The following OCaml code creates a universal type (not mine, from Jane Street's website):
module Univ : sig
type t
val embed: unit -> ('a -> t) * (t -> 'a option)
end = struct
type t = bool -> unit
let embed () =
let r = ref None in
let put x =
let sx = Some x in
fun b -> r := if b then sx else None
in
let get f =
f true;
let res = !r in
f false; res
in
put, get
end
creates a universal type. In it, a call to embed () creates a 2-tuple of functions. The first function in the tuple stores an arbitrary value in the closed-over ref cell; the second function retrieves it. If the wrong value of type t is given to the second function, it returns None instead. However, it appears to me that this would cause the ref cell to be cleared and thus further attempts to retrieve the value to fail, which does not happen. Also, I do not understand in general what happens when passed a wrong value to the projection function get.
When embed is called, a fresh reference r and two fresh closures for the functions put and get are allocated. When some put is called it returns a fresh function (let's call it the setter) that also has r in its closure.
If that setter is given to the corresponding get, the r captured in both closure is the same, hence when get calls f true, it modifies the r in get closure, and let res = !r read a Some value.
let get f =
f true; (* assign the shared reference r *)
let res = !r in (* read r *)
f false; (* clear r *)
res
If the setter is given to a non-corresponding get, the r captured by get and the setter is not the same. (let's call them r1 and r2)
let get f =
f true; (* assign r1 *)
let res = !r in (* read r2 *)
f false; (* clear r1 *)
res
Since the get takes care of always clearing r, we know that before calling
get, r2 is None. Hence get will return None.
Related
for an example, if a function receives a function as a factor and iterates it twice
func x = f(f(x))
I have totally no idea of how the code should be written
You just pass the function as a value. E.g.:
let apply_twice f x = f (f x)
should do what you expect. We can try it out by testing on the command line:
utop # apply_twice ((+) 1) 100
- : int = 102
The (+) 1 term is the function that adds one to a number (you could also write it as (fun x -> 1 + x)). Also remember that a function in OCaml does not need to be evaluated with all its parameters. If you evaluate apply_twice only with the function you receive a new function that can be evaluated on a number:
utop # let add_two = apply_twice ((+) 1) ;;
val add_two : int -> int = <fun>
utop # add_two 1000;;
- : int = 1002
To provide a better understanding: In OCaml, functions are first-class
values. Just like int is a value, 'a -> 'a -> 'a is a value (I
suppose you are familiar with function signatures). So, how do you
implement a function that returns a function? Well, let's rephrase it:
As functions = values in OCaml, we could phrase your question in three
different forms:
[1] a function that returns a function
[2] a function that returns a value
[3] a value that returns a value
Note that those are all equivalent; I just changed terms.
[2] is probably the most intuitive one for you.
First, let's look at how OCaml evaluates functions (concrete example):
let sum x y = x + y
(val sum: int -> int -> int = <fun>)
f takes in two int's and returns an int (Intuitively speaking, a
functional value is a value, that can evaluate further if you provide
values). This is the reason you can do stuff like this:
let partial_sum = sum 2
(int -> int = <fun>)
let total_sum = partial_sum 3 (equivalent to: let total_sum y = 3 + y)
(int = 5)
partial_sum is a function, that takes in only one int and returns
another int. So we already provided one argument of the function,
now one is still missing, so it's still a functional value. If that is
still not clear, look into it more. (Hint: f x = x is equivalent to
f = fun x -> x) Let's come back to your question. The simplest
function, that returns a function is the function itself:
let f x = x
(val f:'a -> 'a = <fun>)
f
('a -> 'a = <fun>)
let f x = x Calling f without arguments returns f itself. Say you
wanted to concatenate two functions, so f o g, or f(g(x)):
let g x = (* do something *)
(val g: 'a -> 'b)
let f x = (* do something *)
(val f: 'a -> 'b)
let f_g f g x = f (g x)
(val f_g: ('a -> 'b) -> ('c -> 'a) -> 'c -> 'b = <fun>)
('a -> 'b): that's f, ('c -> 'a): that's g, c: that's x.
