Comments on: jQuery.makeArray()

By: Maury M. Marques

Maury M. Marques — Thu, 13 Jan 2011 01:27:46 +0000

This works.
$.unique(["a", "b", "c", "d", "a"]);

I do not know why, but this does not work.
$.unique(["a", "b", "c", "d", "a", "b"]);

By: Sanshi Ustc

Sanshi Ustc — Tue, 12 Oct 2010 22:45:40 +0000

That's not true, you can use the $.makeArray to transform arguments inside funtion:
function arrayFromArgs() {
return $.makeArray(arguments);
}

arrayFromArgs('3', '2', '4');

By: Sanshi Ustc

Sanshi Ustc — Tue, 12 Oct 2010 14:45:40 +0000

That's not true, you can use the $.makeArray to transform arguments inside funtion:
function arrayFromArgs() {
return $.makeArray(arguments);
}

arrayFromArgs('3', '2', '4');

By: Jane

Jane — Mon, 13 Sep 2010 08:02:08 +0000

If theres only I single element I get an empty array instead of an array with single element. Any idea why??

By: Nathan Curtis

Nathan Curtis — Thu, 10 Jun 2010 13:32:56 +0000

Technically, you can use makeArguments to make the arguments object into a real array, but it's probably slower than the Array.prototype.slice.call() method. The only requirements for makeArguments to turn an object into an array are that the object has a “length” property, and properties with numerical keys, e.g. { 'length':3, '0':”foo”, '1':”bar”, '2':”baz” }. The arguments object satisfies those requirements.

By: Karl Swedberg

Karl Swedberg — Thu, 14 Jan 2010 19:27:13 +0000

No, but for that you can just do something like this:

function arrayFromArgs() {
return Array.slice.call(arguments);
}

and then call it:

arrayFromArgs('3', '2', '4');

By: Anonymous

Anonymous — Thu, 14 Jan 2010 07:02:00 +0000

Can it be use for $.makeArray(arguments)?