Abhas Bhattacharya
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How does in-built Array iterator work?

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Abhas Bhattacharya
·Jul 12, 2020·

5 min read

We use [...someArray] or array.values() to get multiple values out of a array. And we kind of know how it works - it just returns all the values from the array. Now internally, both of them uses the iterator protocol - which is already defined on array prototype.

But do we know exactly how the in-built Array iterator works? Can you write a precise polyfill to implement iterator protocol on Array?

A little about the iterator protocol

When we use spread operator, like [...someArray] - it treats someArray as a general iterable, not specifically as an array.

  1. It tries to get a iterator from the iterable, using iter = someArray[Symbol.iterator]().

  2. Then it consumes all the values from the iterable by calling iter.next() repeatedly.
    This method .next() returns response in this structure {value: "something", done: true|false}.

  3. While values are available, it returns done:false; after all values are finished, it returns done:true. After this, the consumer (i.e. spread operator) should stop asking for more values.

This is how the iterable/iterator stuff works.

Array Iterator

Array has in-built support for iterable/iterator protocol, through prototype. Array.prototype has a method with key Symbol.iterator, which creates a ArrayIterator.

ArrayIterator internally contains these properties -

  • [[ IteratedArrayLike ]] - which points back to the actual array.
    ( Remember, “this” within the ArrayIterator is the iterator object itself, not the actual array. So, it has to store a link to the array during creation )

  • [[ ArrayLikeNextIndex ]] - the next index whose value should be returned. Default is 0.

  • [[ ArrayLikeIterationKind ]] - For normal purpose, it is “value”.
    In general, it can be “key”, “value” or “key+value”. This allows reuse of the same iterator mechanism for array.keys(), .values() and .entries().


⭐️ Array Iterator.next( )

Now that we have the in-built array iterator, the most important thing is knowing how iter.next() will return the array values one-by-one. Let's look at how that works, in details -

( Say, arr = [[IteratedArrayLike]], index = [[ArrayLikeNextIndex]], kind = [[ArrayLikeIterationKind]] )

  1. If arr is undefined, return {value: undefined, done: true}
    (This is a special step, caused by step 3.a. It will reach here if you call .next() even after iterator has finished.)

  2. If index < arr.length, ( i.e. while values are available)
    a. set key = index

    b. If kind is “key”, return {value: key, done: false}
    (incase of arr.keys(), just return the index. It doesn't even try to access the value.)

    c. Set value = arr[index]

    d. If kind is “value”, return {value: value, done: false}
    (for arr.values() or ...arr, just return value at that index)

    e. If kind is “key+value”, return {value: [key, value], done: false}
    (for arr.entries(), return array of [key, value])

    f. Set [[ArrayLikeNextIndex]] = index + 1
    (⭐️ 1️⃣ Always increment key to next index. This sequential index is used to get the next value, irrespective of holes in that position (whether that index exists or not). It doesn't skip the empty/non-existent indexes.)

  3. Else, (i.e. when index >= arr.length - reached end of array)

    a. Set [[IteratedArrayLike]] = undefined.
    (⭐️ 2️⃣ Yes, once it reaches the end - it sets linked array to undefined. This is to ensure that once the iterator has finished, it will never return any more value. This undefined array is handled in step 1.
    If this was not done and if array length was increased before next call, then it would again return new values after saying done:true earlier.)

    b. Return {value: undefined, done: false}

🤯 Implications

A. [...arr] converts sparse array to dense array. (Reason -⭐️ 1️⃣ above)

foo = [10, , 30, 40]; // [10, <hole>, 30, 40]

bar = [...foo]; // [10, undefined, 30, 40]

// ✅ sparse array
foo.hasOwnProperty(1); // false       
// ⚠️ dense array
bar.hasOwnProperty(1); // true

// ⚠️ array.keys() - returns the empty indexes
Object.keys(foo) // 0, 2, 3
foo.keys(); // 0, 1, 2, 3

B. Array iterator will never return more values after it has finished once - even if the array has more values now. (Reason - ⭐️ 2️⃣ above)

arr = [10]
iter = arr[Symbol.iterator]()

iter.next() // 10 - this was the last value

// but,modify array
// ⭐️ uses updated length, since iterator is not finished
iter.next() // {value: 20, done: false}

iter.next() // {value: undefined, done: true} - end of iterator

// push again after iterator has ended
// ⭐️ doesn't use updated length
iter.next() // {value: undefined, done: true}

That's all. Follow me on LinkedIn for more updates.

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