Idris function to construct empty `List a` where `a` is bound to an instance of `Ord`? - typechecking

I've only read the standard tutorial and fumbled around a bit, so I may be missing something simple.
If this isn't possible in Idris, please explain why. Furthermore, if can be done in another language please provide a code sample and explain what's different about that language's type system that makes it possible.
Here's my approach. Problems first arise in the third section.
Create an empty list of a known type
v : List Nat
v = []
This compiles and manifests in the REPL as [] : List Nat. Excellent.
Generalize to any provided type
emptyList : (t : Type) -> List t
emptyList t = []
v' : List Nat
v' = emptyList Nat
Unsurprisingly, this works and v' == v.
Constrain type to instances of Ord class
emptyListOfOrds : Ord t => (t : Type) -> List t
emptyListOfOrds t = []
v'' : List Nat
v'' = emptyListOfOrds Nat -- !!! typecheck failure
The last line fails with this error:
When elaborating right hand side of v'':
Can't resolve type class Ord t
Nat is an instance of Ord, so what's the problem? I tried replacing the Nats in v'' with Bool (not an instance of Ord), but there was no change in the error.
Another angle...
Does making Ord t an explicit parameter satisfy the type checker? Apparently not, but even if it did requiring the caller to pass redundant information isn't ideal.
emptyListOfOrds' : Ord t -> (t : Type) -> List t
emptyListOfOrds' a b = []
v''' : List Nat
v''' = emptyListOfOrds (Ord Nat) Nat -- !!! typecheck failure
The error is more elaborate this time:
When elaborating right hand side of v''':
When elaborating an application of function stackoverflow.emptyListOfOrds':
Can't unify
Type
with
Ord t
Specifically:
Can't unify
Type
with
Ord t
I'm probably missing some key insights about how values are checked against type declarations.

As other answers have explained, this is about how and where the variable t is bound. That is, when you write:
emptyListOfOrds : Ord t => (t : Type) -> List t
The elaborator will see that 't' is unbound at the point it is used in Ord t and so bind it implicitly:
emptyListOfOrds : {t : Type} -> Ord t => (t : Type) -> List t
So what you'd really like to say is something a bit like:
emptyListOfOrds : (t : Type) -> Ord t => List t
Which would bind the t before the type class constraint, and therefore it's in scope when Ord t appears. Unfortunately, this syntax isn't supported. I see no reason why this syntax shouldn't be supported but, currently, it isn't.
You can still implement what you want, but it's ugly, I'm afraid:
Since classes are first class, you can give them as ordinary arguments:
emptyListOfOrds : (t : type) -> Ord t -> List t
Then you can use the special syntax %instance to search for the default instance when you call emptyListOfOrds:
v'' = emptyListOfOrds Nat %instance
Of course, you don't really want to do this at every call site, so you can use a default implicit argument to invoke the search procedure for you:
emptyListOfOrds : (t : Type) -> {default %instance x : Ord t} -> List t
v'' = emptyListOfOrds Nat
The default val x : T syntax will fill in the implicit argument x with the default value val if no other value is explicitly given. Giving %instance as the default then is pretty much identical to what happens with class constraints, and actually we could probably change the implementation of the Foo x => syntax to do exactly this... I think the only reason I didn't is that default arguments didn't exist yet when I implemented type classes at first.

You could write
emptyListOfOrds : Ord t => List t
emptyListOfOrds = []
v'' : List Nat
v'' = emptyListOfOrds
Or perhaps if you prefer
v'' = emptyListOfOrds {t = Nat}
If you ask for the type of emptyListOfOrds the way you had written, you get
Ord t => (t2 : Type) -> List t2
Turing on :set showimplicits in the repl, and then asking again gives
{t : Type} -> Prelude.Classes.Ord t => (t2 : Type) -> Prelude.List.List t2
It seems specifying an Ord t constraint introduces an an implicit param t, and then your explicit param t gets assigned a new name. You can always explicitly supply a value for that implicit param, e.g. emptyListOfOrds {t = Nat} Nat. As far as if this is the "right" behavior or a limitation for some reason, perhaps you could open an issue about this on github? Perhaps there's some conflict with explicit type params and typeclass constraints? Normally typeclasses are for when you have things implicitly resolved... though I think I remember there being syntax for obtaining an explicit reference to a typeclass instance.

Not an answer, just some thoughts.
The problem here is that (t : Type) introduces new scope that extends to the right but Ord t is outside of this scope:
*> :t emptyListOfOrds
emptyListOfOrds : Ord t => (t2 : Type) -> List t2
You can add class constraint after introducing type variable:
emptyListOfOrds : (t : Type) -> Ord t -> List t
emptyListOfOrds t o = []
But now you need to specify class instance explicitly:
instance [natord] Ord Nat where
compare x y = compare x y
v'' : List Nat
v'' = emptyListOfOrds Nat #{natord}
Maybe it is somehow possible to make Ord t argument implicit.

