JavaScript curry

Question

I`m a newbie at JavaScript trying to understand this tutorial about currying from Oreilly JavaScript Cookbook.

Could someone be kind enough to explain this program i

Answer 1

Squeegy posted a good breakdown, but I figured I'd add mine too.

//Things to note, 'arguments' is a special variable in javascript that holds 
//an array like object full of all the things passed into a function.
//You can test this out with a function like this:
//var alertArgs = function(){alert(arguments);};

function curry(fn, scope) {
    //Either use the passed in 'scope' object, or the window object as your scope
    scope = scope || window;
    //Create a new array for storing the arguments passed into this function
    var args = [];
    //Loop through the extra arguments (we start at '2' because the first two
    //arguments were stored in `fn` and `scope` respectively.
    //We store these in the temporary 'args' array.
    //(in the example, args will end up looking like: [3.0, 4.0])
    for (var i = 2, len = arguments.length; i < len; ++i) {
        args.push(arguments[i]);
    }
    //We return the 'curried' function
    return function() {
        //This array is not used. I assume it is an error.
        var args2 = [];
        //We now have a new set of arguments, passed in to the curried function
        //We loop through these new arguments, (in the example, 6.42 and 8.0)
        //and add them to the arguments we have already saved. In the end, we have
        //the args array looking like: [3.0, 4.0, 6.42, 8.0]
        for (var i = 0; i < arguments.length; i++) {
            args.push(arguments[i]);
        }
        //This line isn't needed, because args2 is always blank.
        var argstotal = args.concat(args2);

        //Finally we call the function, passing in the full array of arguments
        return fn.apply(scope, argstotal);
    };
}

//This function takes 4 arguments
function diffPoint(x1, y1, x2, y2) {
    return [Math.abs(x2 - x1), Math.abs(y2 - y1)];
}

//We partially apply the first 2 arguments, so x1 is always 3.0, 
//and y1 is always 4.0
var diffOrigin = curry(diffPoint, null, 3.0, 4.0);

//We can now call 'diffPoint' indirectly, without having to specify 
//3.0, 4.0 as the first 2 arguments.
var newPt = diffOrigin(6.42, 8.0); //produces array with 3