As I know, explicit type parameters in value definitions is a one way to overcome "value restriction" problem.
Is there another cases when I need to use them?
Upd: I mean "explicitly generic constructs", where type parameter is enclosed in angle brackets, i.e.
let f<'T> x = x
Polymorphic recursion is another case. That is, if you want to use a different generic instantiation within the function body, then you need to use explicit parameters on the definition:
// perfectly balanced tree
type 'a PerfectTree =
| Single of 'a
| Node of ('a*'a) PerfectTree
// need type parameters here
let rec fold<'a,'b> (f:'a -> 'b) (g:'b->'b->'b) : 'a PerfectTree -> 'b = function
| Single a -> f a
| Node t -> t |> fold (fun (a,b) -> g (f a) (f b)) g
let sum = fold id (+)
let ten = sum (Node(Node(Single((1,2),(3,4)))))
This would likely be rare, but when you want to prevent further generalization (ยง14.6.7):
Explicit type parameter definitions on value and member definitions can affect the process of type inference and generalization. In particular, a declaration that includes explicit generic parameters will not be generalized beyond those generic parameters. For example, consider this function:
let f<'T> (x : 'T) y = x
During type inference, this will result in a function of the following type, where '_b is a type inference variable that is yet to be resolved.
f<'T> : 'T -> '_b -> '_b
To permit generalization at these definitions, either remove the explicit generic parameters (if they can be inferred), or use the required number of parameters, as the following example shows:
let throw<'T,'U> (x:'T) (y:'U) = x
Of course, you could also accomplish this with type annotations.
Most obvious example: write a function to calculate the length of a string.
You have to write:
let f (a:string) = a.Length
and you need the annotation. Without the annotation, the compiler can't determine the type of a. Other similar examples exist - particularly when using libraries designed to be used from C#.
Dealing with updated answer:
The same problem applies - string becomes A<string> which has a method get that returns a string
let f (a:A<string>) = a.get().Length
Related
The Kleisli composition operator >=>, also known as the "fish" in Haskell circles, may come in handy in many situations where composition of specialized functions is needed. It works kind of like the >> operator, but instead of composing simple functions 'a -> 'b it confers some special properties on them possibly best expressed as 'a -> m<'b>, where m is either a monad-like type or some property of the function's return value.
Evidence of this practice in the wider F# community can be found e.g. in Scott Wlaschin's Railway oriented programming (part 2) as composition of functions returning the Result<'TSuccess,'TFailure> type.
Reasoning that where there's a bind, there must be also fish, I try to parametrize the canonical Kleisli operator's definition let (>=>) f g a = f a >>= g with the bind function itself:
let mkFish bind f g a = bind g (f a)
This works wonderfully with the caveat that generally one shouldn't unleash special operators on user-facing code. I can compose functions returning options...
module Option =
let (>=>) f = mkFish Option.bind f
let odd i = if i % 2 = 0 then None else Some i
let small i = if abs i > 10 then None else Some i
[0; -1; 9; -99] |> List.choose (odd >=> small)
// val it : int list = [-1; 9]
... or I can devise a function application to the two topmost values of a stack and push the result back without having to reference the data structure I'm operating on explicitly:
module Stack =
let (>=>) f = mkFish (<||) f
type 'a Stack = Stack of 'a list
let pop = function
| Stack[] -> failwith "Empty Stack"
| Stack(x::xs) -> x, Stack xs
let push x (Stack xs) = Stack(x::xs)
let apply2 f =
pop >=> fun x ->
pop >=> fun y ->
push (f x y)
But what bothers me is that the signature val mkFish : bind:('a -> 'b -> 'c) -> f:('d -> 'b) -> g:'a -> a:'d -> 'c makes no sense. Type variables are in confusing order, it's overly general ('a should be a function), and I'm not seeing a natural way to annotate it.
How can I abstract here in the absence of formal functors and monads, not having to define the Kleisli operator explicitly for each type?
You can't do it in a natural way without Higher Kinds.
The signature of fish should be something like:
let (>=>) (f:'T -> #Monad<'U>``) (g:' U -> #Monad<'V>) (x:'T) : #Monad<'V> = bind (f x) g
which is unrepresentable in current .NET type system, but you can replace #Monad with your specific monad, ie: Async and use its corresponding bind function in the implementation.
Having said that, if you really want to use a generic fish operator you can use F#+ which has it already defined by using static constraints. If you look at the 5th code sample here you will see it in action over different types.
Of course you can also define your own, but there is a lot of things to code, in order to make it behave properly in most common scenarios. You can grab the code from the library or if you want I can write a small (but limited) code sample.
The generic fish is defined in this line.
I think in general you really feel the lack of generic functions when using operators, because as you discovered, you need to open and close modules. It's not like functions that you prefix them with the module name, you can do that with operators as well (something like Option.(>=>)) , but then it defeats the whole purpose of using operators, I mean it's no longer an operator.
let mapTuple f (a,b) = (f a, f b)
I'm trying to create a function that applies a function f to both items in a tuple and returns the result as a tuple. F# type inference says that mapTuple returns a 'b*'b tuple. It also assumes that a and b are of the same type.
