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

Question

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

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

        

Answer 1

At the moment you can't join array arguments, because they aren't an array, shown here

so you have to either first turn them into an array like this,

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

or like this, a bit shorter

function displayIt() {
  return 'the list: ' + [].join.call(arguments, ',');
}

if you are using something like babel or a compatible browser to use es6 features, you can also do this using rest arguments.

function displayIt(...args) {
  return 'the list: ' + args.join(',');
}

displayIt('111', '222', '333');

which would let you do even cooler stuff like

function displayIt(start, glue, ...args) {
  return start + args.join(glue);
}

displayIt('the start: ', '111', '222', '333', ',');

Answer 2

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 3

arguments is not a jQuery object, just a regular JavaScript object. Extend it before you try to call .join(). I think you would write:

return 'the list:' + $(arguments)[0];

(I'm not too familiar with jQuery, only Prototype, so I hope this is not completely bogus.)

Edit: It's wrong! But in his response, Doug Neiner describes what I'm trying to accomplish.

Answer 4

You could use this jQuery .joinObj Extension/Plugin I made.

As you'll see in that fiddle, you can use it as follows:

$.joinObj(args, ",");

or

$.(args).joinObj(",");

---

Plugin Code:

(function(c){c.joinObj||(c.extend({joinObj:function(a,d){var b="";if("string"===typeof d)for(x in a)switch(typeof a[x]){case "function":break;case "object":var e=c.joinObj(a[x],d);e!=__proto__&&(b+=""!=b?d+e:e);break;default:"selector"!=x&&"context"!=x&&"length"!=x&&"jquery"!=x&&(b+=""!=b?d+a[x]:a[x])}return b}}),c.fn.extend({joinObj:function(a){return"object"===typeof this&&"string"===typeof a?c.joinObj(this,a):c(this)}}))})(jQuery);

Answer 5

I don't know if there's a simple way to convert arguments into an array, but you can try this:

var toreturn = "the list:";
for(i = 0; i < arguments.length; i++)
{
   if(i != 0) { toreturn += ", "; }
   toreturn += arguments[i];
}

Answer 6

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.