How to write the following higher order Haskell function in Erlang?
applyTwice :: (a -> a) -> a -> a
applyTwice f x = f (f x)
1> Apply2 = fun(F, X) -> F(F(X)) end.
#Fun<erl_eval.12.82930912>
2> F = fun(Arg) -> Arg * 2 end.
#Fun<erl_eval.6.82930912>
3> Apply2(F, 10).
40
The problem is that, I'm not sure if it's what you actually need. Cause in Haskell, you can use applyTwice f as function, but not in Erlang (there is no built-in partial or curry functionality). You also can't do in Erlang something like
applyTwice :: (a -> a) -> a -> a
applyTwice f = f . f
Related
I'm trying to explore the dynamic capabilities of F# for situations where I can't express some function with the static type system. As such, I'm trying to create a mapN function for (say) Option types, but I'm having trouble creating a function with a dynamic number of arguments. I've tried:
let mapN<'output> (f : obj) args =
let rec mapN' (state:obj) (args' : (obj option) list) =
match args' with
| Some x :: xs -> mapN' ((state :?> obj -> obj) x) xs
| None _ :: _ -> None
| [] -> state :?> 'output option
mapN' f args
let toObjOption (x : #obj option) =
Option.map (fun x -> x :> obj) x
let a = Some 5
let b = Some "hi"
let c = Some true
let ans = mapN<string> (fun x y z -> sprintf "%i %s %A" x y z) [a |> toObjOption; b |> toObjOption; c |> toObjOption]
(which takes the function passed in and applies one argument at a time) which compiles, but then at runtime I get the following:
System.InvalidCastException: Unable to cast object of type 'ans#47' to type
'Microsoft.FSharp.Core.FSharpFunc`2[System.Object,System.Object]'.
I realize that it would be more idiomatic to either create a computation expression for options, or to define map2 through map5 or so, but I specifically want to explore the dynamic capabilities of F# to see whether something like this would be possible.
Is this just a concept that can't be done in F#, or is there an approach that I'm missing?
I think you would only be able to take that approach with reflection.
However, there are other ways to solve the overall problem without having to go dynamic or use the other static options you mentioned. You can get a lot of the same convenience using Option.apply, which you need to define yourself (or take from a library). This code is stolen and adapted from F# for fun and profit:
module Option =
let apply fOpt xOpt =
match fOpt,xOpt with
| Some f, Some x -> Some (f x)
| _ -> None
let resultOption =
let (<*>) = Option.apply
Some (fun x y z -> sprintf "%i %s %A" x y z)
<*> Some 5
<*> Some "hi"
<*> Some true
To explain why your approach does not work, the problem is that you cannot cast a function of type int -> int (represented as FSharpFunc<int, int>) to a value of type obj -> obj (represented as FSharpFunc<obj, obj>). The types are the same generic types, but the cast fails because the generic parameters are different.
If you insert a lot of boxing and unboxing, then your function actually works, but this is probably not something you want to write:
let ans = mapN<string> (fun (x:obj) -> box (fun (y:obj) -> box (fun (z:obj) ->
box (Some(sprintf "%i %s %A" (unbox x) (unbox y) (unbox z))))))
[a |> toObjOption; b |> toObjOption; c |> toObjOption]
If you wanted to explore more options possible thanks to dynamic hacks - then you can probably do more using F# reflection. I would not typically use this in production (simple is better - I'd just define multiple map functions by hand or something like that), but the following runs:
let rec mapN<'R> f args =
match args with
| [] -> unbox<'R> f
| x::xs ->
let m = f.GetType().GetMethods() |> Seq.find (fun m ->
m.Name = "Invoke" && m.GetParameters().Length = 1)
mapN<'R> (m.Invoke(f, [| x |])) xs
mapN<obj> (fun a b c -> sprintf "%d %s %A" a b c) [box 1; box "hi"; box true]
It is powerful technique using recursion because its strong describable feature. Tail recursion provides more powerful computation than normal recursion because it changes recursion into iteration. Continuation-Passing Style (CPS) can change lots of loop codes into tail recursion. Continuation Monad provides recursion syntax but in essence it is tail recursion, which is iteration. It is supposed to reasonable use Continuation Monad for 100000 factorial. Here is the code.
type ContinuationBuilder() =
member b.Bind(x, f) = fun k -> x (fun x -> f x k)
member b.Return x = fun k -> k x
member b.ReturnFrom x = x
(*
type ContinuationBuilder =
class
new : unit -> ContinuationBuilder
member Bind : x:(('d -> 'e) -> 'f) * f:('d -> 'g -> 'e) -> ('g -> 'f)
member Return : x:'b -> (('b -> 'c) -> 'c)
member ReturnFrom : x:'a -> 'a
end
*)
let cont = ContinuationBuilder()
//val cont : ContinuationBuilder
let fac n =
let rec loop n =
cont {
match n with
| n when n = 0I -> return 1I
| _ -> let! x = fun f -> f n
let! y = loop (n - 1I)
return x * y
}
loop n (fun x -> x)
let x2 = fac 100000I
There is wrong message: "Process is terminated due to StackOverflowException."