Related

Why are my types not working out? (Use of type 'a pred = 'a -> bool)

I have been stuck on this question for a while. I've been editing and reviewing and changing the types for a while but I can't get the type checker to accept what I am doing, probably because I don't fully understand the error/where I am going wrong on this. I am working with the type:
type 'a pred = 'a -> bool
I believe this means I can use 'a pred as a shortcut to mean 'a -> bool, so an int going to result in a bool in my case, but I don't fully get how to implement it because I can't find many examples on this online which I have checked for.
My latest version is below, but I am getting a few errors from the checker, including Error: operator and operand do not agree. Would someone be able to explain where my error is, and why?
Edit: I now think there is a mismatch between this function and the rest of the code. The rest of the code requires it to be an 'a, polymorphic, while here I am assuming it is an int. However, I'm not sure how to do this function (check if odd) while keeping it a polymorphic type.
fun isOdd (p : int) : bool =
case p
of 1 => true
| 0 => false
| _ => isOdd (p - 2)
I believe this means I can use 'a pred as a shortcut to mean 'a -> bool
That is correct.
In the case of your isOdd predicate, it is an int pred:
> val isOdd = fn : int -> bool
- isOdd : int pred;
> val it = fn : int -> bool
Perhaps your misconception lies in the fact that in spite of expressing : int pred, the result in the REPL is still described as int -> bool? This is because we have only defined a type alias, and those tend reduce to their non-aliased form in SML.
Or perhaps your misconception lies in the 'a reducing to some concrete value? You can operate with 'a pred by not referring to concrete values of 'a. For example, if you want to filter an 'a list for only values that are true for a given 'a pred, then the standard function List.filter will have the type:
- List.filter : 'a pred -> 'a list -> 'a list;
> val 'a it = fn : ('a -> bool) -> 'a list -> 'a list
I'm not sure how to do this function (check if odd) while keeping it a polymorphic type.
I'm not sure, either.
Oddness is a property of integers, not arbitrary types 'a.
You would need to extend the meaning of "odd" to any type first. Then you would need some kind of overloading, since the oddness of every type presumably isn't determined by the same mechanism. I'm pretty sure this is a side-track caused by one or two confusions.

SML: Value restriction error when recursively calling quicksort

I'm writing a quicksort function for an exercise. I already know of the 5-line functional quicksort; but I wanted to improve the partition by having it scan through the list once and return a pair of lists splitting the original list in half. So I wrote:
fun partition nil = (nil, nil)
| partition (pivot :: rest) =
let
fun part (lst, pivot, (lesseq, greater)) =
case lst of
[] => (lesseq, greater)
| (h::t) =>
if h <= pivot then part (t, pivot, (h :: lesseq, greater))
else part (t, pivot, (lesseq, h :: greater))
in
part (rest, pivot, ([pivot], []))
end;
This partitions well enough. It gives me a signature val partition = fn : int list -> int list * int list. It runs as expected.
It's when I use the quicksort below that things start to break.
fun quicksort_2 nil = nil
| quicksort_2 lst =
let
val (lesseq, greater) = partition lst
in
quicksort_2 lesseq # quicksort_2 greater
end;
I can run the above function if I eliminate the recursive calls to quicksort_2; but if I put them back in (to actually go and sort the thing), it will cease to run. The signature will be incorrect as well, giving me val quicksort_2 = fn : int list -> 'a list. The warning I receive when I call the function on a list is:
Warning: type vars not generalized because of value restriction are instantiated to dummy types (X1,X2,...)
What is the problem here? I'm not using any ref variables; the type annotation I've tried doesn't seem to help...
The main issue is that you're lacking the singleton list base case for your quicksort function. It ought to be
fun quicksort [ ] = [ ]
| quicksort [x] = [x]
| quicksort xs =
let
val (l, r) = partition xs
in
quicksort l # quicksort r
end
which should then have type int list -> int list given the type of your partition. We have to add this case as otherwise you'll never hit a base case and instead recurse indefinitely.
For some more detail on why you saw the issues you were having though:
The signature will be incorrect as well, giving me val quicksort_2 = fn : int list -> 'a list
This is because the codomain of your function was never restricted to be less general than 'a list. Taking a look at the possible branches in your original implementation we can see that in the nil branch you return nil (of most general type 'a list) and in the recursive case you get two 'a lists (per our assumptions thus far) and append them, resulting in an 'a list---this is fine so your type is not further restricted.
[Value Restriction Warning]
What is the problem here? I'm not using any ref variables
The value restriction isn't really related to refs (though can often arise when using them). Instead it is the prohibition that anything polymorphic at the top level must be a value by its syntax (and thus precludes the possibility that a computation is behind a type abstractor at the top level). Here it is because given xs : int list we (ignoring the value restriction) have quicksort_2 xs : 'a list---which would otherwise be polymorphic, but is not a syntactic value. Correspondingly it is value restricted.

