## Emulating lisp list operations in js - javascript

What would be the proper way to define the cons, car, and cdr functions if we were to use javascript and functional programming? So far I have:
// CAR = λxy.x
// CDR = λxy.y
// CONS = λxy => ?
const car = a => b => a;
const cdr = a => b => b;
const cons = f => a => b;
let pair = cons(3)(4);
console.log(pair(car));
console.log(pair(cdr));
But my issue is pair(car) seems like an odd way to invoke the function. It seems like car(pair) seems like a better way but I'm not sure how to save the pair so it can be passed like that. How would that be done?

You can write them using continuations -
const cons = (a,b) => k => k(a,b)
const car = k => k((a,b) => a)
const cdr = k => k((a,b) => b)
const l = cons(1, cons(2, cons(3, null)))
console.log(car(l), car(cdr(l)), car(cdr(cdr(l))), cdr(cdr(cdr(l))))
// 1 2 3 null
Define more like list, toString, map, and filter -
const cons = (a,b) => k => k(a,b)
const car = k => k((a,b) => a)
const cdr = k => k((a,b) => b)
const list = (v, ...vs) => v == null ? null : cons(v, list(...vs))
const toString = l => l == null ? "Ø" : car(l) + "->" + toString(cdr(l))
console.log(toString(list(1,2,3,4,5)))
// 1->2->3->4->5->Ø
const square = x => x * x
const map = (f, l) =>
l == null
? null
: cons(f(car(l)), map(f, cdr(l)))
console.log(toString(map(square, list(1,2,3,4,5))))
// 1->4->9->16->25->Ø
const isOdd = x => x & 1
const filter = (f, l) =>
l == null
? null
: Boolean(f(car(l)))
? cons(car(l), filter(f, cdr(l)))
: filter(f, cdr(l))
console.log(toString(filter(isOdd, map(square, list(1,2,3,4,5)))))
// 1->9->25->Ø
Note you could just as easily write cons, car, and cdr abstractions using an array. Lambda is more fun but anything that fulfills the contract is acceptable. Being able to change the underlying representation like this and not require changes in other parts of your program is what makes data abstraction a powerful technique.
const cons = (a,b) => [a,b]
const car = k => k[0]
const cdr = k => k[1]
const l = cons(1, cons(2, cons(3, null)))
console.log(car(l), car(cdr(l)), car(cdr(cdr(l))), cdr(cdr(cdr(l))))
// 1 2 3 null

The following would be one way to define it, where car and cdr take one argument (a pair) instead of two:
// CAR = λxy.x or λp.p[0] where p is xy and p[0] means the first argument (x)
// CDR = λxy.y or λp.p[1] where p is xy and p[1] means the second argument (y)
// CONS = λxyf.xyf -- basically just saving x and y in a closure and deferring a function call
const first = a => b => a;
const second = a => b => b;
const getFirstFromPair = p => p(first), car = getFirstFromPair;
const getSecondFromPair = p => p(second), cdr = getSecondFromPair;
const cons = a => b => f => f(a)(b);
let pair = cons(3)(4);
console.log(car(pair));
console.log(cdr(pair));
Or, if we allow a pair as a single argument, such as (1,2):
const cons = (a,b) => f => f(a,b); // f is like a fake function call to do encapsulation
const car = f => f((a,b) => a); // and we have to un-do the function call here by passing it the first
const cdr = f => f((a,b) => b);
console.log(cdr(cons(1,2)));