I want to be able to pass two different types as parameters. You would think that wouldn't work because they both have to be passed as parameters to f. So I thought if they inherited from the same base class, it might work.
Here is a less generic function for what I am trying to accomplish.
let mapTuple (f:Map<_,_> -> Map<'a,'b>) (a:Map<int,double>,b:Map<double, int>) = (f a, f b)
However, it gives a type mismatch error.
How do I do it? Is what I am trying to accomplish even possible in F#?
Gustavo is mostly right; what you're asking for requires higher-rank types. However,
.NET (and by extension F#) does support (an encoding of) higher-rank types.
Even in Haskell, which supports a "nice" way of expressing such types (once you've enabled the right extension), they wouldn't be inferred for your example.
Digging into point 2 may be valuable: given map f a b = (f a, f b), why doesn't Haskell infer a more general type than map :: (t1 -> t) -> t1 -> t1 -> (t, t)? The reason is that once you include higher-rank types, it's not typically possible to infer a single "most general" type for a given expression. Indeed, there are many possible higher-rank signatures for map given its simple definition above:
map :: (forall t. t -> t) -> x -> y -> (x, y)
map :: (forall t. t -> z) -> x -> y -> (z, z)
map :: (forall t. t -> [t]) -> x -> y -> ([x], [y])
(plus infinitely many more). But note that these are all incompatible with each other (none is more general than another). Given the first one you can call map id 1 'c', given the second one you can call map (\_ -> 1) 1 'c', and given the third one you can call map (\x -> [x]) 1 'c', but those arguments are only valid with each of those types, not with the other ones.
So even in Haskell you need to specify the particular polymorphic signature you want to use - this may be a bit of a surprise if you're coming from a more dynamic language. In Haskell, this is relatively clean (the syntax is what I've used above). However, in F# you'll have to jump through an additional hoop: there's no clean syntax for a "forall" type, so you'll have to create an additional nominal type instead. For example, to encode the first type above in F# I'd write something like this:
type Mapping = abstract Apply : 'a -> 'a
let map (m:Mapping) (a, b) = m.Apply a, m.Apply b
let x, y = map { new Mapping with member this.Apply x = x } (1, "test")
Note that in contrast to Gustavo's suggestion, you can define the first argument to map as an expression (rather than forcing it to be a member of some separate type). On the other hand, there's clearly a lot more boilerplate than would be ideal...
This problem has to do with rank-n types which are supported in Haskell (through extensions) but not in .NET type system.
One way I found to workaround this limitation is to pass a type with a single method instead of a function and then define an inline map function with static constraints, for example let's suppose I have some generic functions: toString and toOption and I want to be able to map them to a tuple of different types:
type ToString = ToString with static member inline ($) (ToString, x) = string x
type ToOption = ToOption with static member ($) (ToOption, x) = Some x
let inline mapTuple f (x, y) = (f $ x, f $ y)
let tuple1 = mapTuple ToString (true, 42)
let tuple2 = mapTuple ToOption (true, 42)
// val tuple1 : string * string = ("True", "42")
// val tuple2 : bool option * int option = (Some true, Some 42)
ToString will return the same type but operating with arbitrary types. ToOption will return two Generics of different types.
By using a binary operator type inference creates the static constraints for you and I use $ because in Haskell it means apply so a nice detail is that for haskellers f $ x reads already apply x to f.
At the risk of stating the obvious, a good enough solution might be to have a mapTuple that takes two functions instead of one:
let mapTuple fa fb (a, b) = (fa a, fb b)
If your original f is generic, passing it as fa and fb will give you two concrete instantiations of the function with the types you're looking for. At worst, you just need to pass the same function twice when a and b are of the same type.
I am trying to write a typed abstract syntax tree datatype that can represent
function application.
So far I have
type Expr<'a> =
| Constant of 'a
| Application of Expr<'b -> 'a> * Expr<'b> // error: The type parameter 'b' is not defined
I don't think there is a way in F# to write something like 'for all b' on that last line - am I approaching this problem wrongly?
In general, the F# type system is not expressive enough to (directly) define a typed abstract syntax tree as the one in your example. This can be done using generalized algebraic data types (GADTs) which are not supported in F# (although they are available in Haskell and OCaml). It would be nice to have this in F#, but I think it makes the language a bit more complex.
Technically speaking, the compiler is complaining because the type variable 'b is not defined. But of course, if you define it, then you get type Expr<'a, 'b> which has a different meaning.
