Could someone please explain the concept of currying to me. I am primarily learning it because we are learning ML in my 'modern programming language' class for a functional language introduction.
In particular you can use this example:
-fun g a = fn b => a+b;
val g = fn: int -> int -> int
-g 2 3;
val it = 5 : int
I'm confused how these parameters are passed or how to even think about it in the first place.
Thank you for any help.
In this case, you make the currying explicit, so it should be easier to understand.
If we read the function definition, it says (paraphrased): "Create a function g, which when given an a returns fn b => a+b."
That is, if we call g 2, we get back the function fn b => 2+b. As such, when we call g 2 3, we actually call (g 2) 3; that is we first get the function stated above back, and then use this function on the value 3, yielding 5.
Currying is simply the concept of making a function in several "stages", each taking an input and producing a new function. SML has syntactic sugar for this, making g equivalent to the following:
fun g a b = a + b;
Related
Apologies if this is considered a dumb question, but how can I make an Isabelle theory recognise ML code? For example, let's say I have an Isabelle file that looks like
ML ‹
type vec = real * real;
fun addvec ((a,b):vec) ((c,d):vec) : vec = (a+b,c+d);
›
lemma "addvec (1,2) (3,4) = (4,6)"
Unfortunately addvec isn't recognised in the lemma. How can I make the function recognised? I've read through How to get ML values from HOL? as well as the Isabelle Cookbook. The former uses a local_setup to assign the constants to Isabelle constants (as far as I can see) using a function called mk_const
fun mk_const c t =
let
val b = Binding.name c
val defb = Binding.name (c ^ "_def")
in (((b, NoSyn), ((defb, []), t)) |> Local_Theory.define) #> snd end
What do the functions Binding.name and Local_Theory.define do, and what is the local_theory type?
Thanks in advance!
The idea is not to define a function written in ML, it is to define in ML a function that you can use in Isabelle.
ML ‹
fun mk_const c t =
let
val b = Binding.name c
val defb = Binding.name (c ^ "_def")
in (((b, NoSyn), ((defb, []), t)) |> Local_Theory.define) #> snd end
›
record point =
coord_x::int
For example, let us define a function that just calls coord_x:
ML ‹
val f = Abs ("x", #{typ "point"}, Const( \<^const_name>‹coord_x›, #{typ "point"} --> #{typ int}) $ Bound 0)
›
Now we have written the definition, we can bind it a name:
local_setup‹mk_const "c" f›
thm c_def
(*c ≡ coord_x*)
local_setup is a keyword to change the theory (so add constants, change the context and so on).
Now obviously you most likely do not want hard coded constants like coord_x.
Some general comments here: I have never used records and I have written a lot of Isabelle. They can be useful (because they are extensible and so on), but they are weird /because they are extensible/. So if you can work on a datatype, do so. It is nicer, it is a proper type (so locales/instances/... just work).
I need a type of tree and a map on those, so I do this:
type 'a grouping =
G of ('a * 'a grouping) list
with
member g.map f =
let (G gs) = g
gs |> List.map (fun (s, g) -> f s, g.map f) |> G
But this makes me wonder:
The map member is boilerplate. In Haskell, GHC would implement fmap for me (... deriving (Functor)). I know F# doesn't have typeclasses, but is there some other way I can avoid writing map myself in F#?
Can I somehow avoid the line let (G gs) = g?
Is this whole construction somehow non-idiomatic? It looks weird to me, but maybe that's just because putting members on sum types is new to me.
I don't think there is a way to derive automatically map, however there's a way to emulate type classes in F#, your code can be written like this:
#r #"FsControl.Core.dll"
#r #"FSharpPlus.dll"
open FSharpPlus
open FsControl.Core.TypeMethods
type 'a grouping =
G of ('a * 'a grouping) list
with
// Add an instance for Functor
static member instance (_:Functor.Map, G gs, _) = fun (f:'b->'c) ->
map (fun (s, g) -> f s, map f g) gs |> G
// TEST
let a = G [(1, G [2, G[]] )]
let b = map ((+) 10) a // G [(11, G [12, G[]] )]
Note that map is really overloaded, the first application you see calls the instance for List<'a> and the second one the instance for grouping<'a>. So it behaves like fmap in Haskell.
