Why does a reversed array equal a regular array in javascript?

Question

IE why would code like

var strArr = ["a", "b"];
console.log(strArr.reverse() === strArr ? true : false);

print true if the reversed array has a different order?

Because it is done by reference and returns a pointer to the array as in the docs — Francisco Presencia
– Francisco Presencia, Commented Jun 5, 2016 at 21:23
Also note that [0,1,2] === [0,1,2] --> false, since === returns true for operands of type object, only if they reference the same object. Contents aren't considered. — 1983
– 1983, Commented Jun 5, 2016 at 21:32

1983 · Accepted Answer · 2016-06-05 21:29:30Z

6

Because reverse reverses the array IN PLACE, so you are comparing the array with itself.

edited Jun 5, 2016 at 21:29

1983

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answered Jun 5, 2016 at 21:22

Amit

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Comments

Redu · Accepted Answer · 2016-06-05 21:36:24Z

3

Actually the Array.prototype.reverse() method morphs the original array to it's reverse and then in addition to that returns a "reference" to this morphed array. Hence the result and the morphed original array becomes the same.

If anyone had asked me before developing this method i would request to keep the original as it is and return a reversed array; alas apparently it wasn't my call.

answered Jun 5, 2016 at 21:36

Redu

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