What is wrong with 100000 factorial using ContinuationMonad?
You need to compile the project in Release mode or check the "Generate tail calls" option in project properties (or use --tailcalls+ if you're running the compiler via command line).
By default, tail call optimization is not enabled in Debug mode. The reason is that, if tail-calls are enabled, you will not see as useful information about stack traces. So, disabling them by default gives you more pleasant debugging experience (even in Debug mode, the compiler optimizes tail-recursive functions that call themselves, which handles most situations).
You probably need to add this memeber to your monad builder:
member this.Delay(mk) = fun c -> mk () c
These are the function definitions.
func1: 'a -> unit
func2: 'b -> 'a
func3: string -> 'b list
The current function
let f = Seq.iter((fun a -> func1(func2 a)) func3(s)
This is as far as I got
let f = func3(s)
|> ((fun a -> func2 a
|> func1)
|> Seq.iter)
I have the feeling it should be possible to loose the lambda and the parens'.
You can do without pipes, simply
Seq.iter (func1 << func2) << func3
(this is a function with some arguments [same than func3] and same output than Seq.iter).
You can test it
let func1 x = printfn "Number: %d" x
let func2 (a, b) = a + b
let func3 = Seq.map (fun n -> (n, 2 * n))
let f : (seq<_> -> unit) = Seq.iter (func1 << func2) << func3
f [1..5]
with output
Number: 3
Number: 6
Number: 9
Number: 12
Number: 15
val func1 : x:int -> unit
val func2 : a:int * b:int -> int
val func3 : (seq<int> -> seq<int * int>)
val f : (seq<int> -> unit)
val it : unit = ()
:)
You can use function composition operator >>:
func3() |> Seq.iter (func2 >> func1)
I think the question is, why do you want to use the pipeline operator?
I find your original code quite readable. You should not try to use pipeline operator (or function composition) just for the sake of using them. Now, in your code, the input s comes at the end, which is a bit unfortunate (you cannot quite see what is the main input for the code). I would probably rewrite it as (also, s is not really a descriptive name):
s |> func3
|> Seq.iter (fun a -> func1 (func2 a))
You can use function composition too - but I do not use it very often, because it does not (always) help with readability. But using it in the argument of Seq.iter is probably quite reasonable.
On a completely unrelated note, you could just use for loop and write:
for a in func3 s do
func1 (func2 a)
I actually find this more readable than any other version of the code here (if F# gives you a language feature for iterating over sequences that does exactly what you need, why not use it?)
I'm converting some F# code to OCaml and I see a lot of uses of this pipeline operator <|, for example:
let printPeg expr =
printfn "%s" <| pegToString expr
The <| operator is apparently defined as just:
# let ( <| ) a b = a b ;;
val ( <| ) : ('a -> 'b) -> 'a -> 'b = <fun>
I'm wondering why they bother to define and use this operator in F#, is it just so they can avoid putting in parens like this?:
let printPeg expr =
Printf.printf "%s" ( pegToString expr )
As far as I can tell, that would be the conversion of the F# code above to OCaml, correct?
Also, how would I implement F#'s << and >> operators in Ocaml?
( the |> operator seems to simply be: let ( |> ) a b = b a ;; )
why they bother to define and use this operator in F#, is it just so they can avoid putting in parens?
It's because the functional way of programming assumes threading a value through a chain of functions. Compare:
let f1 str server =
str
|> parseUserName
|> getUserByName server
|> validateLogin <| DateTime.Now
let f2 str server =
validateLogin(getUserByName(server, (parseUserName str)), DateTime.Now)
In the first snippet, we clearly see everything that happens with the value. Reading the second one, we have to go through all parens to figure out what's going on.
This article about function composition seems to be relevant.
So yes, in a regular life, it is mostly about parens. But also, pipeline operators are closely related to partial function application and point-free style of coding. See Programming is "Pointless", for example.