Also note this way you can decompose G gs without creating the let (G gs) = g
Now regarding what is idiomatic I think many people would agree your solution is more F# idiomatic, but to me new idioms should also be developed in order to get more features and overcome current language limitations, that's why I consider using a library which define clear conventions also idiomatic.
Anyway I agree with #kvb in that it's slightly more idiomatic to define map into a module, in F#+ that convention is also used, so you have the generic map and the specific ModuleX.map
Is it possible to write recursive anonymous functions in SML? I know I could just use the fun syntax, but I'm curious.
I have written, as an example of what I want:
val fact =
fn n => case n of
0 => 1
| x => x * fact (n - 1)
The anonymous function aren't really anonymous anymore when you bind it to a
variable. And since val rec is just the derived form of fun with no
difference other than appearance, you could just as well have written it using
the fun syntax. Also you can do pattern matching in fn expressions as well
as in case, as cases are derived from fn.
So in all its simpleness you could have written your function as
val rec fact = fn 0 => 1
| x => x * fact (x - 1)
but this is the exact same as the below more readable (in my oppinion)
fun fact 0 = 1
| fact x = x * fact (x - 1)
As far as I think, there is only one reason to use write your code using the
long val rec, and that is because you can easier annotate your code with
comments and forced types. For examples if you have seen Haskell code before and
like the way they type annotate their functions, you could write it something
like this
val rec fact : int -> int =
fn 0 => 1
| x => x * fact (x - 1)
As templatetypedef mentioned, it is possible to do it using a fixed-point
combinator. Such a combinator might look like
fun Y f =
let
exception BlackHole
val r = ref (fn _ => raise BlackHole)
fun a x = !r x
fun ta f = (r := f ; f)
in
ta (f a)
end
And you could then calculate fact 5 with the below code, which uses anonymous
functions to express the faculty function and then binds the result of the
computation to res.
val res =
Y (fn fact =>
fn 0 => 1
| n => n * fact (n - 1)
)
5
The fixed-point code and example computation are courtesy of Morten Brøns-Pedersen.
Updated response to George Kangas' answer:
In languages I know, a recursive function will always get bound to a
name. The convenient and conventional way is provided by keywords like
"define", or "let", or "letrec",...
Trivially true by definition. If the function (recursive or not) wasn't bound to a name it would be anonymous.
The unconventional, more anonymous looking, way is by lambda binding.
I don't see what unconventional there is about anonymous functions, they are used all the time in SML, infact in any functional language. Its even starting to show up in more and more imperative languages as well.
Jesper Reenberg's answer shows lambda binding; the "anonymous"
function gets bound to the names "f" and "fact" by lambdas (called
"fn" in SML).
The anonymous function is in fact anonymous (not "anonymous" -- no quotes), and yes of course it will get bound in the scope of what ever function it is passed onto as an argument. In any other cases the language would be totally useless. The exact same thing happens when calling map (fn x => x) [.....], in this case the anonymous identity function, is still in fact anonymous.
The "normal" definition of an anonymous function (at least according to wikipedia), saying that it must not be bound to an identifier, is a bit weak and ought to include the implicit statement "in the current environment".
This is in fact true for my example, as seen by running it in mlton with the -show-basis argument on an file containing only fun Y ... and the val res ..
val Y: (('a -> 'b) -> 'a -> 'b) -> 'a -> 'b
val res: int32
From this it is seen that none of the anonymous functions are bound in the environment.
A shorter "lambdanonymous" alternative, which requires OCaml launched
by "ocaml -rectypes":
(fun f n -> f f n)
(fun f n -> if n = 0 then 1 else n * (f f (n - 1))
7;; Which produces 7! = 5040.
It seems that you have completely misunderstood the idea of the original question:
Is it possible to write recursive anonymous functions in SML?
