order is a first-class property of arrays. Two arrays of varying order whose contents are the same will behave differently. They are unequal.

If you need this behaviour, Array.prototype.sort both arrays. If they contain the same elements they will be deeply equal.

[Lazarencjusz]

It does not solve this problem:
equal([ {keyA:'a', keyB: 'b' }, {keyD:'c', keyD: 'd' } ], [{keyA:'a', keyB: 'b' }, {keyD: 'd', keyD:'c' }] )

[ljharb]

If you want order not to matter, sort your arrays first.