Checking if a function is of type Function in javascript

Question

Yes, I know there are plenty of posts on this, but I do not understand the selected answer of this post. Particularly, why is it necessary to return object?

Why wouldn't something like Object.prototype.toString.call(myFunc) be sufficient as described by MDN?

@DmitriPavlutin: That isn't actually necessary, though; they both work fine. — SLaks
– SLaks, Commented Mar 31, 2016 at 17:04
The point of duck-typing should be that you don't care if it's a Function, only whether it implements apply and call... — ssube
– ssube, Commented Mar 31, 2016 at 17:16

Bergi · Accepted Answer · 2016-03-31 17:05:59Z

1

why is it necessary to return object?

It's not necessary to do object &&, but a shortcut for falsy values that avoids the method call for values such as null. Of course, if you're looking for speed, you probably should go for typeof object == "function"

Why wouldn't something like Object.prototype.toString.call(myFunc) be sufficient?

It is sufficient.

answered Mar 31, 2016 at 17:05

Bergi

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Dmitri Pavlutin · Accepted Answer · 2016-03-31 17:08:12Z

0

return object && getClass.call(object) == '[object Function]'

This construction simply verifies if object is not null or undefined (generally falsy).
This way it's faster, because JavaScript doesn't evaluate the getClass() call on a falsy object.

answered Mar 31, 2016 at 17:08

Dmitri Pavlutin

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