The pipeline |> and function composition >> << operators can produce yet another interesting effect when they are passed to higher-level functions, like here.
Directly from the F# source:
let inline (|>) x f = f x
let inline (||>) (x1,x2) f = f x1 x2
let inline (|||>) (x1,x2,x3) f = f x1 x2 x3
let inline (<|) f x = f x
let inline (<||) f (x1,x2) = f x1 x2
let inline (<|||) f (x1,x2,x3) = f x1 x2 x3
let inline (>>) f g x = g(f x)
let inline (<<) f g x = f(g x)
OCaml Batteries supports these operators, but for reasons of precedence, associativity and other syntactic quirks (like Camlp4) it uses different symbols. Which particular symbols to use has just been settled recently, there are some changes. See: Batteries API:
val (|>) : 'a -> ('a -> 'b) -> 'b
Function application. x |> f is equivalent to f x.
val ( **> ) : ('a -> 'b) -> 'a -> 'b
Function application. f **> x is equivalent to f x.
Note The name of this operator is not written in stone. It is bound to change soon.
val (|-) : ('a -> 'b) -> ('b -> 'c) -> 'a -> 'c
Function composition. f |- g is fun x -> g (f x). This is also equivalent to applying <** twice.
val (-|) : ('a -> 'b) -> ('c -> 'a) -> 'c -> 'b
Function composition. f -| g is fun x -> f (g x). Mathematically, this is operator o.
But Batteries trunk provides:
val ( ## ) : ('a -> 'b) -> 'a -> 'b
Function application. [f ## x] is equivalent to [f x].
val ( % ) : ('a -> 'b) -> ('c -> 'a) -> 'c -> 'b
Function composition: the mathematical [o] operator.
val ( |> ) : 'a -> ('a -> 'b) -> 'b
The "pipe": function application. [x |> f] is equivalent to [f x].
val ( %> ) : ('a -> 'b) -> ('b -> 'c) -> 'a -> 'c
Piping function composition. [f %> g] is [fun x -> g (f x)].
I'm wondering why they bother to define and use this operator in F#, is it just so they can avoid putting in parens like this?
Excellent question. The specific operator you're referring to (<|) is pretty useless IME. It lets you avoid parentheses on rare occasions but more generally it complicates the syntax by dragging in more operators and makes it harder for less experienced F# programmers (of which there are now many) to understand your code. So I've stopped using it.
The |> operator is much more useful but only because it helps F# to infer types correctly in situations where OCaml would not have a problem. For example, here is some OCaml:
List.map (fun o -> o#foo) os
The direct equivalent fails in F# because the type of o cannot be inferred prior to reading its foo property so the idiomatic solution is to rewrite the code like this using the |> so F# can infer the type of o before foo is used:
os |> List.map (fun o -> o.foo)
I rarely use the other operators (<< and >>) because they also complicate the syntax. I also dislike parser combinator libraries that pull in lots of operators.
The example Bytebuster gave is interesting:
let f1 str server =
str
|> parseUserName
|> getUserByName server
|> validateLogin <| DateTime.Now
I would write this as:
let f2 str server =
let userName = parseUserName str
let user = getUserByName server userName
validateLogin user DateTime.Now
There are no brackets in my code. My temporaries have names so they appear in the debugger and I can inspect them and Intellisense can give me type throwback when I hover the mouse over them. These characteristics are valuable for production code that non-expert F# programmers will be maintaining.
I'm implementing a packrat parser in OCaml, as per the Master Thesis by B. Ford. My parser should receive a data structure that represents the grammar of a language and parse given sequences of symbols.
I'm stuck with the memoization part. The original thesis uses Haskell's lazy evaluation to accomplish linear time complexity. I want to do this (memoization via laziness) in OCaml, but don't know how to do it.
So, how do you memoize functions by lazy evaluations in OCaml?
EDIT: I know what lazy evaluation is and how to exploit it in OCaml. The question is how to use it to memoize functions.