Exercise: Think about why the particular signatures have to be like that. Because let f_g f g x = f (g x) is equivalent to let f_g = fun f -> fun g -> fun x -> f (g x), and we do not provide
the argument x, we have created a function concatenation. Play around
with providing partial arguments, look at the signature, and there
will be nothing magical about functions returning functions; or:
functions returning values.
fun count_wcs p =
let
val count = 0
in
g (fn () => count + 1) (fn y => 1) p
end
I'm doing homework and we're not supposed to use mutations, this doesn't reassign the value to anything but it doesn't feel right. Please, don't say what is the right way to do this, because I'm supposed to figure this out.
datatype pattern = Wildcard
| Variable of string
| UnitP
| ConstP of int
| TupleP of pattern list
| ConstructorP of string * pattern
fun g f1 f2 p =
let
val r = g f1 f2
in
case p of
Wildcard => f1 ()
| Variable x => f2 x
| TupleP ps => List.foldl (fn (p,i) => (r p) + i) 0 ps
| ConstructorP(_,p) => r p
| _ => 0
end
This function g has to receive a type unit -> int function as the first argument. I checked the count after I called the function and got 0, so it is ok to write it like this, right? But still, it does feel sloppy.
Added the context (function g and the datatype used). The function count_wcs is supposed to count the number a Wildcard pattern appears in a pattern.
This does not count as a mutation, but it does resemble what you might do if you had them and is probably not going to work, depending on what it is you're doing. Mutations require references and they're made with ref and are de-referenced with !. So just stick away from those. :-)
You're doing something that will be of little benefit:
let
val count = 0
in
...
end
will bind count to 0, but will never cause count to have any other value; I presume that you want to eventually have that count increases. If it had been a reference, val count = ref 0, you could increment it by doing count := !count + 1, but since it's not, you have to make count a variable of some function for it to change.
For example:
fun count_4s ([], count) = count
| count_4s (x::xs, count) = count_4s (xs, if x = 4 then count + 1 else count)
In each call count is constant, but in each subsequent, recursive call it is possibly incremented.
This function g has to receive a type unit -> int function as first argument.
...
g (fn () => count + 1) (fn y => 1) p
...
[...] But still it does feel sloppy.
Assuming that g's first arguments have no side-effects and do not throw exceptions and do not loop forever, all they can do is return the same thing on every call. This makes them rather boring. Typically functions that either take as input or return () : unit do something else, like read or write from a source external to the program.
I wouldn't call it sloppy. Just a bit weird, without knowing the context.
It's very easy to identify mutation in standard ML - if you don't use ref variables and assign them new values with :=, it's not mutation.
And you don't, so you aren't mutating anything.
Also, your function can be simplified to
fun count_wildcards p = g (fn () => 1) (fn _ => 1) p
which makes it even clearer that nothing gets mutated.
(It's not at all clear what it's supposed to accomplish, so it's impossible to tell whether your solution is correct.)
I need to write a pipe function in OCaml such that pipe [f1;...;fn] (where f1,...,fn are functions!) returns a function f such that for any x, f x computes fn(...(f2(f1 x))).
I need to write it using List.fold_left and need to fill in the parameter of the function
let pipe fs =
let f a x = "fill in this part" in
let base = fun x ->x in
List.fold_left f base fs;;
I already filled in the base. If the first parameter to pipe is an empty list, it returns the second parameter. Ex: pipe [] 3 = 3.
i know for the let f a x part I want to perform function x of the accumulated functions a.
I'm just not sure how to write that. I wrote let f a x = x a but that gave me an error when i tested
pipe [(fun x -> x+x); (fun x -> x + 3)] 3
it should run x+x on 3 then run x+3 on the result and give me 9 but it gave it a type error when i tried to use let f a x = x a
for the fill in part
# let _ = pipe [(fun x -> x+x); (fun x -> x + 3)] 3;;
File "", line 1, characters 24-25:
Error: This expression has type 'a -> 'a
but an expression was expected of type int
What is the correct format to create a function that takes in 2 functions and runs them on each other. Ex: make a function that takes in functions a and b and runs b on the result of a.
To evaluate (fold_left g init fs) x:
when fs is empty, (fold_left g init fs) x = init x. In your case, you want it to be x.
when fs = fs' # [fn]: according to what you would like to be true, the expression should evaluate to
fn (fold_left g init fs' x) but using the definition of fold_left it evaluates also to (g (fold_left g init fs') fn) x.
Hence if the following equations are true:
init x = x
(g k f) x = f (k x)
the problem is solved. Hence, let us define init = fun x -> x and
g k f = fun x -> f (k x).
Well, base is a function like this fun x -> ....
Similarly, your function f needs to return a function, so assume it returns something that looks like this:
fun z -> ...
You have to figure out what this function should be doing with its argument z.
figured it out just needed that z in there for a and x to call
I want to define a function check_char_fun: (char -> 'a) -> (char ->' a) -> bool that, given two functions
on char, return true when both functions are the same (ie, when they are exactly the same behavior on every one of the possible values of a char) and false otherwise.
let check_char_fun f1 f2 =
let aux = true
for i=0 to 255 do
if (f1 (char_of_int i))=(f2 (char_of_int i))
then aux=false;
done;
if aux=true
then true
else false;;
I am learning OCaml, so I don't know what can I do.