And the simple answer is yes. The complex answer is (among others?) an example of this done using a fix point combinator, not a "lambdanonymous" (what ever that is supposed to mean) example done in another language using features not even remotely possible in SML.
All you have to do is put rec after val, as in
val rec fact =
fn n => case n of
0 => 1
| x => x * fact (n - 1)
Wikipedia describes this near the top of the first section.
let fun fact 0 = 1
| fact x = x * fact (x - 1)
in
fact
end
This is a recursive anonymous function. The name 'fact' is only used internally.
Some languages (such as Coq) use 'fix' as the primitive for recursive functions, while some languages (such as SML) use recursive-let as the primitive. These two primitives can encode each other:
fix f => e
:= let rec f = e in f end
let rec f = e ... in ... end
:= let f = fix f => e ... in ... end
In languages I know, a recursive function will always get bound to a name. The convenient and conventional way is provided by keywords like "define", or "let", or "letrec",...
The unconventional, more anonymous looking, way is by lambda binding. Jesper Reenberg's answer shows lambda binding; the "anonymous" function gets bound to the names "f" and "fact" by lambdas (called "fn" in SML).
A shorter "lambdanonymous" alternative, which requires OCaml launched by "ocaml -rectypes":
(fun f n -> f f n)
(fun f n -> if n = 0 then 1 else n * (f f (n - 1))
7;;
Which produces 7! = 5040.
I'm new to ocaml and tryin to write a continuation passing style function but quite confused what value i need to pass into additional argument on k
for example, I can write a recursive function that returns true if all elements of the list is even, otherwise false.
so its like
let rec even list = ....
on CPS, i know i need to add one argument to pass function
so like
let rec evenk list k = ....
but I have no clue how to deal with this k and how does this exactly work
for example for this even function, environment looks like
val evenk : int list -> (bool -> ’a) -> ’a = <fun>
evenk [4; 2; 12; 5; 6] (fun x -> x) (* output should give false *)
The continuation k is a function that takes the result from evenk and performs "the rest of the computation" and produces the "answer". What type the answer has and what you mean by "the rest of the computation" depends on what you are using CPS for. CPS is generally not an end in itself but is done with some purpose in mind. For example, in CPS form it is very easy to implement control operators or to optimize tail calls. Without knowing what you are trying to accomplish, it's hard to answer your question.
For what it is worth, if you are simply trying to convert from direct style to continuation-passing style, and all you care about is the value of the answer, passing the identity function as the continuation is about right.
A good next step would be to implement evenk using CPS. I'll do a simpler example.
If I have the direct-style function
let muladd x i n = x + i * n
and if I assume CPS primitives mulk and addk, I can write
let muladdk x i n k =
let k' product = addk x product k in
mulk i n k'
And you'll see that the mulptiplication is done first, then it "continues" with k', which does the add, and finally that continues with k, which returns to the caller. The key idea is that within the body of muladdk I allocated a fresh continuation k' which stands for an intermediate point in the multiply-add function. To make your evenk work you will have to allocate at least one such continuation.
I hope this helps.
Whenever I've played with CPS, the thing passed to the continuation is just the thing you would normally return to the caller. In this simple case, a nice "intuition lubricant" is to name the continuation "return".
let rec even list return =
if List.length list = 0
then return true
else if List.hd list mod 2 = 1
then return false
else even (List.tl list) return;;
let id = fun x -> x;;
Example usage: "even [2; 4; 6; 8] id;;".
Since you have the invocation of evenk correct (with the identity function - effectively converting the continuation-passing-style back to normal style), I assume that the difficulty is in defining evenk.
k is the continuation function representing the rest of the computation and producing a final value, as Norman said. So, what you need to do is compute the result of v of even and pass that result to k, returning k v rather than just v.
You want to give as input the result of your function as if it were not written with continuation passing style.
Here is your function which tests whether a list has only even integers:
(* val even_list : int list -> bool *)
let even_list input = List.for_all (fun x -> x mod 2=0) input
Now let's write it with a continuation cont:
(* val evenk : int list -> (bool -> 'a) -> 'a *)
let evenk input cont =
let result = even_list input in
(cont result)
You compute the result your function, and pass resultto the continuation ...