EDIT: The data structure I wrote that represents grammars is:
type ('a, 'b, 'c) expr =
| Empty of 'c
| Term of 'a * ('a -> 'c)
| NTerm of 'b
| Juxta of ('a, 'b, 'c) expr * ('a, 'b, 'c) expr * ('c -> 'c -> 'c)
| Alter of ('a, 'b, 'c) expr * ('a, 'b, 'c) expr
| Pred of ('a, 'b, 'c) expr * 'c
| NPred of ('a, 'b, 'c) expr * 'c
type ('a, 'b, 'c) grammar = ('a * ('a, 'b, 'c) expr) list
The (not-memoized) function that parse a list of symbols is:
let rec parse g v xs = parse' g (List.assoc v g) xs
and parse' g e xs =
match e with
| Empty y -> Parsed (y, xs)
| Term (x, f) ->
begin
match xs with
| x' :: xs when x = x' -> Parsed (f x, xs)
| _ -> NoParse
end
| NTerm v' -> parse g v' xs
| Juxta (e1, e2, f) ->
begin
match parse' g e1 xs with
| Parsed (y, xs) ->
begin
match parse' g e2 xs with
| Parsed (y', xs) -> Parsed (f y y', xs)
| p -> p
end
| p -> p
end
( and so on )
where the type of the return value of parse is defined by
type ('a, 'c) result = Parsed of 'c * ('a list) | NoParse
For example, the grammar of basic arithmetic expressions can be specified as g, in:
type nt = Add | Mult | Prim | Dec | Expr
let zero _ = 0
let g =
[(Expr, Juxta (NTerm Add, Term ('$', zero), fun x _ -> x));
(Add, Alter (Juxta (NTerm Mult, Juxta (Term ('+', zero), NTerm Add, fun _ x -> x), (+)), NTerm Mult));
(Mult, Alter (Juxta (NTerm Prim, Juxta (Term ('*', zero), NTerm Mult, fun _ x -> x), ( * )), NTerm Prim));
(Prim, Alter (Juxta (Term ('<', zero), Juxta (NTerm Dec, Term ('>', zero), fun x _ -> x), fun _ x -> x), NTerm Dec));
(Dec, List.fold_left (fun acc d -> Alter (Term (d, (fun c -> int_of_char c - 48)), acc)) (Term ('0', zero)) ['1';'2';'3';])]
The idea of using lazyness for memoization is use not functions, but data structures, for memoization. Lazyness means that when you write let x = foo in some_expr, foo will not be evaluated immediately, but only as far as some_expr needs it, but that different occurences of xin some_expr will share the same trunk: as soon as one of them force computation, the result is available to all of them.
This does not work for functions: if you write let f x = foo in some_expr, and call f several times in some_expr, well, each call will be evaluated independently, there is not a shared thunk to store the results.
So you can get memoization by using a data structure instead of a function. Typically, this is done using an associative data structure: instead of computing a a -> b function, you compute a Table a b, where Table is some map from the arguments to the results. One example is this Haskell presentation of fibonacci:
fib n = fibTable !! n
fibTable = [0,1] ++ map (\n -> fib (n - 1) + fib (n - 2)) [2..]
(You can also write that with tail and zip, but this doesn't make the point clearer.)
See that you do not memoize a function, but a list: it is the list fibTable that does the memoization. You can write this in OCaml as well, for example using the LazyList module of the Batteries library:
open Batteries
module LL = LazyList
let from_2 = LL.seq 2 ((+) 1) (fun _ -> true)
let rec fib n = LL.at fib_table (n - 1) + LL.at fib_table (n - 2)
and fib_table = lazy (LL.Cons (0, LL.cons 1 <| LL.map fib from_2))
However, there is little interest in doing so: as you have seen in the example above, OCaml does not particularly favor call-by-need evaluation -- it's reasonable to use, but not terribly convenient as it was forced to be in Haskell. It is actually equally simple to directly write the cache structure by direct mutation:
open Batteries
let fib =
let fib_table = DynArray.of_list [0; 1] in
let get_fib n = DynArray.get fib_table n in
fun n ->
for i = DynArray.length fib_table to n do
DynArray.add fib_table (get_fib (i - 1) + get_fib (i - 2))
done;
get_fib n
This example may be ill-chosen, because you need a dynamic structure to store the cache. In the packrat parser case, you're tabulating parsing on a known input text, so you can use plain arrays (indexed by the grammar rules): you would have an array of ('a, 'c) result option for each rule, of the size of the input length and initialized to None. Eg. juxta.(n) represents the result of trying the rule Juxta from input position n, or None if this has not yet been tried.
Lazyness is a nice way to present this kind of memoization, but is not always expressive enough: if you need, say, to partially free some part of your result cache to lower memory usage, you will have difficulties if you started from a lazy presentation. See this blog post for a remark on this.
Why do you want to memoize functions? What you want to memoize is, I believe, the parsing result for a given (parsing) expression and a given position in the input stream. You could for instance use Ocaml's Hashtables for that.
The lazy keyword.
Here you can find some great examples.
If it fits your use case, you can also use OCaml streams instead of manually generating thunks.