Can I annotate the complete type of a `fun` declaration?

In a learning environment, what are my options to provide type signatures for functions?
Standard ML doesn't have top-level type signatures like Haskell. Here are the alternatives I have considered:
Module signatures, which require either a separate signature file, or the type signature being defined in a separate block inside the same file as the module itself. This requires the use of modules, and in any production system that would be a sane choice.
Modules may seem a little verbose in a stub file when the alternative is a single function definition. They both introduce the concept of modules, perhaps a bit early,
Using val and val rec I can have the complete type signature in one line:
val incr : int -> int =
fn i => i + 1
val rec map : ('a -> 'b) -> 'a list -> 'b list =
fn f => fn xs => case xs of
[] => []
| x::ys => f x :: map f ys
Can I have this and also use fun?
If this is possible, I can't seem to get the syntax right.
Currently the solution is to embed the argument types and the result type as such:
fun map (f : 'a -> 'b) (xs : 'a list) : 'b list =
raise Fail "'map' is not implemented"
But I have experienced that this syntax gives the novice ML programmer the impression that the solution either cannot or should not be updated to the model solution:
fun map f [] = []
| map f (x::xs) = f x :: map f xs
It seems then that the type signatures, which are supposed to aid the student, prevents them from pattern matching. I cannot say if this is because they think that the type signatures cannot be removed or if they should not be removed. It is, of course, a matter of style whether they should (and where), but the student should be enabled to explore a style of type inference.
By using a let or local bound function, and shadowing
you can declare the function, and then assign it to a value.
using local for this is more convenient, since it has the form:
local decl in decl end, rather than let decl in expr end,
meaning let's expr, wants a top-level argument f
val map = fn f => let fun map = ... in map end
I don't believe people generally use local, anymore primarily because modules can do anything that local can, and more, but perhaps it is worth considering it as an anonymous module, when you do not want to explain modules yet.
local
fun map (f : 'a -> 'b) (x::rest : 'a list) : 'b list
= f x :: map f rest
| map _ ([]) = []
in
val (map : ('a -> 'b) -> 'a list -> 'b list) = map;
end
Then when it comes time to explain modules, you can declare the structure inside the local, around all of the declarations,
and then remove the local, and try to come up with a situation, where they have coded 2 functions, and it's more appropriate to replace 2 locals, with 1 structure.
local
structure X = struct
fun id x = x
end
in val id = X.id
end
perhaps starting them off with something like the following:
exception ReplaceSorryWithYourAnswer
fun sorry () = raise ReplaceSorryWithYourAnswer
local
(* Please fill in the _'s with the arguments
and the call to sorry() with your answer *)
fun map _ _ = sorry ()
in
val map : ('a -> 'b) -> ('a list) -> ('b list) = map
end

SML functor expose a type without exposing implementation (implementing sets)

I'm writing a functor to implement sets in standard ML. Since sets don't allow duplicates and I don't want it to be constrained to equality types, it's declared like this:
signature SET = sig
type t
type 'a set
val add : t -> t set -> t set
...
end
functor ListSet (EQ : sig type t val equal : t * t -> bool end) :> SET = struct
type t = EQ.t
type 'a set = 'a list
fun add x s = ...
...
end
I use :> so that list operations cannot be used on sets, hiding the internal implementation and allowing to change the representation (e.g. to a BST)
However, this also hides type t, therefore function add when used like this gives an error:
structure IntSet = ListSet (struct type t = int val equal = op= end);
val s0 = IntSet.empty
val s1 = IntSet.add 0 s0
Function: IntSet.add : IntSet.t -> IntSet.t IntSet.set -> IntSet.t IntSet.set
Argument: 0 : int
Reason:
Can't unify int (*In Basis*) with
IntSet.t (*Created from applying functor ListEqSet*)
(Different type constructors)
Is there a way to keep the implementation hidden but somehow expose the type t? Or is there a better approach to implementing sets?
P.S. The main reason I can't have equality types is to allow sets of sets, and while I can keep the lists sorted and define eqtype 'a set, it adds unnecessary complexity.
You need what's sometimes called a translucent signature ascription, that is, you hide some of the types and expose the others:
functor ListSet (Eq : EQ) :> SET where type t = Eq.t = ...
You have to expose the type t using a type refinement:
functor ListSet (Eq : sig type t val equal : t * t -> bool end) :> SET where type t = Eq.t =
struct
...
end
This is equivalent to an expansion of the signature SET where the type t is specified transparently as
type t = Eq.t

Does "Value Restriction" practically mean that there is no higher order functional programming?