Can the following polymorphic functions
let id x = x;;
let compose f g x = f (g x);;
let rec fix f = f (fix f);; (*laziness aside*)
be written for types/type constructors or modules/functors? I tried
type 'x id = Id of 'x;;
type 'f 'g 'x compose = Compose of ('f ('g 'x));;
type 'f fix = Fix of ('f (Fix 'f));;
for types but it doesn't work.
Here's a Haskell version for types:
data Id x = Id x
data Compose f g x = Compose (f (g x))
data Fix f = Fix (f (Fix f))
-- examples:
l = Compose [Just 'a'] :: Compose [] Maybe Char
type Natural = Fix Maybe -- natural numbers are fixpoint of Maybe
n = Fix (Just (Fix (Just (Fix Nothing)))) :: Natural -- n is 2
-- up to isomorphism composition of identity and f is f:
iso :: Compose Id f x -> f x
iso (Compose (Id a)) = a
Haskell allows type variables of higher kind. ML dialects, including Caml, allow type variables of kind "*" only. Translated into plain English,
In Haskell, a type variable g can correspond to a "type constructor" like Maybe or IO or lists. So the g x in your Haskell example would be OK (jargon: "well-kinded") if for example g is Maybe and x is Integer.
In ML, a type variable 'g can correspond only to a "ground type" like int or string, never to a type constructor like option or list. It is therefore never correct to try to apply a type variable to another type.
As far as I'm aware, there's no deep reason for this limitation in ML. The most likely explanation is historical contingency. When Milner originally came up with his ideas about polymorphism, he worked with very simple type variables standing only for monotypes of kind *. Early versions of Haskell did the same, and then at some point Mark Jones discovered that inferring the kinds of type variables is actually quite easy. Haskell was quickly revised to allow type variables of higher kind, but ML has never caught up.
The people at INRIA have made a lot of other changes to ML, and I'm a bit surprised they've never made this one. When I'm programming in ML, I might enjoy having higher-kinded type variables. But they aren't there, and I don't know any way to encode the kind of examples you are talking about except by using functors.
You can do something similar in OCaml, using modules in place of types, and functors (higher-order modules) in place of higher-order types. But it looks much uglier and it doesn't have type-inference ability, so you have to manually specify a lot of stuff.
module type Type = sig
type t
end
module Char = struct
type t = char
end
module List (X:Type) = struct
type t = X.t list
end
module Maybe (X:Type) = struct
type t = X.t option
end
(* In the following, I decided to omit the redundant
single constructors "Id of ...", "Compose of ...", since
they don't help in OCaml since we can't use inference *)
module Id (X:Type) = X
module Compose
(F:functor(Z:Type)->Type)
(G:functor(Y:Type)->Type)
(X:Type) = F(G(X))
let l : Compose(List)(Maybe)(Char).t = [Some 'a']
module Example2 (F:functor(Y:Type)->Type) (X:Type) = struct
(* unlike types, "free" module variables are not allowed,
so we have to put it inside another functor in order
to scope F and X *)
let iso (a:Compose(Id)(F)(X).t) : F(X).t = a
end
Well... I'm not an expert of higher-order-types nor Haskell programming.
But this seems to be ok for F# (which is OCaml), could you work with these:
type 'x id = Id of 'x;;
type 'f fix = Fix of ('f fix -> 'f);;
type ('f,'g,'x) compose = Compose of ('f ->'g -> 'x);;
The last one I wrapped to tuple as I didn't come up with anything better...
You can do it but you need to make a bit of a trick:
newtype Fix f = In{out:: f (Fix f)}
You can define Cata afterwards:
Cata :: (Functor f) => (f a -> a) -> Fix f -> a
Cata f = f.(fmap (cata f)).out
That will define a generic catamorphism for all functors, which you can use to build your own stuff. Example:
data ListFix a b = Nil | Cons a b
data List a = Fix (ListFix a)
instance functor (ListFix a) where
fmap f Nil = Nil
fmap f (Cons a lst) = Cons a (f lst)