How to migrate Control.ST code to Idris 2 (Control.App)? - migration

Idris 2 doesn't have Control.ST, only Control.Monad.ST which is a completely different beast (it is basically the same as Haskell's Control.Monad.ST, i.e. mutable references behind a safe, pure interface). It seems that Control.App is roughly-vaguely what is supposed to replace it. However, the Control.App documentation isn't written with Control.ST in mind, and I can't figure out what the migration path is supposed to be.
For example, in this Js frontend library, we have the following Idris 1 API:
public export
interface Dom (m : Type -> Type) where
DomRef : (a:Type) -> (f : a -> Type) -> (g : a -> Type) -> a -> Type
initBody : List (DomOption a f g) -> ((x:a) -> f x -> Html (g x)) -> (z:a) -> f z -> ST m Var [add (DomRef a f g z)]
clearDom : (dom : Var) -> ST m () [remove dom (DomRef a f g z)]
domPut : (dom : Var) -> {x:a} -> f x -> ST m () [dom ::: (DomRef a f g x)]
It is not at all clear what the App version of this interface would be. Is it supposed to simply use App {l} e () everywhere, and the relationship between e and DomRef {e} would be tracked somehow differently? Is initBody / clearDom supposed to use the with pattern, i.e. something like withDom : (App {l} e ()) -> App {l} e ()? If domPut : {x:a} -> f x -> App {l} e (), how is f's type connected to DomRef {e}?

Related

Defining groups in Idris

I defined monoid in Idris as
interface Is_monoid (ty : Type) (op : ty -> ty -> ty) where
id_elem : () -> ty
proof_of_left_id : (a : ty) -> ((op a (id_elem ())) = a)
proof_of_right_id : (a : ty) -> ((op (id_elem ())a) = a)
proof_of_associativity : (a, b, c : ty) -> ((op a (op b c)) = (op (op a b) c))
then tried to define groups as
interface (Is_monoid ty op) => Is_group (ty : Type) (op : ty -> ty -> ty) where
inverse : ty -> ty
proof_of_left_inverse : (a : ty) -> (a = (id_elem ()))
but during compilation it showed
When checking type of Group.proof_of_left_inverse:
Can't find implementation for Is_monoid ty op
Is there a way around it.
The error message is a bit misleading, but indeed, the compiler does not know which implementation of Is_monoid to use for your call to id_elem in your definition of proof_of_left_inverse. You can make it work by making it making the call more explicit:
proof_of_left_inverse : (a : ty) -> (a = (id_elem {ty = ty} {op = op} ()))
Now, why is this necessary? If we have a simple interface like
interface Pointed a where
x : a
we can just write a function like
origin : (Pointed b) => b
origin = x
without specifying any type parameters explicitly.
One way to understand this is to look at interfaces and implementations through the lens of other, in a way more basic Idris features. x can be thought of as a function
x : {a : Type} -> {auto p : PointedImpl a} -> a
where PointedImpl is some pseudo type that represents the implementations of Pointed. (Think a record of functions.)
Similarly, origin looks something like
origin : {b : Type} -> {auto j : PointedImpl b} -> b
x notably has two implicit arguments, which the compiler tries to infer during type checking and unification. In the above example, we know that origin has to return a b, so we can unify a with b.
Now i is also auto, so it is not only subject to unification (which does not help here), but in addition, the compiler looks for "surrounding values" that can fill that hole if no explicit one was specified. The first place to look after local variables which we don't have is the parameter list, where we indeed find j.
Thus, our call to origin resolves without us having to explicitly specify any additional arguments.
Your case is more akin to this:
interface Test a b where
x : a
y : b
test : (Test c d) => c
test = x
This will error in the same manner your example did. Going through the same steps as above, we can write
x : {a : Type} -> {b -> Type} -> {auto i : TestImpl a b} -> a
test : {c : Type} -> {d -> Type} -> {auto j : TestImpl c d} -> c
As above, we can unify a and c, but there is nothing that tells us what d is supposed to be. Specifically, we can't unify it with b, and consequently we can't unify TestImpl a b with TestImpl c d and thus we can't use j as value for the auto-parameter i.
Note that I don't claim that this is how things are implemented under the covers. This is just an analogy in a sense, but one that holds up to at least some scrutiny.