## Related

### How to use the Trampoline type as the base monad of a transformer more efficiently?

I have an array transformer type that exhibits interleaved effect layers to ensure a lawful effect implementation. You can easily read the structure from the type's of operation const arrOfT = of => x => of([of(x)]). The type implements an effectful fold as its basic operation. I use a left fold, because the underlying array type is inherently strict: const arrFoldT = chain => f => init => mmx => chain(mmx) (mx => { const go = (acc, i) => i === mx.length ? acc : chain(mx[i]) (x => go(f(acc) (x), i + 1)) // ^^^^^^^^^^^^^^^^^^^^^ non-tail call position return go(init, 0); }); As you can see the implementation is not stack safe. However, stack safety is just another computational effect that can be encoded through a monad. I implemented one for the Trampoline type: const monadRec = o => { while (o.tag === "Chain") o = o.f(o.x); return o.tag === "Of" ? o.x : _throw(new TypeError("unknown case")); }; const recChain = mx => fm => mx.tag === "Chain" ? Chain(mx.x) (x => recChain(mx.f(x)) (fm)) : mx.tag === "Of" ? fm(mx.x) : _throw(new TypeError("unknown case")); const Chain = x => f => ({tag: "Chain", f, x}); const Of = x => ({tag: "Of", x}); While the implementations are straightforward the application is not. I am pretty sure I am applying it all wrong: const mmx = Of( Array(1e5) .fill(Chain(1) (x => Of(x)))); // ^^^^^^^^^^^^ redundant continuation const main = arrFoldT(recChain) (acc => y => recMap(x => x + y) (acc)) (Of(0)) (mmx); monadRec(main); // 100000 I need to use Chain when creating the large effectful array, because Of signals the the control flow to break out of the trampoline. With Chain on the other hand I have to specifiy a redundant continuation. My first idea was to flip Chain's arguments and rely on partial application, but this doesn't work with the current implemenetation. Is there a way to use the type more efficiently? Here is a working example: // ARRAYT const arrFoldT = chain => f => init => mmx => chain(mmx) (mx => { const go = (acc, i) => i === mx.length ? acc : chain(mx[i]) (x => go(f(acc) (x), i + 1)) return go(init, 0); }); // TRAMPOLINE const monadRec = o => { while (o.tag === "Chain") o = o.f(o.x); return o.tag === "Of" ? o.x : _throw(new TypeError("unknown case")); }; const Chain = x => f => ({tag: "Chain", f, x}); const Of = x => ({tag: "Of", x}); // Functor const recMap = f => tx => Of(f(tx.x)); // Monad const recChain = mx => fm => mx.tag === "Chain" ? Chain(mx.x) (x => recChain(mx.f(x)) (fm)) : mx.tag === "Of" ? fm(mx.x) : _throw(new TypeError("unknown case")); const recOf = Of; // MAIN const mmx = Of( Array(1e5) .fill(Chain(1) (x => Of(x)))); const main = arrFoldT(recChain) (acc => y => recMap(x => x + y) (acc)) (Of(0)) (mmx); console.log( monadRec(main)); // 100000

First, the definition of your array monad transformer is wrong. ArrayT m a = m (Array (m a)) The above type definition does not correctly interleave the underlying monad. Following is an example value of the above data type. of([of(1), of(2), of(3)]) There are several problems with this data type. There is no effect for the end of the array. The effects are not ordered. Hence, they can be executed in any order. The underlying monad wraps the individual elements as well as the entire array. This is just wrong. Following is an example value of the correct array monad transformer type. of([1, of([2, of([3, of([])])])]) Note that. There is an effect for the end of the array. The effects are ordered. This is because the data type is defined recursively. The underlying monad wraps the individual steps of the array. It doesn't wrap the entire array again. Now, I understand why you want to define ArrayT m a = m (Array (m a)). If m = Identity then you get back an actual Array a, which supports random access of elements. of([of(1), of(2), of(3)]) === [1, 2, 3] On the other hand, the recursive array monad transformer type returns a linked list when m = Identity. of([1, of([2, of([3, of([])])])]) === [1, [2, [3, []]]] However, there's no way to create a lawful array monad transformer type which also returns an actual array when the underlying monad is Identity. This is because monad transformers are inherently algebraic data structures, and arrays are not algebraic. The closest you can get is by defining ArrayT m a = Array (m a). However, this would only satisfy the monad laws when the underlying monad is commutative. Just remember, when defining a monad transformer data type. The underlying monad must wrap at most one value at a time. The underlying monad must be nested, to correctly order and interleave effects. Coming back, the Trampoline monad is just the Free monad. We can define it as follows. // pure : a -> Free a const pure = value => ({ constructor: pure, value }); // bind : Free a -> (a -> Free b) -> Free b const bind = monad => arrow => ({ constructor: bind, monad, arrow }); // thunk : (() -> Free a) -> Free a const thunk = eval => ({ constructor: thunk, eval }); // MonadFree : Monad Free const MonadFree = { pure, bind }; // evaluate : Free a -> a const evaluate = expression => { let expr = expression; let stack = null; while (true) { switch (expr.constructor) { case pure: if (stack === null) return expr.value; expr = stack.arrow(expr.value); stack = stack.stack; break; case bind: stack = { arrow: expr.arrow, stack }; expr = expr.monad; break; case thunk: expr = expr.eval(); } } }; I'll also copy my implementation of the array monad transformer from my previous answer. // Step m a = null | { head : a, tail : ListT m a } // ListT m a = m (Step m a) // nil : Monad m -> ListT m a const nil = M => M.pure(null); // cons : Monad m -> a -> ListT m a -> ListT m a const cons = M => head => tail => M.pure({ head, tail }); // foldr : Monad m -> (a -> m b -> m b) -> m b -> ListT m a -> m b const foldr = M => f => a => m => M.bind(m)(step => step ? f(step.head)(foldr(M)(f)(a)(step.tail)) : a); Thus, when the underlying monad is Free then the operations are stack safe. // replicate :: Number -> a -> ListT Free a const replicate = n => x => n ? cons(MonadFree)(x)(thunk(() => replicate(n - 1)(x))) : nil(MonadFree); // map : (a -> b) -> Free a -> Free b const map = f => m => bind(m)(x => pure(f(x))); // add : Number -> Free Number -> Free Number const add = x => map(y => x + y); // result : Free Number const result = foldr(MonadFree)(add)(pure(0))(replicate(1000000)(1)); console.log(evaluate(result)); // 1000000 Putting it all together. // pure : a -> Free a const pure = value => ({ constructor: pure, value }); // bind : Free a -> (a -> Free b) -> Free b const bind = monad => arrow => ({ constructor: bind, monad, arrow }); // thunk : (() -> Free a) -> Free a const thunk = eval => ({ constructor: thunk, eval }); // MonadFree : Monad Free const MonadFree = { pure, bind }; // evaluate : Free a -> a const evaluate = expression => { let expr = expression; let stack = null; while (true) { switch (expr.constructor) { case pure: if (stack === null) return expr.value; expr = stack.arrow(expr.value); stack = stack.stack; break; case bind: stack = { arrow: expr.arrow, stack }; expr = expr.monad; break; case thunk: expr = expr.eval(); } } }; // Step m a = null | { head : a, tail : ListT m a } // ListT m a = m (Step m a) // nil : Monad m -> ListT m a const nil = M => M.pure(null); // cons : Monad m -> a -> ListT m a -> ListT m a const cons = M => head => tail => M.pure({ head, tail }); // foldr : Monad m -> (a -> m b -> m b) -> m b -> ListT m a -> m b const foldr = M => f => a => m => M.bind(m)(step => step ? f(step.head)(foldr(M)(f)(a)(step.tail)) : a); // replicate :: Number -> a -> ListT Free a const replicate = n => x => n ? cons(MonadFree)(x)(thunk(() => replicate(n - 1)(x))) : nil(MonadFree); // map : (a -> b) -> Free a -> Free b const map = f => m => bind(m)(x => pure(f(x))); // add : Number -> Free Number -> Free Number const add = x => map(y => x + y); // result : Free Number const result = foldr(MonadFree)(add)(pure(0))(replicate(1000000)(1)); console.log(evaluate(result)); // 1000000

