Why doesn't indexOf work after pushing object in array

Question

Im trying to get indexOf of a object after its pushed inside a array. This is not returning the same value back as i do indexOf whenever objext is allready in the array.

SCENARIO

var arr = [];
setInterval(function() {
	var path = { one: "f00"};
    if (typeof path !== "undefined") {
        if (arr.indexOf(path) === -1) {
            console.log("Not Exists!!")
            arr.push(path)
        } else {
            console.log("Exists!!")
        }
    }
	console.log(arr)
}, 2000)

What is the different between the working of

You're using an object in the indexOf, it might not know how to compare them — Patrick Barr
– Patrick Barr, Commented Mar 13, 2017 at 16:51
@JohnSmiths because you're passing a direct reference to the object instead of another object that has the same data — Patrick Barr
– Patrick Barr, Commented Mar 13, 2017 at 16:52
Possible duplicate of Javascript oddness with array of objects and indexOf — Heretic Monkey
– Heretic Monkey, Commented Mar 13, 2017 at 16:53
@MikeMcCaughan no its not, the first console.log returns correct value "0" and the second console shows -1 — John Smiths
– John Smiths, Commented Mar 13, 2017 at 16:54
Possible duplicate of Javascript array.indexOf doesn't search objects — Sebastian Simon
– Sebastian Simon, Commented Mar 13, 2017 at 16:54

Matthew Herbst · Accepted Answer · 2017-03-13 18:12:44Z

1

The issue is that JavaScript doesn't do a deep compare of objects, so it doesn't recognize them as the same.

var a = { name: 'foo' }
var b = { name: 'foo' }
a === b // false

However, since you have access to the object before the insert, you can save a reference to it, and then search for that reference:

var arr = []
var obj = { path: 'foo' }
arr.push(obj)
arr.indexOf(obj) // 0

This is because indexOf uses the strict equality === comparison. So in this case, the references to obj and the object at arr[0] are the same.

Edit

Based on your changed question, here is a way to write your function to do what you want:

var arr = [];

function findAdnSet(obj) {
  var index = arr.indexOf(obj);

  if (index !== -1) {
    return index;
  } else {
    arr.push(obj);
    return arr.length - 1; // No reason to use indexOf here, you know the location since you pushed it, meaning it HAS to be the last element in the array
  }
}

var path = { name: 'foo' };
findAndSet(path);

A more robust option than using indexOf since your function might not always have a good reference available is to use find/findIndex:

var arr = [];

function findAndSet(obj) {
  var index = arr.findIndex(function(item) {
    if (item.name === 'foo') {
      return true;
    }
  });

  if (index) { // findIndex returns `undefined` if nothing is found, not -1
    return index;
  } else {
    arr.push(obj);
    return arr.length - 1;
  }
}

// You don't need a reference anymore since our method is doing a "deep" compare of the objects
findAndSet({ name: 'foo' });

edited Mar 13, 2017 at 18:12

answered Mar 13, 2017 at 16:55

Matthew Herbst

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5 Comments

John Smiths Over a year ago

I edited my code, can you show example with that one. Why i dont get ! -1

Matthew Herbst Over a year ago

The reference to path changes each time the function is run, since you redefine path each time the function is run. You'll have to save path before you define the setTimeout call.

John Smiths Over a year ago

Is it because i'm using setInterval and forEach() to get the path ?

Matthew Herbst Over a year ago

Well, I don't see a forEach anywhere in your code, but yes to the setInterval. Every time you call the function everything inside is happening for essentially the first time. When the function ends in-between runs the internal scope (memory) of the function is lost.

Matthew Herbst Over a year ago

I have added two new examples that should help you.

user7491506 · Accepted Answer · 2017-03-13 16:55:26Z

0

The first time you do indexOf you push and search for the object 'path' so it is found. The second time you create an object and add push it to the array, and then search for another new object (which happens to have the same values), but since it is not the same object that you pushed it is not found.

answered Mar 13, 2017 at 16:55

user7491506

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