BST using modules - OCaml

I am trying to create module/interface (i dont exactly know how its called, i am new to the language) for basic operations on BST in OCaml. My goal is to have an implementation that lets me doing something like this:
T.create();;
T.push(2);;
T.push(3);;
T.push(5);;
in order to get a bst tree consisting of 2,3,5.
But at the moment to achieve this i have to write something like this:
let teeBst = T.push(2)(T.push(3)(T.push(5)(T.create())));;
So when I am checking/using my code I have to do it like this:
let tee2 = T.push(2)(T.push(3)(T.push(5)(T.create())));;
T.postorder(tee2);;
The output is fine:
# val tee2 : T.bt = <abstr>
# - : int list = [2; 3; 5]
But, as I said before, I would like to achieve this doing as below:
T.push(2);;
T.push(3);;
T.push(5);;
T.postorder();;
(I realise this requires some changes to my postorder function but the one I am currently using is a temporary one so I can check the tree I have atm )
Below is my implementation. If you see the solution, please let me know ;)
module type Tree =
sig
type bt
val create: unit -> bt
val push: int -> bt -> bt
val find: int -> bt -> bool
val preorder: bt -> int list
val postorder: bt -> int list
val inorder: bt -> int list
end;;
module T : Tree =
struct
type bt = E | B of bt * int * bt
let create () = E
let rec push x = function
| E -> B(E, x, E)
| B (l, y, r) when x<y -> B(push x l, y, r)
| B (l, y, r) when x>y -> B(l, y, push x r)
| xs -> xs;;
let rec find x = function
| E -> false
| B(l, y,_) when x< y -> find x l
| B(_,y,r) when x>y -> find x r
| _ -> true;;
let rec preorder = function
| B(l,v,r) -> v::(preorder r) # (preorder l)
| E -> [];;
let rec inorder = function
| B(l,v,r) ->(inorder r) # v::(inorder l)
| E -> []
let rec postorder = function
| B(l,v,r) -> (postorder r) # (postorder l) # [v]
| E -> []
end;;
It seems like you want modules to be classes, but I'd advise you to consider more idiomatic solutions. Have you considered using the pipe operator?
T.create()
|> T.push(2)
|> T.push(3)
|> T.push(5)
|> T.postorder;;
Or with local open (which makes more sense if you have a module with a longer name than just T of course) you can even do
T.(
create()
|> push(2)
|> push(3)
|> push(5)
|> postorder
);
What you're asking for would require introducing global mutable state, which isn't just "some changes" but an entirely different paradigm. And one that is generally frowned upon because it makes your code unpredictable and hard to debug since it relies on state that might change at any moment from anywhere.
Another possibility is to actually use classes, since OCaml has those too. Then you'd still have mutable state, but it would at least be contained.

Unsolved metavariable for function that has no inhabited arguments

I am getting an unsolved metavariable for foo in the code below:
namespace Funs
data Funs : Type -> Type where
Nil : Funs a
(::) : {b : Type} -> (a -> List b) -> Funs (List a) -> Funs (List a)
data FunPtr : Funs a -> Type -> Type where
here : FunPtr ((::) {b} _ bs) b
there : FunPtr bs b -> FunPtr (_ :: bs) b
total foo : FunPtr [] b -> Void
How do I convince Idris that foo has no valid patterns to match on?
I've tried adding
foo f = ?foo
and then doing a case split in Emacs on f (just to see what might come up), but that just removes the line, leaving foo as an unsolved meta.
It turns out all I need to do is enumerate all possible patterns for foo's argument, and then Idris is able to figure out, one by one, that they are un-unifyable with foo's type:
foo : FunPtr [] b -> Void
foo here impossible
foo (there _) impossible

Doing rank-n quantification in Idris

I can only do rank-n types in Idris 0.9.12 in a rather clumsy way:
tupleId : ((a : Type) -> a -> a) -> (a, b) -> (a, b)
tupleId f (a, b) = (f _ a, f _ b)
I need the underscores wherever there's a type application, because Idris throws parse errors when I try to make the (nested) type arguments implicit:
tupleId : ({a : Type} -> a -> a) -> (a, b) -> (a, b) -- doesn't compile
A probably bigger issue is that I can't do class constraints in higher-rank types at all. I can't translate the following Haskell function to Idris:
appShow :: Show a => (forall a. Show a => a -> String) -> a -> String
appShow show x = show x
This also prevents me from using Idris functions as type synonyms for types such as Lens, which is Lens s t a b = forall f. Functor f => (a -> f b) -> s -> f t in Haskell.
Any way to remedy or circumvent the above issues?
I've just implemented this in master, allowing implicits in arbitrary scopes, and it'll be in the next hackage release. It's not well tested yet though! I have at least tried the following simple examples, and a few others:
appShow : Show a => ({b : _} -> Show b => b -> String) -> a -> String
appShow s x = s x
AppendType : Type
AppendType = {a, n, m : _} -> Vect n a -> Vect m a -> Vect (n + m) a
append : AppendType
append [] ys = ys
append (x :: xs) ys = x :: append xs ys
tupleId : ({a : _} -> a -> a) -> (a, b) -> (a, b)
tupleId f (a, b) = (f a, f b)
Proxy : Type -> Type -> Type -> Type -> (Type -> Type) -> Type -> Type
Producer' : Type -> (Type -> Type) -> Type -> Type
Producer' a m t = {x', x : _} -> Proxy x' x () a m t
yield : Monad m => a -> Producer' a m ()
The main constraint at the minute is that you can't give values for implicit arguments directly, except at the top level. I'll do something about that eventually...

