1

When I set arugments[0] = chewie; arguments[1] = luke it prints:

Chewbacca

Chewbacca

Han Solo

Chewbacca

And what I expected was the following:

Chewbacca

Luke Skywalker

Han Solo

Chewbacca 

function starWars(luke, darth, han, chewie) {

   arguments[0] = chewie;
   arguments[1] = luke;

   document.write(arguments[0] + "<br>");
   document.write(arguments[1] + "<br>");
   document.write(arguments[2] + "<br>");
   document.write(arguments[3] + "<br>");
 
} 

starWars("Luke Skywalker", "Darth Vader", "Han Solo", "Chewbacca");

Can anyone explain me in details why it is not working as I expected?

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2
  • 1
    You should not be assigning values to arguments manually. In almost every case, it's no longer necessary to use the arguments object in ES2015 or later
    – Mulan
    Commented May 7, 2017 at 10:11
  • 1
    Because arguments[0] and luke are the exact same thing. Not two references to the same value, but basically both are kind of the same variable. If you change arguments[0], luke will also be changed; and vice versa. So much for the knowledge; no reason to explit or even use this. Listen to @naomik, avoid the arguments object as much as possible.
    – Thomas
    Commented May 7, 2017 at 10:29

3 Answers 3

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3

In loose mode, the entries in arguments are linked to the named parameters. So by setting arguments[0], you're setting luke. Literally, the lines arguments[0] = chewie; and luke = chewie; are interchangeable. So later when you use arguments[1] from luke, you're using the updated value (chewie).

Strict mode removes this magic link:

"use strict";
function starWars(luke, darth, han, chewie) {

   arguments[0] = chewie;
   arguments[1] = luke;

   console.log(arguments[0]);
   console.log(arguments[1]);
   console.log(arguments[2]);
   console.log(arguments[3]);
 
} 

starWars("Luke Skywalker", "Darth Vader", "Han Solo", "Chewbacca");

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1

You're re-assigning the luke variable with chewie since argument[0] is a reference to luke.

So after arguments[0] = chewie, what happens is luke = chewie.

A workaround could be :

var tempLuke = argument[0];
arguments[0] = chewie;
arguments[1] = tempLuke;
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1

When you change arguments object, you also change the argument variables of the function, because it holds references to the variables.

To summarize

1) arguments[0] = chewie;

Here, you also set variable luke to chewie ('chewbacca')

2) arguments[1] = luke;

Here, luke is already became 'chewbacca' in the first step. So, you are making argument dart also 'chewbacca'.

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