Simple way to recursively curry javascript functions http://jsfiddle.net/amatiasq/osrsomq0/

curry.js

/**
 * @param {Function} fn Function to curry.
 * @param {Number} lenght The arguments required to invoke the function. Optional. By default is fn.length
 * @returns {Function} The currified function.
 */
function curry(fn, length) {
  length = length || fn.length;
  return function currified() {
    var args = [].slice.call(arguments);

    if (args.length === 0)
      return currified;

    if (args.length >= length)
      return fn.apply(this, args);

    var child = fn.bind.apply(fn, [this].concat(args));
    return curry(child, length - args.length);
  };
}

usage.js

var log = console.log.bind(console);

// empty calls are ignored
curry(log, 1) ()()()()(1); // Logs: 1

// extra args are passed
curry(log, 1) (1, 2, 3, 4); // Logs: 1 2 3 4

// recursive currification
curry(log, 5) (1)(2)(3)(4)(5); // 1 2 3 4 5

// works with all combination
curry(log, 4) (1, 2, 3, 4); // Logs: 1 2 3 4
curry(log, 4) (1, 2, 3)(4); // Logs: 1 2 3 4
curry(log, 4) (1, 2)(3, 4); // Logs: 1 2 3 4
curry(log, 4) (1)(2, 3, 4); // Logs: 1 2 3 4
curry(log, 4) (1, 2)(3)(4); // Logs: 1 2 3 4
curry(log, 4) (1)(2)(3, 4); // Logs: 1 2 3 4
curry(log, 4) (1)(2, 3)(4); // Logs: 1 2 3 4
curry(log, 4) (1)(2)(3)(4); // Logs: 1 2 3 4

// function is not invoked until argument lenght is reached
console.log(curry(log, 4) (1)(2)(3)); // returns function currified() { ... }

// when argument length is reached function is invoked and its result is returned
try {
  curry(log, 4) (1)(2)(3)(4)(5); // logs 1 2 3 4 & throws undefined is not a function
} catch(err) {
  console.log(err);
}

function sum(a, b) {
  return a + b;
}

// argument length detected automatically
console.log(curry(sum)(1)(2));