How would I translate a Haskell type class into F#?

I'm trying to translate the Haskell core library's Arrows into F# (I think it's a good exercise to understanding Arrows and F# better, and I might be able to use them in a project I'm working on.) However, a direct translation isn't possible due to the difference in paradigms. Haskell uses type-classes to express this stuff, but I'm not sure what F# constructs best map the functionality of type-classes with the idioms of F#. I have a few thoughts, but figured it best to bring it up here and see what was considered to be the closest in functionality.
For the tl;dr crowd: How do I translate type-classes (a Haskell idiom) into F# idiomatic code?
For those accepting of my long explanation:
This code from the Haskell standard lib is an example of what I'm trying to translate:
class Category cat where
id :: cat a a
comp :: cat a b -> cat b c -> cat a c
class Category a => Arrow a where
arr :: (b -> c) -> a b c
first :: a b c -> a (b,d) (c,d)
instance Category (->) where
id f = f
instance Arrow (->) where
arr f = f
first f = f *** id
Attempt 1: Modules, Simple Types, Let Bindings
My first shot at this was to simply map things over directly using Modules for organization, like:
type Arrow<'a,'b> = Arrow of ('a -> 'b)
let arr f = Arrow f
let first f = //some code that does the first op
That works, but it loses out on polymorphism, since I don't implement Categories and can't easily implement more specialized Arrows.
Attempt 1a: Refining using Signatures and types
One way to correct some issues with Attempt 1 is to use a .fsi file to define the methods (so the types enforce easier) and to use some simple type tweaks to specialize.
type ListArrow<'a,'b> = Arrow<['a],['b]>
//or
type ListArrow<'a,'b> = LA of Arrow<['a],['b]>
But the fsi file can't be reused (to enforce the types of the let bound functions) for other implementations, and the type renaming/encapsulating stuff is tricky.
Attempt 2: Object models and interfaces
Rationalizing that F# is built to be OO also, maybe a type hierarchy is the right way to do this.
type IArrow<'a,'b> =
abstract member comp : IArrow<'b,'c> -> IArrow<'a,'c>
type Arrow<'a,'b>(func:'a->'b) =
interface IArrow<'a,'b> with
member this.comp = //fun code involving "Arrow (fun x-> workOn x) :> IArrow"
Aside from how much of a pain it can be to get what should be static methods (like comp and other operators) to act like instance methods, there's also the need to explicitly upcast the results. I'm also not sure that this methodology is still capturing the full expressiveness of type-class polymorphism. It also makes it hard to use things that MUST be static methods.
Attempt 2a: Refining using type extensions
So one more potential refinement is to declare the interfaces as bare as possible, then use extension methods to add functionality to all implementing types.
type IArrow<'a,'b> with
static member (&&&) f = //code to do the fanout operation
Ah, but this locks me into using one method for all types of IArrow. If I wanted a slightly different (&&&) for ListArrows, what can I do? I haven't tried this method yet, but I would guess I can shadow the (&&&), or at least provide a more specialized version, but I feel like I can't enforce the use of the correct variant.
Help me
So what am I supposed to do here? I feel like OO should be powerful enough to replace type-classes, but I can't seem to figure out how to make that happen in F#. Were any of my attempts close? Are any of them "as good as it gets" and that'll have to be good enough?
My brief answer is:
OO is not powerful enough to replace type classes.
The most straightforward translation is to pass a dictionary of operations, as in one typical typeclass implementation. That is if typeclass Foo defines three methods, then define a class/record type named Foo, and then change functions of
Foo a => yadda -> yadda -> yadda
to functions like
Foo -> yadda -> yadda -> yadda