### How to share intermediate results of continuations?

Please note that even though the example in this question is encoded in Javascript, the underlying concepts are common in Haskell and I while I prefer to express myself in Javascript I also appreciate answers in Haskell. In Javascript I use CPS to handle asynchronous computations according to monadic principles. For the sake of simplicity, however, I will use the normal continuation monad for this question. As soon as my continuation compositions grow, I keep finding myself in a situation where I need access to intermediate results of these compositions. Since Javascript is imperative it is easy to store such results in variables and access them later. But since we're talking about continuations accessing intermediate results means calling functions and accessing them several times means a lot of re-evaluation. This seems to be well suited for memoization. But how can I memoize a function's return value if that very function doesn't return anything but merely calls its continuation (and btw. as I mentioned before I use asynchronous functions that also don't return anything in the current cycle of Javascript's event loop). It seems as if I have to extract the right continuation. Maybe this is possible with delimited continuations through shift/reset, but I don't know how to apply these combinators. This issue is probably not that hard to solve and I'm just confused by the magical land of continuation passing style...so please be indulgent with me. Here is a simplified example of Cont without memoization in Javascript: const taggedLog = tag => s => (console.log(tag, s), s); const id = x => x; const Cont = k => ({ runCont: k, [Symbol.toStringTag]: "Cont" }); const contAp = tf => tx => Cont(k => tf.runCont(f => tx.runCont(x => k(f(x))))); const contLiftA2 = f => tx => ty => contAp(contMap(f) (tx)) (ty); const contOf = x => Cont(k => k(x)); const contMap = f => tx => Cont(k => tx.runCont(x => k(f(x)))); const contReset = tx => // delimited continuations contOf(tx.runCont(id)); const contShift = f => // delimited continuations Cont(k => f(k).runCont(id)); const inc = contMap(x => taggedLog("eval inc") (x + 1)); const inc2 = inc(contOf(2)); const inc3 = inc(contOf(3)); const add = contLiftA2(x => y => taggedLog("eval add") (x + y)); const mul = contLiftA2(x => y => taggedLog("eval mul") (x * y)); const intermediateResult = add(inc2) (inc3); mul(intermediateResult) (intermediateResult).runCont(id); /* should only log four lines: eval inc 3 eval inc 4 eval add 7 eval mul 49 */

Your problems seems to be that your Cont has no monad implementation yet. With that, it's totally simple to access previous results - they're just in scope (as constants) of nested continuation callbacks: const contChain = tx => f => Cont(k => tx.runCont(r => f(r).runCont(k))); contChain( add(inc2) (inc3), intermediateResult => { const intermediateCont = contOf(intermediateResult); return mul(intermediateCont) (intermediateCont); }).runCont(id); (Of course it's a little weird that all your functions are already lifted and take Cont values as arguments - they shouldn't do that and simply be functions that return Cont values) Your code in Haskell: import Control.Monad.Cont import Control.Applicative let inc = liftA (+1) let inc2 = inc $ return 2 let inc3 = inc $ return 3 let add = liftA2 (+) let mul = liftA2 (*) (`runCont` id) $ add inc2 inc3 >>= \intermediateResult -> let intermediateCont = return intermediateResult in mul intermediateCont intermediateCont -- 49 {- or with do notation: -} (`runCont` id) $ do intermediateResult <- add inc2 inc3 let intermediateCont = return intermediateResult mul intermediateCont intermediateCont -- 49 (I haven't used monad transformers to make a taggedLog side effect)

