Why can't I successfully use 'call' on Array.isArray?

Question

So I'm used to automatically writing things such as this:

Array.prototype.slice.call(arguments);

--and similar such applications of various methods via Array.prototype.XXX.call or Object.prototype.XXX.call([]).

So I went to try out the Array.isArray method which I didn't know existed, and automatically wrote Array.prototype.isArray.call, which of course failed since isArray isn't defined on the prototype.

But then I tried Array.isArray.call([]) slightly unthinkingly and got false, which confused me.

It worked fine on the third try when I went the plain ol' boring way (the correct usage) and simply wrote Array.isArray(thing_to_be_tested), but why did I get false on the previous attempt? I don't understand why that doesn't work.

Because you were basically calling Array.isArray() (only in a [] context), and without arguments it tells you that undefined is not an array. — Bergi
– Bergi, Commented Feb 1, 2015 at 21:59
@Adeneo When you're in the middle of working with the native APIs and playing with them and bending what they can do.... — temporary_user_name
– temporary_user_name, Commented Feb 1, 2015 at 22:00
Well, isArray is not part of the prototype chain, and it only accepts one argument, and doesn't care about the this value or scope, so don't ! — adeneo
– adeneo, Commented Feb 1, 2015 at 22:02
And you should generally avoid converting arguments, nodelist etc. to array that way as well, even if it is useful sometimes, and I must confess to do doing it quite a bit myself, even knowing the proper way is to use a for loop — adeneo
– adeneo, Commented Feb 1, 2015 at 22:04

elclanrs · Accepted Answer · 2015-02-01 21:57:46Z

4

Because Array.isArray has one argument, which you are not passing along. You are simply passing the this value, which is not used in the function:

Array.isArray.call(null, [])

answered Feb 1, 2015 at 21:57

elclanrs

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1 Comment

temporary_user_name Over a year ago

I knew that. Really I swear I did. Not enough coffee. 12 minutes til I can accept answer, lol.

jAndy · Accepted Answer · 2015-02-01 22:04:54Z

2

The first thing .call() expects is the context. So you're basically calling .isArray() with the context of an Array, but an undefined value.

Call instead:

Array.isArray.call( null, [] );

~~or to make it more realistic:~~

var isarr = Function.prototype.call.bind( Array.isArray, null );

Now you can call that like

isarr( [] );

edited Feb 1, 2015 at 22:04

answered Feb 1, 2015 at 21:58

jAndy

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5 Comments

Bergi Over a year ago

or to make it more simple: var isArr = Array.isArray;

Bergi Over a year ago

If at all, use var isArr = Array.isArray.bind(Array);.

jAndy Over a year ago

haha yes indeed. For that method the context has no meaning and it should just be referenced directly.

Bergi Over a year ago

Ah I though with ES6 and subclassable arrays that might change, but apparently it doesn't.

jAndy Over a year ago

Still, the method is not on the prototype chain. So contrary to for instance Object.prototype.toString we can just reference the method in a variable whilst that would not work for the toString example.

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