How are non-string literals evaluated in Javascript?

Question

When checking out the length property on a string I made a mistake and forgot the quotes around length. Instead of an error, one of the characters appeared from the string.

const string = 'name'
s[length]
>> "a"

I have done other combinations and usually get the second letter returned. So is the word length being converted into a Boolean, which is then converted to a number? Or, is something else going on?

Update. this is all I did:

const a = [1,2,3]
>> undefined
a["length"]
>> 3
const s = 'name'
>> undefined
s[length]
>> "a"
s['length']
>> 4
length
>> 1
a[length]
>> 2

Could you be using the variable length anywhere else in your code? — Sebastian Speitel
– Sebastian Speitel, Commented Mar 22, 2018 at 9:00
It was just in the console (Firefox), and I hadn't done much else. — cham
– cham, Commented Mar 22, 2018 at 9:01
Check it @SebastianSpeitel said because ("name")[undefined] = 'n' because it cast undefined to 0 — dloeda
– dloeda, Commented Mar 22, 2018 at 9:01

visibleman · Accepted Answer · 2018-03-22 09:59:05Z

4

In the default scope, length will evaluate to window.length, which is the number of frames or iframes in the window (usually 0, but in your case maybe 1?)

Edit: A quick look at the meaning of the different examples in your updated question.

const a = [1,2,3]; // Array of length 3
a["length"];       // equals a.length , 3
const s = 'name';  // String of length 4
s[length];         // s[window.length], s[1], 'a'
s['length'];       // equals s.length, 4
length;            // equals window.length, 1
a[length]          // equals a[window.length], a[1] , 2

edited Mar 22, 2018 at 9:59

answered Mar 22, 2018 at 9:01

visibleman

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4 Comments

Naramsim Over a year ago

Good one!, Do you know other similar "shortcuts"?

visibleman Over a year ago

@Naramsim, not entirely sure what you mean. In the 'default' or top level scope, this evaluates to 'window' which means that any property of 'window' can be accessed by it's short name, e.g. location points to window.location and so on

Sebastian Speitel Over a year ago

In the same manner, you can mostly use setInterval and don't have to use window.setInterval

Lain Over a year ago

@Naramsim: console.dir(window);

Marc Schroth · Accepted Answer · 2018-03-22 09:06:09Z

1

Not an answer, but a general piece of advice:

use s.length instead of s['length']

You avoid the mistake you just ran into.

I personally only use square brackets when I'm accessing values in a plain old array or when I need to use a variable or a property name that compilers don't like (ie. s[myVar] or s['some-property']).

edited Mar 22, 2018 at 9:06

answered Mar 22, 2018 at 9:04

Marc Schroth

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3 Comments

Jai Over a year ago

Use the comment area for your valueable advice as this is the answer area.

Naramsim Over a year ago

He cant, I think, look at his reputation

Marc Schroth Over a year ago

Would if i could comment :)

Justinas · Accepted Answer · 2018-03-22 09:02:16Z

0

Usually you do s.length as length is property of String.

If you have length variable somewhere in your code, then it's typecasted to integer.

If you do not have it, then undefined is converted to integer.

s[length] converts to s[0] in your case

answered Mar 22, 2018 at 9:02

Justinas

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1 Comment

Kaiido Over a year ago

That's completely wrong. Bracket notation casts to String, not to integer (and there is not even an integer type in js...).

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