It seems that I can't avoid getting impure to obtain the desired behavior. The impurity is only local though, because I just replace the continuation chain with its result value. I can do this without changing the behavior of my program, because this is exactly what referential transparency guarantees us. Here is the transformation of the Cont constructor: const Cont = k => ({ runCont: k, [Symbol.toStringTag]: "Cont" }); // becomes const Cont = k => thisify(o => { // A o.runCont = (res, rej) => k(x => { // B o.runCont = l => l(x); // C return res(x); // D }, rej); // E o[Symbol.toStringTag] = "Cont"; return o; }); thisify in line A merely mimics this context, so that the Object to be constructed is aware of itself. Line B is the decisive change: Instead of just passing res to the continuation k I construct another lambda that stores the result x wrapped in a continuation under the runTask property of the current Task object (C), before it calls res with x (D). In case of an error rej is just applied to x, as usual (E). Here is the runnning example from above, now working as expected: const taggedLog = pre => s => (console.log(pre, s), s); const id = x => x; const thisify = f => f({}); // mimics this context const Cont = k => thisify(o => { o.runCont = (res, rej) => k(x => { o.runCont = l => l(x); return res(x); }, rej); o[Symbol.toStringTag] = "Cont"; return o; }); const contAp = tf => tx => Cont(k => tf.runCont(f => tx.runCont(x => k(f(x))))); const contLiftA2 = f => tx => ty => contAp(contMap(f) (tx)) (ty); const contOf = x => Cont(k => k(x)); const contMap = f => tx => Cont(k => tx.runCont(x => k(f(x)))); const inc = contMap(x => taggedLog("eval inc") (x + 1)); const inc2 = inc(contOf(2)); const inc3 = inc(contOf(3)); const add = contLiftA2(x => y => taggedLog("eval add") (x + y)); const mul = contLiftA2(x => y => taggedLog("eval mul") (x * y)); const intermediateResult = add(inc2) (inc3); mul(intermediateResult) (intermediateResult).runCont(id); /* should merely log eval inc 3 eval inc 4 eval add 7 eval add 49 */

### How to implement a coroutine based on multi-shot delimited continuations?

I recently implemented delimited continuations in CPS with reset/shift: // reset :: ((a -> a) -> a) -> (a -> r) -> r reset = k => f => f(k(id)); // shift :: ((a -> r) -> (r -> r) -> r) -> (a -> r) -> r shift = f => k => f(k) (id); Studying the theory I realized the following connections: reset ~ function* // scope of the generator function shift ~ yield reset ~ async // scope of the asyn function shift ~ await As far as I understand the theory, Javascript's generators are a asymmetric, stackless, one-shot and first class coroutines. asymmetric means that the called generator can only yield to its caller stackless means a generator cannot yield from within nested functions one-shot means that a generator can only resume from a specific position once first class means a generator object can be passed around like normal data Now I want to implement a coroutine based on reset/shift with the following traits: asymmetric stackful multi-shot first class When looking at the following contrived example const id = x => x; const mul = x => y => x * y; const add = x => y => x + y; const sub = x => y => x - y; const reset = k => f => f(k(id)); const shift = f => k => f(k) (id); const of = x => k => k(x); const lift2 = f => tx => ty => k => tx(x => ty(y => k(f(x) (y)))); const k0 = lift2(sub) (reset (lift2(add) (of(3)) (shift (k => of(mul(5) (2)))))) (of(1)); // 9 const k1 = lift2(sub) (reset (lift2(add) (of(3)) (shift (k => of(k(mul(5) (2))))))) (of(1)); // 12 const k2 = lift2(sub) (reset (lift2(add) (of(3)) (shift (k => of(k(k(mul(5) (2)))))))) (of(1)); // 15 console.log(k0(id)); console.log(k1(id)); console.log(k2(id)); it seems that reset/shift already meet the last two criteria, because delimited continuations are just first class, composable functions and I can invoke the continuation k as often as required. To answer the why, I want to bypass the following limitation in connection with the list monad. Are these assumptions correct? Even if I haven't make any mistakes so far I am overwhelmed by the complexity of the task at this point. I have no clue how to implement the desired coroutine or even where to begin. I didn't found an example implementation either. I don't expect a complete implementation but maybe some guidance to achieve my goal. Goal I want to bypass the following limitation of coroutines implemented by Javascript's generators: const arrMap = f => xs => xs.map(x => f(x)); const arrAp = fs => xs => fs.reduce((acc, f) => acc.concat(xs.map(x => f(x))), []); const arrChain = xs => fm => xs.reduce((acc, x) => acc.concat(fm(x)), []); const arrOf = x => [x]; const do_ = (of, chain) => it => { const loop = ({done, value}) => done ? value : chain(value) (x => loop(it.next(x))); return loop(it.next()); }; const z = function*() { const x = yield [1,2,3] return [x, x]; } console.log( arrChain([1,2,3]) (x => [x, x])); console.log( do_(arrOf, arrChain) (z()));