You're nearly there:
let check_char_fun f1 f2 =
let aux = ref true in
for i = 0 to 255 do
if (f1 (char_of_int i)) = (f2 (char_of_int i)) then aux := false
else ()
done;
!aux
Unlike the variables in imperative languages, bindings in OCaml are immutable by default. To create a real variable, we create a bool ref which is mutable and can be changed from within the loop.
OCaml does not have a distinction between statements and expressions like the imperative languages: There are only expressions! Thats why you always need the else clause to an if; this ways the resulting expression always returns a value (in both if and else case) the type of which must be the same -- in this case of type unit (the type of the value () -- which would be void in C).
Your code is not very OCaml-like, but thats what I personally love about OCaml: The functional style is not forced down your throat and you can implement algorithms in an imperative style without entering the academic ivory tower.
First, you have to have a definition for what a "behavior" is. If your functions can raise exceptions the problem gets harder. Your code assumes the functions always return a value, which seems like a good simplification for a beginning problem.
You're also using the (somewhat out-of-date) definition of character that OCaml uses, in that codes are limited to the range 0 .. 255. This also seems OK.
So the only problem I see in your code is that you're expecting to be able to change the value of the aux variable. Variables in OCaml are immutable: you can't change the value that they're bound to.
If you want to keep your code mostly as it is, you can change aux so its value is a reference to a bool. Then you can change the boolean value inside the reference (while aux remains bound to the same reference).
To make a reference to a bool and change the value:
# let x = ref true;;
val x : bool ref = {contents = true}
# !x;;
- : bool = true
# x := false;;
- : unit = ()
# !x;;
- : bool = false
(One of the reasons to study OCaml is to learn how to work with immutable values. So I'd suggest looking for other ways to solve the problem that don't require the use of references.)
let rec range i j =
if i > j then [] else i :: (range (i+1) j);;
let check_char_fun f1 f2 =
let lc = List.map char_of_int (range 0 255) in
List.for_all (fun c -> (f1 c) = (f2 c)) lc;;
test:
#let id x =x;;
val id : 'a -> 'a = <fun>
# check_char_fun id id;;
- : bool = true
# check_char_fun id (fun x -> 'a');;
- : bool = false
Or:
exception Fails_in of char;;
let check_char_fun f1 f2 =
let lc = List.map char_of_int (range 0 255) in
List.iter (fun c ->
if (f1 c) <> (f2 c) then raise (Fails_in c)
) lc;;
# try (
check_char_fun id id
) with Fails_in c -> Printf.printf "Diff(%d)(%c)" (int_of_char c) c
;;
- : unit = ()
# try (
check_char_fun id (fun x -> 'a')
) with Fails_in c -> Printf.printf "Diff(%d)(%c)" (int_of_char c) c
;;
Diff(0)()- : unit = ()
The following applies each function to each character value in the 0 .. 255 range and compares their results, but it does not check for cases where a function raises an exception or causes a side effect elsewhere:
open Core.Std
let check_char_fun f1 f2 =
let chars = List.map ~f:char_of_int (List.range 0 256) in
List.for_all ~f:(fun c -> (f1 c) = (f2 c)) chars
I want to write Ocaml function, that takes two parameters: other function (int->int) and int value, and than check somehow if it was used with these to parameters earlier. how to do it?
so other way of looking at that problem is how to identify function with the identification that can be variable?
The problem is: Make function g, that takes functions f and int value n, than check if g was already used for f for that value n, if yes return previously got result, otherwise count f for n value. f is int->int
You can compare functions with the == operator.
# let f x = x + 2;;
val f : int -> int = <fun>
# let g x = x + 5;;
val g : int -> int = <fun>
# f == g;;
- : bool = false
# f == f;;
- : bool = true
#
Using the == operator is very dangerous, however. Comparing things for physical equality is inadvisable because it pierces the veil of referential transparency. I would personally look for another way to solve whatever problem you're working on. (If you'll forgive the suggestion.)
You need to flip your idea around: instead of keeping the function f and g separately, have g turn f into a memoizing version of itself:
module IntMap = Map.Make (struct type t = int let compare a b = a - b end)
let g f =
let m = ref (IntMap.empty) in
fun x ->
try IntMap.find x !m
with Not_found ->
let r = f x in
m := IntMap.add x r !m;
r
It's obviously worth doing benchmarks to see if the cost of computation is worse that the one of memoization. Also, it could be better to use a Hashtbl instead of a Map (left as an exercise).