and at each call site you know the concrete 'instance' to pass based on the type at the call-site.
Here's a short example of what I mean:
// typeclass
type Showable<'a> = { show : 'a -> unit; showPretty : 'a -> unit } //'
// instances
let IntShowable =
{ show = printfn "%d"; showPretty = (fun i -> printfn "pretty %d" i) }
let StringShowable =
{ show = printfn "%s"; showPretty = (fun s -> printfn "<<%s>>" s) }
// function using typeclass constraint
// Showable a => [a] -> ()
let ShowAllPretty (s:Showable<'a>) l = //'
l |> List.iter s.showPretty
// callsites
ShowAllPretty IntShowable [1;2;3]
ShowAllPretty StringShowable ["foo";"bar"]
See also
https://web.archive.org/web/20081017141728/http://blog.matthewdoig.com/?p=112
Here's the approach I use to simulate Typeclasses (from http://code.google.com/p/fsharp-typeclasses/ ).
In your case, for Arrows could be something like this:
let inline i2 (a:^a,b:^b ) =
((^a or ^b ) : (static member instance: ^a* ^b -> _) (a,b ))
let inline i3 (a:^a,b:^b,c:^c) =
((^a or ^b or ^c) : (static member instance: ^a* ^b* ^c -> _) (a,b,c))
type T = T with
static member inline instance (a:'a ) =
fun x -> i2(a , Unchecked.defaultof<'r>) x :'r
static member inline instance (a:'a, b:'b) =
fun x -> i3(a, b, Unchecked.defaultof<'r>) x :'r
type Return = Return with
static member instance (_Monad:Return, _:option<'a>) = fun x -> Some x
static member instance (_Monad:Return, _:list<'a> ) = fun x -> [x]
static member instance (_Monad:Return, _: 'r -> 'a ) = fun x _ -> x
let inline return' x = T.instance Return x
type Bind = Bind with
static member instance (_Monad:Bind, x:option<_>, _:option<'b>) = fun f ->
Option.bind f x
static member instance (_Monad:Bind, x:list<_> , _:list<'b> ) = fun f ->
List.collect f x
static member instance (_Monad:Bind, f:'r->'a, _:'r->'b) = fun k r -> k (f r) r
let inline (>>=) x (f:_->'R) : 'R = T.instance (Bind, x) f
let inline (>=>) f g x = f x >>= g
type Kleisli<'a, 'm> = Kleisli of ('a -> 'm)
let runKleisli (Kleisli f) = f
type Id = Id with
static member instance (_Category:Id, _: 'r -> 'r ) = fun () -> id
static member inline instance (_Category:Id, _:Kleisli<'a,'b>) = fun () ->
Kleisli return'
let inline id'() = T.instance Id ()
type Comp = Comp with
static member instance (_Category:Comp, f, _) = (<<) f
static member inline instance (_Category:Comp, Kleisli f, _) =
fun (Kleisli g) -> Kleisli (g >=> f)
let inline (<<<) f g = T.instance (Comp, f) g
let inline (>>>) g f = T.instance (Comp, f) g
type Arr = Arr with
static member instance (_Arrow:Arr, _: _ -> _) = fun (f:_->_) -> f
static member inline instance (_Arrow:Arr, _:Kleisli<_,_>) =
fun f -> Kleisli (return' <<< f)
let inline arr f = T.instance Arr f
type First = First with
static member instance (_Arrow:First, f, _: 'a -> 'b) =
fun () (x,y) -> (f x, y)
static member inline instance (_Arrow:First, Kleisli f, _:Kleisli<_,_>) =
fun () -> Kleisli (fun (b,d) -> f b >>= fun c -> return' (c,d))
let inline first f = T.instance (First, f) ()
let inline second f = let swap (x,y) = (y,x) in arr swap >>> first f >>> arr swap
let inline ( *** ) f g = first f >>> second g
let inline ( &&& ) f g = arr (fun b -> (b,b)) >>> f *** g
Usage:
> let f = Kleisli (fun y -> [y;y*2;y*3]) <<< Kleisli ( fun x -> [ x + 3 ; x * 2 ] ) ;;
val f : Kleisli<int,int list> = Kleisli <fun:f#4-14>
> runKleisli f <| 5 ;;
val it : int list = [8; 16; 24; 10; 20; 30]
> (arr (fun y -> [y;y*2;y*3])) 3 ;;
val it : int list = [3; 6; 9]
> let (x:option<_>) = runKleisli (arr (fun y -> [y;y*2;y*3])) 2 ;;
val x : int list option = Some [2; 4; 6]
> ( (*) 100) *** ((+) 9) <| (5,10) ;;
val it : int * int = (500, 19)
> ( (*) 100) &&& ((+) 9) <| 5 ;;
val it : int * int = (500, 14)
> let x:List<_> = (runKleisli (id'())) 5 ;;
val x : List<int> = [5]
Note: use id'() instead of id
Update: you need F# 3.0 to compile this code, otherwise here's the F# 2.0 version.
And here's a detailed explanation of this technique which is type-safe, extensible and as you can see works even with some Higher Kind Typeclasses.