### Reversing the order of default arguments in JavaScript

I have the following recursive compose function: const compose = (f, n = 1) => n > 1 ? compose(compose(f), n - 1) : g => x => f(g(x)); const length = a => a.length; const filter = p => a => a.filter(p); const countWhere = compose(length, 2)(filter); const odd = n => n % 2 === 1; console.log(countWhere(odd)([1,2,3,4,5,6,7,8,9])); // 5 Now, what I'd like to do is flip the arguments of compose so that the default argument is first: const compose = (n = 1, f) => n > 1 ? // wishful thinking compose(n - 1, compose(f)) : // compose(f) is the same as compose(1, f) g => x => f(g(x)); const length = a => a.length; const filter = p => a => a.filter(p); const countWhere = compose(2, length)(filter); // I want to call it like this const odd = n => n % 2 === 1; console.log(countWhere(odd)([1,2,3,4,5,6,7,8,9])); // 5 What's the most elegant way to write such functions where the default arguments come first? Edit: I actually want to create the map and ap methods of functions of various arities, so that I can write: const length = a => a.length; const filter = p => a => a.filter(p); const countWhere = length.map(2, filter); // length <$> filter const pair = x => y => [x, y]; const add = x => y => x + y; const mul = x => y => x * y; const addAndMul = pair.map(2, add).ap(2, mul); // (,) <$> (+) <*> (*) Hence, I'd rather not curry the methods as Bergi suggested in his answer. For more information, read: Is implicit wrapping and unwrapping of newtypes in Haskell a sound idea?

I would recommend to just not overload your functions or use default parameters: const compose = n => f => n > 1 ? compose(n - 1)(composeOne(f)) : g => x => f(g(x)); const composeOne = compose(1); In this case you could probably also just inline it, as it seems composeOne wouldn't be called anywhere else: const compose = n => f => n > 1 ? compose(n - 1)(compose(1)(f)) : g => x => f(g(x)); Or even not do a recursive call at all, but always create the g => x => … lambda and transform it conditionally: const compose = n => f => { const l = g => x => f(g(x)); return n > 1 ? compose(n - 1)(l) : l; }; // same without temporary variables: const id = x => x; const compose = n => f => (n > 1 ? compose(n-1) : id)(g => x => f(g(x)))

What's the most elegant way to write such functions where the default arguments come first? Using only default initialisers requires some arcane hackery: function demo(n, f = [n, n = 1][0]) { console.log(n, f); } demo(2, "f"); // 2 f demo("g"); // 1 g console.log(demo.length) // 1 The most straightforward way would be destructuring with a conditional operator: function demo(...args) { const [n, f] = args.length < 2 ? [1, ...args] : args; console.log(n, f); } demo(2, "f"); // 2 f demo("g"); // 1 g console.log(demo.length) // 0 More in the spirit of "reversing the order of arguments" might be doing that literally: function demo(...args) { const [f, n = 1] = args.reverse(); console.log(n, f); } demo(2, "f"); // 2 f demo("g"); // 1 g console.log(demo.length) // 0 The latter two attempts have the drawback of requiring an extra declaration (preventing us from using concise arrow functions) and also don't reflect the actual number or required parameters in the .length.

### Extracting data from a function chain without arrays

This is an advanced topic of How to store data of a functional chain of Monoidal List? I am pretty sure we can somehow extract data from a function chain without using an array storing data. The basic structure is : L = a => L very simple, but this structure generates a list: L(1)(2)(3)(4)(5)(6)() This may be related to What is a DList? , but this structure strictly depends on function chain only. So, what is the way to pull out the whole values? Current achievement of mine merely pulling out the head and tail, and I don't know how to fix this. EDIT: I forgot to mention what I try to do is List.fold(f) / reduce(f) operation. So, if one choses f as Array.concat which means you can extract data as an array, but simply fold is not limited to array concatenation. and f can be add/sum etc. So, currently, so far, to visualize the internal behavior, in a sense, I write log as f. EDIT2 I must clarify more. The specification can be presented: const f = (a) => (b) => a + b;//binary operation A(a)(b)(f) = f(a)(b) // a + b A(a)(b)(c)(f) = f(f(a)(b))(c) // a + b + c So this is exactly (a b c).reduce(f) thing, and when f = (a) => (b) => a.concat(b) The result would be [a, b, c]. Array.concat is merely a member of generalized binary operations f. At first this challenge is easy for my skill, but turned out hard and felt it's better to ask smarter coder. Thanks. const A = a => { const B = b => (b === undefined) ? (() => { log("a " + a); return A(); })() : c => (c === undefined) ? (() => { log("b " + b); return B()(); })() : B; return B; }; A(1)(2)(3)(4)(5)(6)() function log(m) { console.log((m)); //IO return m; }; result: b 6 a 1 a undefined

