Why doesn't .join() work with function arguments?

Question

Why does this work (returns "one, two, three"):

var words = ['one', 'two', 'three']; $("#main").append('<p>' + words.join(", ") + '</p>'); 

and this work (returns "the list: 111"):

var displayIt = function() {     return 'the list: ' + arguments[0]; }    $("#main").append('<p>' + displayIt('111', '222', '333') + '</p>'); 

but not this (returns blank):

var displayIt = function() {     return 'the list: ' + arguments.join(","); }    $("#main").append('<p>' + displayIt('111', '222', '333') + '</p>'); 

What do I have to do to my "arguments" variable to be to use .join() on it?

Answer 1

It doesn't work because the arguments object is not an array, although it looks like it. It has no join method:

>>> var d = function() { return '[' + arguments.join(",") + ']'; } >>> d("a", "b", "c") TypeError: arguments.join is not a function 

To convert arguments to an array, you can do:

var args = Array.prototype.slice.call(arguments); 

Now join will work:

>>> var d = function() {   var args = Array.prototype.slice.call(arguments);   return '[' + args.join(",") + ']'; } >>> d("a", "b", "c"); "[a,b,c]" 

Alternatively, you can use jQuery's makeArray, which will try to turn "almost-arrays" like arguments into arrays:

var args = $.makeArray(arguments); 

Here's what the Mozilla reference (my favorite resource for this sort of thing) has to say about it:

The arguments object is not an array. It is similar to an array, but does not have any array properties except length. For example, it does not have the pop method. ...

The arguments object is available only within a function body. Attempting to access the arguments object outside a function declaration results in an error.

Answer 2

If you are not interested on other Array.prototype methods, and you want simply to use join, you can invoke it directly, without converting it to an array:

var displayIt = function() {     return 'the list: ' + Array.prototype.join.call(arguments, ','); }; 

Also you might find useful to know that the comma is the default separator, if you don't define a separator, by spec the comma will be used.