How Does Jquery Chain Functions While Still Returning An Array?

Having read How does jQuery accomplish chaining of commands? and how does jquery chaining work? I am still left to wonder how something like this is possible in jQuery var a = $('

Solution 1:

Using the jQuery source code viewer site the definition for $ (http://james.padolsey.com/jquery/#v=1.9.1&fn=$) is as follows:

$

function (selector, context) {
    // The jQuery object is actually just the init constructor 'enhanced'returnnew jQuery.fn.init(selector, context, rootjQuery);
}

Notice it is returning from init(). Inside of init (http://james.padolsey.com/jquery/#v=1.9.1&fn=init) we see that at the end it calls a function called makeArray:

init

function (selector, context, rootjQuery) {
    var match, elem;

// rest of function defintionreturn jQuery.makeArray(selector, this);
}

Which brings us to the answer to the question 'How does jQuery chain functions while still returning an array?'

makeArray (http://james.padolsey.com/jquery/#v=1.9.1&fn=jQuery.makeArray) looks like this:

makeArray

function (arr, results) {
    var ret = results || [];

    if (arr != null) {
        if (isArraylike(Object(arr))) {
            jQuery.merge(ret, typeof arr === "string" ? [arr] : arr);
        } else {
            core_push.call(ret, arr);
        }
    }

    return ret;
}