Quite the series of questions you have here. Here's my take on it: We start with a way to construct lists nil is a constant which represents the empty list cons (x, list) constructs a new list with x added to the front of list // nil : List a const nil = (c, n) => n // cons : (a, List a) -> List a const cons = (x, y = nil) => (c, n) => c (y (c, n), x) // list : List Number const myList = cons (1, cons (2, cons (3, cons (4, nil)))) console.log (myList ((x, y) => x + y, 0)) // 10 And to satisfy your golfy variadic curried interface, here is autoCons const autoCons = (init, n) => { const loop = acc => (x, n) => isFunction (x) ? acc (x, n) : loop (cons (x, acc)) return loop (nil) (init, n) } const isFunction = f => f != null && f.constructor === Function && f.length === 2 const nil = (c, n) => n const cons = (x, y = nil) => (c, n) => c (y (c, n), x) console.log ( autoCons (1) ((x,y) => x + y, 0) // 1 , autoCons (1) (2) ((x,y) => x + y, 0) // 3 , autoCons (1) (2) (3) ((x,y) => x + y, 0) // 6 , autoCons (1) (2) (3) (4) ((x,y) => x + y, 0) // 10 ) Our encoding makes it possible to write other generic list functions, like isNil // isNil : List a -> Bool const isNil = l => l ((acc, _) => false, true) console.log ( isNil (autoCons (1)) // false , isNil (autoCons (1) (2)) // false , isNil (nil) // true ) Or like length // length : List a -> Int const length = l => l ((acc, _) => acc + 1, 0) console.log ( length (nil) // 0 , length (autoCons (1)) // 1 , length (autoCons (1) (2)) // 2 , length (autoCons (1) (2) (3)) // 3 ) Or nth which fetches the nth item in the list // nth : Int -> List a -> a const nth = n => l => l ( ([ i, res ], x) => i === n ? [ i + 1, x ] : [ i + 1, res] , [ 0, undefined ] ) [1] console.log ( nth (0) (autoCons ("A") ("B") ("C")) // "A" , nth (1) (autoCons ("A") ("B") ("C")) // "B" , nth (2) (autoCons ("A") ("B") ("C")) // "C" , nth (3) (autoCons ("A") ("B") ("C")) // undefined ) We can implement functions like map and filter for our list // map : (a -> b) -> List a -> List b const map = f => l => l ( (acc, x) => cons (f (x), acc) , nil ) // filter : (a -> Bool) -> List a -> List a const filter = f => l => l ( (acc, x) => f (x) ? cons (x, acc) : acc , nil ) We can even make a program using our list which takes a list as an argument // rcomp : (a -> b) -> (b -> c) -> a -> c const rcomp = (f, g) => x => g (f (x)) // main : List String -> String const main = letters => autoCons (map (x => x + x)) (filter (x => x !== "dd")) (map (x => x.toUpperCase())) (rcomp, x => x) (letters) ((x, y) => x + y, "") main (autoCons ("a") ("b") ("c") ("d") ("e")) // AABBCCEE Run the program in your browser below const nil = (c, n) => n const cons = (x, y = nil) => (c, n) => c (y (c, n), x) const isFunction = f => f != null && f.constructor === Function && f.length === 2 const autoCons = (init, n) => { const loop = acc => (x, n) => isFunction (x) ? acc (x, n) : loop (cons (x, acc)) return loop (nil) (init, n) } const map = f => l => l ( (acc, x) => cons (f (x), acc) , nil ) const filter = f => l => l ( (acc, x) => f (x) ? cons (x, acc) : acc , nil ) const rcomp = (f, g) => x => g (f (x)) const main = letters => autoCons (map (x => x + x)) (filter (x => x !== "dd")) (map (x => x.toUpperCase())) (rcomp, x => x) (letters) ((x, y) => x + y, "") console.log (main (autoCons ("a") ("b") ("c") ("d") ("e"))) // AABBCCEE Sorry, my bad Let's rewind and look at our initial List example // list : List Number const myList = cons (1, cons (2, cons (3, cons (4, nil)))) console.log ( myList ((x, y) => x + y, 0) // 10 ) We conceptualize myList as a list of numbers, but we contradict ourselves by calling myList (...) like a function. This is my fault. In trying to simplify the example, I crossed the barrier of abstraction. Let's look at the types of nil and cons – // nil : List a // cons : (a, List a) -> List a Given a list of type List a, how do we get a value of type a out? In the example above (repeated below) we cheat by calling myList as a function. This is internal knowledge that only the implementer of nil and cons should know // myList is a list, not a function... this is confusing... console.log ( myList ((x, y) => x + y, 0) // 10 ) If you look back at our original implementation of List, // nil : List a const nil = (c, n) => n // cons : (a, List a) -> List a const cons = (x, y = nil) => (c, n) => c (y (c, n), x) I also cheated you giving simplified type annotations like List a. What is List, exactly? We're going to address all of this and it starts with our implementation of List List, take 2 Below nil and cons have the exact same implementation. I've only fixed the type annotations. Most importantly, I added reduce which provides a way to get values "out" of our list container. The type annotation for List is updated to List a r – this can be understood as "a list containing values of type a that when reduced, will produce a value of type r." // type List a r = (r, a) -> r // nil : List a r const nil = (c, n) => n // cons : (a, List a r) -> List a r const cons = (x, y = nil) => (c, n) => c (y (c, n), x) // reduce : ((r, a) -> r, r) -> List a -> r const reduce = (f, init) => l => l (f, init) Now we can maintain List as a sane type, and push all the wonky behavior you want into the autoCons function. Below we update autoCons to work with our list acc using our new reduce function const autoCons = (init, n) => { const loop = acc => (x, n) => isFunction (x) // don't break the abstraction barrier ? acc (x, n) // extract the value using our appropriate list module function ? reduce (x, n) (acc) : loop (cons (x, acc)) return loop (nil) (init, n) } So speaking of types, let's examine the type of autoCons – autoCons (1) // "lambda (x,n) => isFunction (x) ... autoCons (1) (2) // "lambda (x,n) => isFunction (x) ... autoCons (1) (2) (3) // "lambda (x,n) => isFunction (x) ... autoCons (1) (2) (3) (add, 0) // 6 Well autoCons always returns a lambda, but that lambda has a type that we cannot determine – sometimes it returns another lambda of its same kind, other times it returns a completely different result; in this case some number, 6 Because of this, we cannot easily mix and combine autoCons expressions with other parts of our program. If you drop this perverse drive to create variadic curried interfaces, you can make an autoCons that is type-able // autoCons : (...a) -> List a r const autoCons = (...xs) => { const loop = (acc, x = nil, ...xs) => x === nil ? acc : loop (cons (x, acc), ...xs) return loop (nil, ...xs) } Because autoCons now returns a known List (instead of the mystery unknown type caused by variadic currying), we can plug an autoCons list into the various other functions provided by our List module. const c = autoCons (1, 2, 3) const d = autoCons (4, 5, 6) console.log ( toArray (c) // [ 1, 2, 3 ] , toArray (map (x => x * x) (d)) // [ 16, 25, 36 ] , toArray (filter (x => x != 5) (d)) // [ 4, 6 ] , toArray (append (c, d)) // [ 1, 2, 3, 4, 5, 6 ] ) These kind of mix-and-combine expressions is not possible when autoCons returns a type we cannot rely upon. Another important thing to notice is the List module gives us a place to expand its functionality. We can easily add functions used above like map, filter, append, and toArray – you lose this flexibility when trying to shove everything through the variadic curried interface Let's look at those additions to the List module now – as you can see, each function is well-typed and has behavior we can rely upon // type List a r = (r, a) -> r // nil : List a r // cons : (a, List a r) -> List a r // reduce : ((r, a) -> r, r) -> List a r -> r // length : List a r -> Int const length = reduce ( (acc, _) => acc + 1 , 0 ) // map : (a -> b) -> List a r -> List b r const map = f => reduce ( (acc, x) => cons (f (x), acc) , nil ) // filter : (a -> Bool) -> List a r -> List a r const filter = f => reduce ( (acc,x) => f (x) ? cons (x, acc) : acc , nil ) // append : (List a r, List a r) -> List a r const append = (l1, l2) => (c, n) => l2 (c, l1 (c, n)) // toArray : List a r -> Array a const toArray = reduce ( (acc, x) => [ ...acc, x ] , [] ) Even autoCons makes sense as part of our module now // autoCons : (...a) -> List a r const autoCons = (...xs) => { const loop = (acc, x = nil, ...xs) => x === nil ? acc : loop (cons (x, acc), ...xs) return loop (nil, ...xs) } Add any other functions to the List module // nth: Int -> List a r -> a // isNil : List a r -> Bool // first : List a r -> a // rest : List a r -> List a r // ...

