Why are my JavaScript function names clashing?

JavascriptFunction

Javascript Problem Overview


I wrote the following script just to see what happens when a variable and a function that has a function assigned to it have their names clash:

var f = function() {
	console.log("Me original.");
}

function f() {
	console.log("Me duplicate.");
}

f();

The output I'm getting is "Me original." Why was the other function not called?

Also, if I change my original assignment to var f = new function() {, I get "Me original", followed by a TypeError saying object is not a function. Can someone please explain?

Javascript Solutions


Solution 1 - Javascript

Function declarations are hoisted (moved to the top) in JavaScript. While incorrect in terms of parsing order, the code you have is semantically the same as the following since function declarations are hoisted:

function f() {
    console.log("Me duplicate.");
}
var f = function() {
    console.log("Me original.");
}


f();

Which in turn, with the exception of the function's name is the same as:

var f = function() {
    console.log("Me duplicate.");
}
var f = function() {
    console.log("Me original.");
}


f();

Which in turn, because of variable hoisting is the same as:

var f;
f = function() {
    console.log("Me duplicate.");
}
f = function() {
    console.log("Me original.");
}

f();

Which explains what you're getting, you're overriding the function. More generally, multiple var declarations are allowed in JavaScript - var x = 3; var x = 5 is perfectly legal. In the new ECMAScript 6 standard, let statements forbid this.

This article by @kangax does a fantastic job in demystifying functions in javascript

Solution 2 - Javascript

If doesn't look like anyone answered your follow-up question so I'll answer it here, though you should generally ask follow-up questions as separate questions.

You asked why this:

var f = new function() {
    console.log("Me original.");
}

function f() {
    console.log("Me duplicate.");
}

f();

prints out "Me original." and then an error.

What is happening here is that the new causes the function to be used as a constructor. So this is equivalent to the following:

function myConstructor() {
    console.log("Me original.");
}
var f = new myConstructor();

function f() {
    console.log("Me duplicate.");
}

f();

And thanks to the function hoisting that Benjamin explained, the above is essentially equivalent to this:

var myConstructor = function() {
    console.log("Me original.");
};
var f = function() {
    console.log("Me duplicate.");
};

f = new myConstructor();

f();

This expression:

var f = new function() {
    console.log("Me original.");
}

causes a new object to be constructed and assigned to f, using an anonymous function as the constructor. "Me original." is printed out as the constructor executes. But the object that is constructed is not itself a function, so when this eventually executes:

f();

you get an error, because f is not a function.

Solution 3 - Javascript

Forgive me if this is the wrong way to approach adding a point. I haven't been around here all the much, and would welcome constructive direction and/or criticism.

Benjamin's answer addresses the OP's question excellently, but I'd like to add one tweak that'll give us a full tour of hoisting and its oddities.

If we begin the original code with a call to f, like so:

f();

var f = function() {
   console.log("Me original.");
};

function f() {
   console.log("Me duplicate.");
}

f();

The output will then be:

Me duplicate.
Me original.

The reason being that var and function statements are hoisted in slightly different ways.

For var the declaration is moved to the top of the current scope*, but any assignment is not hoisted. As far as the value of the declared var goes, it's undefined until the original assignment line is reached.

For functionstatements, both the declaration and definition are hoisted. Function expressions, as used in the var f = function() {... construct, are not hoisted.

So after hoisting, execution is as if the code were:

var f; // declares var f, but does not assign it.

// name and define function f, shadowing the variable
function f() { 
  console.log("Me duplicate.");
}

// call the currently defined function f
f(); 

// assigns the result of a function expression to the var f,
// which shadows the hoisted function definition once past this point lexically
f = function() { 
  console.log("Me original."); 
}

// calls the function referenced by the var f
f();

*All JavaScript scope is lexical, or function, scope, but it seemed like it would just confuse things to use the f word at that point.

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questionankush981View Question on Stackoverflow
Solution 1 - JavascriptBenjamin GruenbaumView Answer on Stackoverflow
Solution 2 - JavascriptJLRisheView Answer on Stackoverflow
Solution 3 - JavascriptcodelahomaView Answer on Stackoverflow