How to correctly curry a function in JavaScript?

@Aadit,

I’m posting this because you shared a comment on my answer to To “combine” functions in javascript in a functional way? I didn’t specifically cover currying in that post because it’s a very contentious topic and not really a can of worms I wanted to open there.

I’d be wary using the phrasing “how to correctly curry” when you seem to be adding your own sugar and conveniences into your implementation.

Anyway, all of that aside, I truly don’t intend for this to be an argumentative/combative post. I’d like to be able to have an open, friendly discussion about currying in JavaScript while emphasizing some of the differences between our approaches.

Without further ado…

To clarify:

Given f is a function and f.length is n. Let curry(f) be g. We call g with m arguments. What should happen? You say:

1. If m === 0 then just return g.
2. If m < n then partially apply f to the m new arguments, and return a new curried function which accepts the remaining n - m arguments.
3. If m === n then apply f to the m arguments. If the result is a function then curry the result. Finally, return the result.
4. If m > n then apply f to the first n arguments. If the result is a function then curry the result. Finally, apply the result to the remaining m - n arguments and return the new result.

Let’s see a code example of what @Aadit M Shah‘s code actually does

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

var z = add(1, 2, 3);
console.log(z(4)); // 10

There are two things happening here:

1. You’re attempting to support calling curried functions with variadic arguments.
2. You’re automatically currying returned functions

I don’t believe there’s a lot of room for debate here, but people seem to miss what currying actually is

via: Wikipedia
In mathematics and computer science, currying is the technique of translating the evaluation of a function that takes multiple arguments (or a tuple of arguments) into evaluating a sequence of functions, each with a single argument…

I’m bolding that last bit, because it’s so important; each function in the sequence only takes a single argument; not variadic (0, 1, or more) arguments like you suggest.

You mention haskell in your post, too, so I assume you know that Haskell has no such thing as functions that take more than one argument. (Note: a function that takes a tuple is still just a function that takes one argument, a single tuple). The reasons for this are profound and afford you a flexibility in expressiveness not afforded to you by functions with variadic arguments.

So let’s re-ask that original question: What should happen?

Well, it’s simple when each function only accepts 1 argument. At any time, if more than 1 argument is given, they’re just dropped.

function id(x) {
  return x;
}

What happens when we call id(1,2,3,4)? Of course we only get the 1 back and 2,3,4 are completely disregarded. This is:

1. how JavaScript works
2. how Wikipedia says currying should work
3. how we should implement our own curry solution

Before we go further, I’m going to use ES6-style arrow functions but I will also include the ES5 equivalent at the bottom of this post. (Probably later tonight.)

another currying technique

In this approach, we write a curry function that continuously returns single-parameter functions until all arguments have been specified

As a result of this implementation we have 6 multi-purpose functions.

// no nonsense curry
const curry = f => {
  const aux = (n, xs) =>
    n === 0 ? f (...xs) : x => aux (n - 1, [...xs, x])
  return aux (f.length, [])
}
   
// demo
let sum3 = curry(function(x,y,z) {
  return x + y + z;
});
    
console.log (sum3 (3) (5) (-1)); // 7

OK, so we’ve seen a curry technique that is implemented using a simple auxiliary loop. It has no dependencies and a declarative definition that is under 5 lines of code. It allows functions to be partially applied, 1 argument at a time, just as a curried function is supposed to work.

No magic, no unforeseen auto-currying, no other unforeseen consequences.

But what really is the point of currying anyway?

Well, as it turns out, I don’t really curry functions that I write. As you can see below, I generally define all of my reusable functions in curried form. So really, you only need curry when you want to interface with some functions that you don’t have control over, perhaps coming from a lib or something; some of which might have variadic interfaces!

I present curryN

// the more versatile, curryN
const curryN = n => f => {
  const aux = (n, xs) =>
    n === 0 ? f (...xs) : x => aux (n - 1, [...xs, x])
  return aux (n, [])
};

// curry derived from curryN
const curry = f => curryN (f.length) (f);

// some caveman function
let sumN = function() {
  return [].slice.call(arguments).reduce(function(a, b) {
    return a + b;
  });
};

// curry a fixed number of arguments
let g = curryN (5) (sumN);
console.log (g (1) (2) (3) (4) (5)); // 15

To curry or not to curry? That is the question

We’ll write some examples where our functions are all in curried form. Functions will be kept extremely simple. Each with 1 parameter, and each with a single return expression.

// composing two functions
const comp = f => g => x => f (g (x))
const mod  = y => x => x % y
const eq   = y => x => x === y
const odd  = comp (eq (1)) (mod (2))

console.log (odd(1)) // true
console.log (odd(2)) // false

Your countWhere function

// comp :: (b -> c) -> (a -> b) -> (a -> c)
const comp = f => g => x =>
  f(g(x))

// mod :: Int -> Int -> Int
const mod = x => y =>
  y % x

// type Comparable = Number | String
// eq :: Comparable -> Comparable -> Boolean
const eq = x => y =>
  y === x

// odd :: Int -> Boolean
const odd =
  comp (eq(1)) (mod(2))

// reduce :: (b -> a -> b) -> b -> ([a]) -> b
const reduce = f => y => ([x,...xs]) =>
  x === undefined ? y : reduce (f) (f(y)(x)) (xs)

// filter :: (a -> Boolean) -> [a] -> [a]
const filter = f =>
  reduce (acc => x => f (x) ? [...acc,x] : acc) ([])

// length :: [a] -> Int
const length = x =>
  x.length

// countWhere :: (a -> Boolean) -> [a] -> Int
const countWhere = f =>
  comp (length) (filter(f));

console.log (countWhere (odd) ([1,2,3,4,5]))
// 3

Remarks

So to curry or not to curry?

// to curry
const add3 = curry((a, b, c) =>
  a + b + c
)

// not to curry
const add3 = a => b => c =>
 a + b + c

With ES6 arrow functions being the go-to choice for today’s JavaScripter, I think the choice to manually curry your functions is a no-brainer. It’s actually shorter and has less overhead to just write it out in curried form.

That said, you’re still going to be interfacing with libs that do not offer curried forms of the functions they expose. For this situation, I’d recommend

• curry and curryN (defined above)
• partial (as defined here)

@Iven,

Your curryN implementation is very nice. This section exists solely for you.

const U = f=> f (f)
const Y = U (h=> f=> f(x=> h (h) (f) (x)))

const curryN = Y (h=> xs=> n=> f=>
  n === 0 ? f(...xs) : x=> h ([...xs, x]) (n-1) (f)
) ([])

const curry = f=> curryN (f.length) (f)

const add3 = curry ((x,y,z)=> x + y + z)

console .log (add3 (3) (6) (9))