Does "Value Restriction" practically mean that there is no higher order functional programming?
I have a problem that each time I try to do a bit of HOP I get caught by a VR error. Example:
let simple (s:string)= fun rq->1
let oops= simple ""
type 'a SimpleType= F of (int ->'a-> 'a)
let get a = F(fun req -> id)
let oops2= get ""
and I would like to know whether it is a problem of a prticular implementation of VR or it is a general problem that has no solution in a mutable type-infered language that doesn't include mutation in the type system.
Does “Value Restriction” mean that there is no higher order functional programming?
Absolutely not! The value restriction barely interferes with higher-order functional programming at all. What it does do is restrict some applications of polymorphic functions—not higher-order functions—at top level.
Let's look at your example.
Your problem is that oops and oops2 are both the identity function and have type forall 'a . 'a -> 'a. In other words each is a polymorphic value. But the right-hand side is not a so-called "syntactic value"; it is a function application. (A function application is not allowed to return a polymorphic value because if it were, you could construct a hacky function using mutable references and lists that would subvert the type system; that is, you could write a terminating function type type forall 'a 'b . 'a -> 'b.
Luckily in almost all practical cases, the polymorphic value in question is a function, and you can define it by eta-expanding:
let oops x = simple "" x
This idiom looks like it has some run-time cost, but depending on the inliner and optimizer, that can be got rid of by the compiler—it's just the poor typechecker that is having trouble.
The oops2 example is more troublesome because you have to pack and unpack the value constructor:
let oops2 = F(fun x -> let F f = get "" in f x)
This is quite a but more tedious, but the anonymous function fun x -> ... is a syntactic value, and F is a datatype constructor, and a constructor applied to a syntactic value is also a syntactic value, and Bob's your uncle. The packing and unpacking of F is all going to be compiled into the identity function, so oops2 is going to compile into exactly the same machine code as oops.
Things are even nastier when you want a run-time computation to return a polymorphic value like None or []. As hinted at by Nathan Sanders, you can run afoul of the value restriction with an expression as simple as rev []:
Standard ML of New Jersey v110.67 [built: Sun Oct 19 17:18:14 2008]
- val l = rev [];
stdIn:1.5-1.15 Warning: type vars not generalized because of
value restriction are instantiated to dummy types (X1,X2,...)
val l = [] : ?.X1 list
-
Nothing higher-order there! And yet the value restriction applies.
In practice the value restriction presents no barrier to the definition and use of higher-order functions; you just eta-expand.
I didn't know the details of the value restriction, so I searched and found this article. Here is the relevant part:
Obviously, we aren't going to write the expression rev [] in a program, so it doesn't particularly matter that it isn't polymorphic. But what if we create a function using a function call? With curried functions, we do this all the time:
- val revlists = map rev;
Here revlists should be polymorphic, but the value restriction messes us up:
- val revlists = map rev;
stdIn:32.1-32.23 Warning: type vars not generalized because of
value restriction are instantiated to dummy types (X1,X2,...)
val revlists = fn : ?.X1 list list -> ?.X1 list list
Fortunately, there is a simple trick that we can use to make revlists polymorphic. We can replace the definition of revlists with
- val revlists = (fn xs => map rev xs);
val revlists = fn : 'a list list -> 'a list list
and now everything works just fine, since (fn xs => map rev xs) is a syntactic value.
(Equivalently, we could have used the more common fun syntax:
- fun revlists xs = map rev xs;
val revlists = fn : 'a list list -> 'a list list
with the same result.) In the literature, the trick of replacing a function-valued expression e with (fn x => e x) is known as eta expansion. It has been found empirically that eta expansion usually suffices for dealing with the value restriction.
To summarise, it doesn't look like higher-order programming is restricted so much as point-free programming. This might explain some of the trouble I have when translating Haskell code to F#.
Edit: Specifically, here's how to fix your first example:
let simple (s:string)= fun rq->1
let oops= (fun x -> simple "" x) (* eta-expand oops *)
type 'a SimpleType= F of (int ->'a-> 'a)
let get a = F(fun req -> id)
let oops2= get ""
I haven't figured out the second one yet because the type constructor is getting in the way.
Here is the answer to this question in the context of F#.
To summarize, in F# passing a type argument to a generic (=polymorphic) function is a run-time operation, so it is actually type-safe to generalize (as in, you will not crash at runtime). The behaviour of thusly generalized value can be surprising though.
For this particular example in F#, one can recover generalization with a type annotation and an explicit type parameter:
type 'a SimpleType= F of (int ->'a-> 'a)
let get a = F(fun req -> id)
let oops2<'T> : 'T SimpleType = get ""

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