Set Comparability for Objects in ECMA6 JS [duplicate]

Question

I've been using a set to store and retrieve simple objects which represent coordinates.

// Creates a point in hex coordinates function hexPoint(q, r) { this.q = q; this.r = r; }

I am generating these points multiple times, so that I have easy coordinate pairs to pass around, but I wish to be able to store them in such a way that I do not store duplicate coordinates.

The ECMA 6 Set object relies on references to test object equality, so I am wondering if there is a way to supply this set with a comparable function so that I may allow it to test equality for new objects with the same fields. Otherwise, is there something else I can do to avoid re-implementing this data structure?

plain objects don't have repeats, and you can use your numbers as keys, sticking q+r together to make a "unique" key that you can look for in hexPoint() before returning a new object. — dandavis
– dandavis, Commented Jun 23, 2015 at 19:41

Community · Accepted Answer · 2017-05-23 11:58:30Z

0

Why not add an isEqual(otherPoint)? It might look something like:

function HexPoint(q, r) { this.q = q; this.r = r; } HexPoint.prototype.isEqual = function(otherPoint) { return this.q === otherPoint.q && this.r === otherPoint.r; }

Then you can make instances of HexPoint with:

var point = new HexPoint(...); if (point.isEqual(someOtherPoint)) { ... }

A similar Q&A points out no native option exists yet.

edited May 23, 2017 at 11:58

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answered Jun 23, 2015 at 19:30

Cymen

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3 Comments

Michael Von Hippel Over a year ago

Will the Set be able to use the isEqual function to perform its equality checks? As I understand it, the implementation of set only checks the reference and does not perform any other checks. I want the set to be able to maintain its own uniqueness without me having to check for it.

Cymen Over a year ago

@MichaelVonHippel I think you have to build a say a UniqueSet class if you want that. It would check on add if the Object was already present. If using HexPoint, you'd probably want to add an isEqual like above and use it.

Michael Von Hippel Over a year ago

Unfortunately, doing that check seems like needless complexity to add(since I would have to iterate over all existing elements anyway.) I guess the alternative is to use a factory for all hexpoints or simply make my own set that implements hashing.

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