Given an expression like A(a)(b)(f) where f is a function, it's impossible to know whether f is supposed to be added to the list or whether it's the reducing function. Hence, I'm going to describe how to write expressions like A(a)(b)(f, x) which is equivalent to [a, b].reduce(f, x). This allows us to distinguish when the list ends depending upon how many arguments you provide: const L = g => function (x, a) { switch (arguments.length) { case 1: return L(k => g((f, a) => k(f, f(a, x)))); case 2: return g((f, a) => a)(x, a); } }; const A = L(x => x); const xs = A(1)(2)(3)(4)(5); console.log(xs((x, y) => x + y, 0)); // 15 console.log(xs((x, y) => x * y, 1)); // 120 console.log(xs((a, x) => a.concat(x), [])); // [1,2,3,4,5] It works due to continuations. Every time we add a new element, we accumulate a CPS function. Each CPS function calls the previous CPS function, thereby creating a CPS function chain. When we give this CPS function chain a base function, it unrolls the chain and allows us to reduce it. It's the same idea behind transducers and lenses. Edit: user633183's solution is brilliant. It uses the Church encoding of lists using right folds to alleviate the need for continuations, resulting in simpler code which is easy to understand. Here's her solution, modified to make foldr seem like foldl: const L = g => function (x, a) { switch (arguments.length) { case 1: return L((f, a) => f(g(f, a), x)); case 2: return g(x, a); } }; const A = L((f, a) => a); const xs = A(1)(2)(3)(4)(5); console.log(xs((x, y) => x + y, 0)); // 15 console.log(xs((x, y) => x * y, 1)); // 120 console.log(xs((a, x) => a.concat(x), [])); // [1,2,3,4,5] Here g is the Church encoded list accumulated so far. Initially, it's the empty list. Calling g folds it from the right. However, we also build the list from the right. Hence, it seems like we're building the list and folding it from the left because of the way we write it. If all these functions are confusing you, what user633183 is really doing is: const L = g => function (x, a) { switch (arguments.length) { case 1: return L([x].concat(g)); case 2: return g.reduceRight(x, a); } }; const A = L([]); const xs = A(1)(2)(3)(4)(5); console.log(xs((x, y) => x + y, 0)); // 15 console.log(xs((x, y) => x * y, 1)); // 120 console.log(xs((a, x) => a.concat(x), [])); // [1,2,3,4,5] As you can see, she is building the list backwards and then using reduceRight to fold the backwards list backwards. Hence, it looks like you're building and folding the list forwards.

