Functional JavaScript: Closure & Recursion. Why does this fail?

Question

desired functionality:

mult(3);
 //(x) => 3 * mult(x)
mult(3)(4);
 //(x) => 3 * (4 * mult(x))
mult(3)(4)();
//12

attempt:

function mult(x){
    if(x === undefined){
        return 1;
    }else{
        return (y => x * mult(y));
    }
}

result:

mult(3)
//y => x * mult(y)
//looks pretty good

mult(3)()
//3
//exactly what I want so far. 

mult(3)(4)()
//Uncaught TypeError: mult(...)(...) is not a function

sure enough,

mult(3)(4)
//NaN

Yet mult(3) looks good and typeof mult(3) === "function".

What gives? Can I not be this fancy in JS? Any why not?

Think about what happens when you call mult(3)(4). You are calling (y => 3 * mult(y)), where y is 4. What does mult(4) return? A function. What does 3 * <a function> return? NaN. — Felix Kling
– Felix Kling, Commented Feb 19, 2018 at 19:11
Probably take a look at the answers here: JS Curry function with Recursion — Herohtar
– Herohtar, Commented Feb 19, 2018 at 19:14

melpomene · Accepted Answer · 2018-02-19 19:20:34Z

2

In

mult(3)(4)

mult(3) yields y => 3 * mult(y).

Thus

(y => 3 * mult(y))(4)

becomes

3 * mult(4)

mult(4) yields y => 4 * mult(y).

3 * (y => 4 * mult(y))

is nonsense because you're trying to multiply 3 by a function. This is why you're getting NaN here, and NaN can't itself be further applied.

Possible solution:

function mkmult(acc) {
    return x =>
        x === undefined
            ? acc
            : mkmult(acc * x);
}

const mult = mkmult(1);

console.log(mult(3)());
console.log(mult(3)(4)());
console.log(mult(3)(4)(5)());

edited Feb 19, 2018 at 19:20

answered Feb 19, 2018 at 19:12

melpomene

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5 Comments

Patrick Roberts Over a year ago

Wait actually, my comment was correct. Why are you doing const mult = mkmult(1) instead of just defining it as mult?

melpomene Over a year ago

@PatrickRoberts Because otherwise mult() doesn't give you 1.

Patrick Roberts Over a year ago

That wasn't a requirement though?

melpomene Over a year ago

@PatrickRoberts It's in OP's code. And it makes sense anyway: We can generalize our function to something like a left fold: function fold_left(acc, op) { return x => x === undefined ? acc : fold_left(op(acc, x), op); } and in this case acc and x could even have different types. What you're suggesting is akin to foldl1, which 1) forces acc and x to have the same type and 2) doesn't work on empty lists.

Paul Over a year ago

@melpomene Thank you for showing me the error of my ways. You are my light in the darkness; the lighthouse of truth. Tonight, I will rest easy.

user9366559 · Accepted Answer · 2018-02-19 19:20:28Z

2

If you think about it, the mult should never be passed undefined since it represents the "current" or left side of the multiplication. The inner function receiving y is the one that should handle providing the result.

function mult(x){
    return y => {
        if (y === undefined) return x;
        else return mult(x * y);
    };
}

console.log(mult(3))
console.log(mult(3)())
console.log(mult(3)(4)())

answered Feb 19, 2018 at 19:20

user9366559

2 Comments

Paul Over a year ago

It's beautiful. Utter beauty. I went with @melpomene 's answer because it clearly showed the error of my sinful ways in addition to how to provide the functionality I desire.

Mulan Over a year ago

@doodlemeister, doesn’t work for empty set mult () due to improper assumptions about x

Mulan · Accepted Answer · 2018-02-19 20:54:17Z

2

Here's another way that uses a secondary parameter with a default value

base case – x is undefined, return the accumulator acc
inductive case - x is not undefined – return a new lambda that asks for the next input y and recur using y and acc * x

const mult = (x, acc = 1) =>
  x === undefined
    ? acc
    : y => mult (y, acc * x)

console.log (mult ())             // 1
console.log (mult (2) ())         // 2
console.log (mult (2) (3) ())     // 6
console.log (mult (2) (3) (4) ()) // 24

Personally, I would use a known (explicit) empty value rather than relying upon the implicit undefined

const Empty =
  Symbol ()

const mult = (x = Empty, acc = 1) =>
  x === Empty
    ? acc
    : y => mult (y, acc * x)

console.log (mult ())             // 1
console.log (mult (2) ())         // 2
console.log (mult (2) (3) ())     // 6
console.log (mult (2) (3) (4) ()) // 24

Really though, mult as it is defined above is a toy. A more practical implementation would use variadic interface instead of the weirder currying with () used to signal a return value

// fixed arity
const mult = (x, y) =>
  x * y

// variadic interface
const multiply = (x = 1, ...xs) =>
  xs.reduce (mult, x)
  
console.log (multiply ())        // 1
console.log (multiply (2))       // 2
console.log (multiply (2, 3))    // 6
console.log (multiply (2, 3, 4)) // 24

edited Feb 19, 2018 at 20:54

answered Feb 19, 2018 at 19:39

Mulan

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2 Comments

Paul Over a year ago

I think I'm in love.

Mulan Over a year ago

@Paul all of my answers on recursion are answered from a functional programming perspective. I hope you find some of them helpful ^_^

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