Why can't I .call() Function.call?

Question

In Javascript, Function.call() can call Function given a this value and zero or more arguments.

Function.call itself is a function. So in theory, Function.call should be the same (or similarly acting) function as Function.call.call.

In V8, this seems to be the case:

> Function.call === Function.call.call
true

When we call Function.call(), we get an anonymous function

> Function.call()
[Function: anonymous]

However, I can't call .call() on Function.call.

> Function.call.call()
TypeError: undefined is not a function
at repl:1:21
at REPLServer.defaultEval (repl.js:132:27)
at bound (domain.js:291:14)
at REPLServer.runBound [as eval] (domain.js:304:12)
at REPLServer.<anonymous> (repl.js:279:12)
at REPLServer.emit (events.js:107:17)
at REPLServer.Interface._onLine (readline.js:214:10)
at REPLServer.Interface._line (readline.js:553:8)
at REPLServer.Interface._ttyWrite (readline.js:830:14)
at ReadStream.onkeypress (readline.js:109:10)

What is going on here? Function.call is clearly a function - it's not undefined as this error message suggests.

Nice question and answer, even though pointless in a normal programming context :) — nicovank, Nov 30 '16 at 09:42
@nicovank Of course, but I was exploring some "what if's" of Javascript and encountered this accidentally. :) — Qix - MONICA WAS MISTREATED, Nov 30 '16 at 09:42
[Of course you can](http://stackoverflow.com/q/31074664/1048572)! — Bergi, Dec 01 '16 at 23:21
@Bergi If you pass in a `this` value, sure. The confusion stemmed from the error message. — Qix - MONICA WAS MISTREATED, Dec 01 '16 at 23:22

score 4 · Accepted Answer · answered Nov 30 '16 at 09:40

4

Short answer: The error message is very misleading. It is the same error message you get when you do

(undefined)();

Longer answer:

The second .call() is being invoked with a this of Function.call.

Calling it with no parameters causes it to call this with undefined as the this value.

Therefore, you're really doing

Function.call.call(undefined)

which means you're (metaphorically) doing

undefined.call()

which is really just

undefined()

Passing nothing (or undefined) to the this parameter of Function.call.call() is essentially negating the this context of the first Function.call() (which would be just Function itself), causing .call() to be invoked on undefined.

This yields the error message that is produced: undefined is not a function.

answered Nov 30 '16 at 09:40

Qix - MONICA WAS MISTREATED

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…so working examples would be `Function.call.call(Function)` or `Function.call.call(console.log, console, "hello")` – Bergi Dec 01 '16 at 23:22
@Bergi Correct. Again, the confusion stemmed from the _error message_. – Qix - MONICA WAS MISTREATED Dec 01 '16 at 23:29
2

Ah, I didn't realise you self-answered your question. I just thought that adding an example of working double-`call` usage might benefit the explanation. – Bergi Dec 01 '16 at 23:32

Why can't I .call() Function.call?

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