CODEWITHSUNDEEP
javascript
temporary_user_name
temporary_user_name

Reputation: 37078

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

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.

Upvotes: 1

Views: 123

Answers (2)

jAndy
jAndy

Reputation: 236062

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( [] );

Upvotes: 2

elclanrs
elclanrs

Reputation: 94121

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, [])

Upvotes: 4

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