Puzzle: JS Function that returns itself until there are no arguments

Question

I\'m trying to solve a puzzle, and am at my wit\'s end trying to figure it out.

I\'m supposed to make a function that works like this:

add(1);                


        

Answer 1

Here's a variation using bind:

var add = function _add(a, b) {
    var boundAdd = _add.bind(null, a + b);

    boundAdd.valueOf = function() { 
        return a + b; 
    }

    return boundAdd;
}.bind(null, 0);

We're taking advantage of a feature of bind that lets us set default arguments on the function we're binding to. From the docs:

bind() also accepts leading default arguments to provide to the target function when the bound function is called.

So, _add acts as a sort of master function which takes two parameters a and b. It returns a new function boundAdd which is created by binding the original _add function's a parameter to a + b; it also has an overridden valueOf function which returns a + b (the valueOf function was explained quite well in @RobG's answer).

To get the initial add function, we bind _add's a parameter to 0.

Then, when add(1) is called, a = 0 (from our initial bind call) and b = 1 (passed argument). It returns a new function where a = 1 (bound to a + b).

If we then call that function with (2), that will set b = 2 and it'll return a new function where a = 3.

If we then call that function with (3), that will set b = 3 and it'll return a new function where a = 6.

And so on until valueOf is called, at which point it'll return a + b. Which, after add(1)(2)(3), would be 3 + 3.