Why does "[[1, 2], [3, 4]].indexOf([1, 2])" return -1?

Question

I spent the last hour and a half trying to find a bug in my code, when I finally realized that this JavaScript code:

[[1, 2], [3, 4]].indexOf([1, 2]);

returns -1, even though something such as [1, 2, 3].indexOf(1); correctly returns 0...

Why does this happen, and how can I find the correct index of the subarray?

Here you can find more ways to compare arrays: http://stackoverflow.com/questions/7837456/how-to-compare-arrays-in-javascript — sinisake, Dec 27 '16 at 17:36

score 1 · Accepted Answer · answered Dec 27 '16 at 17:31

1

The indexOf takes only primitive arguments and you cannot match:

[1, 2] == [1, 2]

Which obviously gives false.

answered Dec 27 '16 at 17:31

Praveen Kumar Purushothaman

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`indexOf` does a `===` instead of `==`. Now try to explain to the OP why `[42, NaN].indexOf(NaN)` returns `-1`. :D – Ionică Bizău Dec 27 '16 at 17:33
@IonicăBizău Both `NaN == NaN` and `NaN === NaN` returns `false`. `NaN` is not a primitive type. – Praveen Kumar Purushothaman Dec 27 '16 at 17:33
@IonicăBizău — "indexOf does a === instead of ==" — That doesn't make a difference here. – Quentin Dec 27 '16 at 17:34
@Quentin You're right, but the OP can say: Why does `["42"].indexOf(42)` return `-1`? That's my point. It should be `===` instead of `==`. – Ionică Bizău Dec 27 '16 at 17:35
When `==` itself gives a `false`, `===` we can be very sure, it gives `false`. – Praveen Kumar Purushothaman Dec 27 '16 at 17:35
@IonicăBizău I am not telling that `indexOf` does a `==`. I am just telling that `NaN == NaN` gives a false. `#Difference` – Praveen Kumar Purushothaman Dec 27 '16 at 17:36
"_The indexOf takes only primitive arguments_" Nope, you can check a reference with `indexOf` too. – Teemu Dec 27 '16 at 17:37
@Teemu Lemme check. – Praveen Kumar Purushothaman Dec 27 '16 at 17:37
2

@PraveenKumar *The indexOf takes only primitive arguments* — `a = [1, 2]; [a, [1, 2]].indexOf(a) // => 0` – Ionică Bizău Dec 27 '16 at 17:38
@Teemu From MDN: `searchValue` *A string representing the value to search for.* AFAIK, `string` is a primitive data type... `:)` – Praveen Kumar Purushothaman Dec 27 '16 at 17:38
@PraveenKumar Exactly. Is it primitive? – Ionică Bizău Dec 27 '16 at 17:40
Hmm... Your last comment wasn't probably pointed to me? [Here you can see](https://jsfiddle.net/gqs7ndtp/) an object found by `indexOf` method. – Teemu Dec 27 '16 at 17:40
@IonicăBizău According to MDN, yes, but according to reality, no. – Praveen Kumar Purushothaman Dec 27 '16 at 17:41
1

@PraveenKumar Reference? MDN says: *In JavaScript, there are 6 primitive data types: string, number, boolean, null, undefined, symbol (new in ECMAScript 2015).* Where would you put my `a`? – Ionică Bizău Dec 27 '16 at 17:42
3

@PraveenKumar String.prototype.indexOf != Array.prototype.indexOf – Jonas Wilms Dec 27 '16 at 17:49
1

You can certainly pass any value to `Array#indexOf`, not just primitive values. – Felix Kling Dec 27 '16 at 17:55

Nina Scholz · Answer 2 · 2016-12-27T17:45:16.740

You could iterate the array and check every item of the pattern and return the index.

function getIndexOf(array, pattern) {
    var index = -1;
    array.some(function (a, i) {
        if (a.length !== pattern.length) {
            return false;
        }
        if (a.every(function (b, j) { return b === pattern[j]; })) {
            index = i;
            return true;
        }
    });
    return index;
}

console.log(getIndexOf([[1, 2], [3, 4], [7, 8]], [7, 8]));

Another way with JSON and indexOf.

function getIndexOf(array, pattern) {
    return array.map(function (a) { return JSON.stringify(a); }).indexOf(JSON.stringify(pattern));
}

console.log(getIndexOf([[1, 2], [3, 4], [7, 8]], [7, 8]));

or one can use **a.join("")** instead of **JSON.stringify(a)** — kevin ternet, Dec 27 '16 at 18:15

Why does "[[1, 2], [3, 4]].indexOf([1, 2])" return -1?

2 Answers2