callback, err: is not a function

Question

So I am going back to some of the basics and found that I am having trouble with the simplest of things. I have a callback that is throwing me an error, I suspect this is a hoisting thing, but have been googling and can't seem to find what I am looking for.

function add (a,b){return a+b;}
function multiply (a,b){return a*b;}


function applyFn(callback) {
    return callback;
};

var f = applyFn(add);

console.log(f); //logs => function add(a,b){return a+b;}

f(3)(4); // returns => Uncaught TypeError: f(...) is not a function(…)

I know that var f gets hoisted to the top with a value of undefined, while the named functions are hoisted intact. When I console log console.log(typeof f) it returns function - so I am a little confused about what is going on...am I on the right track, or is it a different issue completely?

EDIT: for more clarity the applyFn was supposed to be able to use either the add or multiplyfunction, and so it would look more like this:

function add (a,b){return a+b;}
function multiply (a,b){return a*b;}

//I needed to write this function, but nothing else
function applyFn(callback) {
    return function(){};
};

var f = applyFn(add);
var x = applyFn(multiply)

f(3)(4);
x(6)(8);

Answer 1

To answer your question, you need to look at what applyFn returns. It takes a function, and simply returns that function. If you assign what it returns, the variable that you assigned to is now simply a reference to the function you originally passed in. In this case, that's the function add and it simply adds two numbers together. When you call f, you're really calling add, which means 1. it needs two arguments and 2. it's going to return the result of adding the two arguments, not another function that you can then call. If your goal is to do currying, then your function needs to return another function, so that it can be invoked:

function adder(num1) {
  return function (num2) {
    return num1 + num2;
  };
}

adder(2)(3);

Here is what I think you're looking for:

function applyFn(func) {
  return function (x) {
    return function (y) {
      return func(x, y);
    };
  };
}

This implementation of applyFn should do what you expect with add and multiply

Answer 2

your code is not failing at f(3), but at the second call. f(3) is not a function, so f(3)(4) fails.
applyFn should something be more like this:

var applyFn = fn => a => b => fn(a,b);
//aka.
function applyFn(fn){
    return function(a){
        return function(b){
            return fn(a, b);
        }
    }
}

but this is a pretty limited implementation, since it only works on functions with exactly two arguments. Better would be something more dynamically, like:

function _curry(n, f, a){
    var c = function(...args){
        var b = a.concat(args);
        return b.length < n? _curry(n, f, b): f.apply(this, b);
    }
    Object.defineProperty(c, "length", { configurable: true, value: Math.max(0, n-a.length) });
    return c;
}

function curry(n, f, a){
    if(typeof n === "function") a=f, f=n, n=f.length;
    return _curry(n, f, Array.from(Object(a)));
}

var add = curry(function(a, b){ return a + b });

console.log(add, add(3), add(3)(4), add(3, 4));

but on the other hand, the add and multiply-methods don't make sense, the way they are implemented at the moment.
I would never favor add(foo)(bar) or even add(foo, bar) over foo + bar.

the only place where they make sense is in a context like:

arr.map( add(3) );
//or
promise.then( add(3) );

so why not implement them directly as

var add = b => a => a+b;
var multiply = b => a => a*b;

and skip the part with applyFn?

//why b=>a instead of a=>b? consistency:
var subtract = b => a => a-b;
//so that subtract(3) also does exactly what you would think it does

Answer 3

Your syntax for executing f is wrong. f(3)(4) should be f(3, 4). Checked it and it works. Check this jsfiddle