How does delete exactly work?

Question

Please check the jsfiddle - http://jsfiddle.net/du8svaym/

var a = [2, 4, "bang", undefined, NaN, 5]; for (i in a) alert(a[i]); //alerting a[3]=undefined delete a[1]; for (i in a) alert(a[i]); //Why not alerting a[1]= undefined? alert(a[1]); //it is undefined! but not alerted, what is happening under the hood?

If you notice, the first loop alert alerts a value which is undefined. In the 2nd loop alert a[1] is undefined since we deleted it, but is not alerted. What is the difference between the two undefined, how exactly or differently is delete setting the undefined?

Because its deleting from array. And the loop you are using it will traverse array by elements of array and not from 0 to 5 — Ravi Dhoriya ツ
– Ravi Dhoriya ツ, Commented Dec 30, 2014 at 11:24
Please go through this i think it will help you to understand the concept 2ality.com/2012/06/dense-arrays.html — Arunprasanth K V
– Arunprasanth K V, Commented Dec 30, 2014 at 11:32
deleting does not change the array length. You can confirm. So it is not removing the element. — Nikhil Pai
– Nikhil Pai, Commented Dec 30, 2014 at 11:36
@NikhilPai delete does change the array length BUT it does not update the .length property. See developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — David Hughes
– David Hughes, Commented Dec 30, 2014 at 11:46

David Hughes · Accepted Answer · 2014-12-30 11:43:49Z

The delete operator removes a property from an object. Arrays are JavaScript objects like so:

var foo = { '0': 2, '1': 4, '2': "bang", '3': undefined, '4': NaN, '5': 5 } delete foo[1]; // completely removes the property '1' from foo. console.log(foo); // outputs: { '0': 2, '2': "bang", '3': undefined, '4': NaN, '5': 5 }

Sorry for delayed edit, got pulled away straight after I noticed crucial typo but before I could rectify it! :)

dr_dev · Accepted Answer · 2014-12-30 12:02:35Z

For the first iteration, the array looks like this.

[0: 2, 1: 4, 2: "bang", 3: undefined, 4: NaN, 5: 5]

Now, after you delete a1, it removes 4. So, now the array looks like

[0: 2, 2: "bang", 3: undefined, 4: NaN, 5: 5]

Now, as you can see a1 does not exist. So it gives undefined.

Deleting an element will not affect the array length, as it does not change the indexes. If you want the array index to be affected, use splice.

unobf · Accepted Answer · 2014-12-30 11:25:12Z

0

JavaScript Arrays are sparse, that means that they do not have to store a value at every position. Delete removes the value at the position, making it sparse.

Try this

a = []; // empty array a[1000] = 'hello'; for (i in a) { console.log(i); } console.log(a.length)

You will see that it will only print 1000 in the loop whereas the length of the array is 1001

answered Dec 30, 2014 at 11:25

unobf

7,2641 gold badge25 silver badges36 bronze badges

4 Comments

Nikhil Pai Over a year ago

I know this. What is the difference between a[1] which becomes undefined after deleting and a[3] which is undefined

Niet the Dark Absol Over a year ago

a[3] is defined to be undefined. a[1] really is undefined.

Nikhil Pai Over a year ago

Right, so a[1] really is undefined. How is that set to undefined

dr_dev Over a year ago

As there is no value, for the index 1 in the array, it shows 'undefined'. Look at my answer.

Collectives™ on Stack Overflow

How does delete exactly work?

3 Answers 3

1 Comment

Comments

4 Comments

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

Comments

4 Comments

Related