I'll have to admit I haven't read through your linked questions and I'm mainly here for the fun puzzle... but does this help in any way? I figured you want to differentiate between adding an element (calling with a new value) and running a function on the list (calling with a function). Since I had to somehow pass the function to run, I couldn't get the (1) vs () syntax to work. This uses an interface that returns an object with concat to extend the list, and fold to run a reducer on the list. Again, not sure if it's a complete answer, but it might help you explore other directions. const Empty = Symbol(); const L = (x, y = Empty) => ({ concat: z => L(z, L(x, y)), fold: (f, seed) => f(x, y === Empty ? seed : y.fold(f, seed)) }); const sum = (a, b) => a + b; console.log( L(1) .concat(2).concat(3).concat(4).concat(5).concat(6) .fold(sum, 0) )

Work in progress Thanks to the stunning contribution by #user3297291 , I somehow could refactor the code to fit my specification, but not working because I am lost the concept during the implementation :( The point is whole thing must be curried, and no object.method is involved. Can anyone "debug" please :) The initial value is set to the first element, in this example as 1 I think this is almost done. const isFunction = f => (typeof f === 'function'); const Empty = Symbol(); const L = (x = Empty) => (y = Empty) => z => isFunction(z) ? (() => { const fold = f => seed => f(x)(y) === Empty ? seed : (L)(y)(f); return fold(z)(x); })() : L(z)(L(x)(y)); const sum = a => b => a + b; console.log( L(1)(2)(3)(4)(5)(6)(sum) ); Output z => isFunction(z) ? (() => { const fold = f => seed => f(x)(y) === Empty ? seed : (L)(y)(f); return fold(z)(x); })() : L(z)(L(x)(y))

I've gone through the various questions you have but I'm still not sure I entirely understand what you're looking for. On the off chance you're simply looking to represent a linked list, here is a "dumb" representation that does not use clever tricks like overloaded arguments or default parameter values: const List = (() => { const nil = Symbol() // ADT const Nil = nil const Cons = x => xs => ({ x, xs }) const match = ({ Nil, Cons }) => l => l === nil ? Nil : Cons(l.x)(l.xs) // Functor const map = f => match({ Nil, Cons: x => xs => Cons(f(x))(map(f)(xs)) }) // Foldable const foldr = f => z => match({ Nil: z, Cons: x => xs => f(x)(foldr(f)(z)(xs)) // danger of stack overflow! // https://wiki.haskell.org/Foldr_Foldl_Foldl%27 }) return { Nil, Cons, match, map, foldr } })() const { Nil, Cons, match, map, foldr } = List const toArray = foldr(x => xs => [x, ...xs])([]) const l = Cons(1)(Cons(2)(Cons(3)(Nil))) const l2 = map(x => x * 2)(l) const l3 = map(x => x * 3)(l2) const a = toArray(l3) console.log(a) // => [6, 12, 18]