Why are certain function calls termed "illegal invocations" in JavaScript?

JavascriptInvocation

Javascript Problem Overview


For example, if I do this:

var q = document.querySelectorAll;

q('body');

I get an "Illegal invocation" error in Chrome. I can't think of any reason why this is necessary. For one, it's not the case with all native code functions. In fact I can do this:

var o = Object; // which is a native code function

var x = new o();

And everything works just fine. In particular I've discovered this problem when dealing with document and console. Any thoughts?

Javascript Solutions


Solution 1 - Javascript

It's because you've lost the "context" of the function.

When you call:

document.querySelectorAll()

the context of the function is document, and will be accessible as this by the implementation of that method.

When you just call q there's no longer a context - it's the "global" window object instead.

The implementation of querySelectorAll tries to use this but it's no longer a DOM element, it's a Window object. The implementation tries to call some method of a DOM element that doesn't exist on a Window object and the interpreter unsurprisingly calls foul.

To resolve this, use .bind in newer versions of Javascript:

var q = document.querySelectorAll.bind(document);

which will ensure that all subsequent invocations of q have the right context. If you haven't got .bind, use this:

function q() {
    return document.querySelectorAll.apply(document, arguments);
}

Solution 2 - Javascript

In my case Illegal invocation occurred due to passing undeclared variable to function as argument. Make sure to declare variable before passing to function.

Solution 3 - Javascript

you can use like this :

let qsa = document.querySelectorAll;
qsa.apply(document,['body']);

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questionuser1152187View Question on Stackoverflow
Solution 1 - JavascriptAlnitakView Answer on Stackoverflow
Solution 2 - JavascriptFawadView Answer on Stackoverflow
Solution 3 - Javascriptlove-for-codingView Answer on Stackoverflow