If you wanted to express this in F#, you'd have to use a workaround based on interfaces (an interface can have generic method, which give you a way to express constraint like exists 'b which you need here). This will probably get very ugly very soon, so I do not think it is a good approach, but it would look something like this:
// Represents an application that returns 'a but consists
// of an argument 'b and a function 'b -> 'a
type IApplication<'a> =
abstract Appl<'b> : Expr<'b -> 'a> * Expr<'b> -> unit
and Expr<'a> =
// Constant just stores a value...
| Constant of 'a
// An application is something that we can call with an
// implementation (handler). The function then calls the
// 'Appl' method of the handler we provide. As this method
// is generic, it will be called with an appropriate type
// argument 'b that represents the type of the argument.
| Application of (IApplication<'a> -> unit)
To represent an expression tree of (fun (n:int) -> string n) 42, you could write something like:
let expr =
Application(fun appl ->
appl.Appl(Constant(fun (n:int) -> string n),
Constant(42)))
A function to evaluate the expression can be written like this:
let rec eval<'T> : Expr<'T> -> 'T = function
| Constant(v) -> v // Just return the constant
| Application(f) ->
// We use a bit of dirty mutable state (to keep types simpler for now)
let res = ref None
// Call the function with a 'handler' that evaluates function application
f { new IApplication<'T> with
member x.Appl<'A>(efunc : Expr<'A -> 'T>, earg : Expr<'A>) =
// Here we get function 'efunc' and argument 'earg'
// The type 'A is the type of the argument (which can be
// anything, depending on the created AST)
let f = eval<'A -> 'T> efunc
let a = eval<'A> earg
res := Some <| (f a) }
res.Value.Value
As I said, this is a bit really extreme workaround, so I do not think it is a good idea to actually use it. I suppose the F# way of doing this would be to use untyped Expr type. Can you write a bit more about the overall goal of your project (perhaps there is another good approach)?
I am trying to write a typed abstract syntax tree datatype that can represent
function application.
So far I have
type Expr<'a> =
| Constant of 'a
| Application of Expr<'b -> 'a> * Expr<'b> // error: The type parameter 'b' is not defined
I don't think there is a way in F# to write something like 'for all b' on that last line - am I approaching this problem wrongly?
In general, the F# type system is not expressive enough to (directly) define a typed abstract syntax tree as the one in your example. This can be done using generalized algebraic data types (GADTs) which are not supported in F# (although they are available in Haskell and OCaml). It would be nice to have this in F#, but I think it makes the language a bit more complex.
Technically speaking, the compiler is complaining because the type variable 'b is not defined. But of course, if you define it, then you get type Expr<'a, 'b> which has a different meaning.
If you wanted to express this in F#, you'd have to use a workaround based on interfaces (an interface can have generic method, which give you a way to express constraint like exists 'b which you need here). This will probably get very ugly very soon, so I do not think it is a good approach, but it would look something like this:
// Represents an application that returns 'a but consists
// of an argument 'b and a function 'b -> 'a
type IApplication<'a> =
abstract Appl<'b> : Expr<'b -> 'a> * Expr<'b> -> unit
and Expr<'a> =
// Constant just stores a value...
| Constant of 'a
// An application is something that we can call with an
// implementation (handler). The function then calls the
// 'Appl' method of the handler we provide. As this method
// is generic, it will be called with an appropriate type
// argument 'b that represents the type of the argument.
| Application of (IApplication<'a> -> unit)
To represent an expression tree of (fun (n:int) -> string n) 42, you could write something like:
let expr =
Application(fun appl ->
appl.Appl(Constant(fun (n:int) -> string n),
Constant(42)))
A function to evaluate the expression can be written like this:
let rec eval<'T> : Expr<'T> -> 'T = function
| Constant(v) -> v // Just return the constant
| Application(f) ->
// We use a bit of dirty mutable state (to keep types simpler for now)
let res = ref None
// Call the function with a 'handler' that evaluates function application
f { new IApplication<'T> with
member x.Appl<'A>(efunc : Expr<'A -> 'T>, earg : Expr<'A>) =
// Here we get function 'efunc' and argument 'earg'
// The type 'A is the type of the argument (which can be
// anything, depending on the created AST)
let f = eval<'A -> 'T> efunc
let a = eval<'A> earg
res := Some <| (f a) }
res.Value.Value
As I said, this is a bit really extreme workaround, so I do not think it is a good idea to actually use it. I suppose the F# way of doing this would be to use untyped Expr type. Can you write a bit more about the overall goal of your project (perhaps there is another good approach)?
I was chatting with Sadek Drobi on twitter when be brought up that F# didn't seem to support Infinite Types. It turns out that in C# you can do something along these lines:
delegate RecDelegate<T> RecDelegate<T>(T x);
However, after some experimentation on both our parts, we determined that the same in F# seems impossible both implicit and explicitly.
Explicit:
type 'a specialF = 'a->specialF<'a>
error FS0191: This type definition
involves an immediate cyclic reference
through an abbreviation, struct field
or inheritance relation.
Implicit:
let rec specialF (x: 'a) = specialF
Type mismatch. Expecting a 'b but
given a 'a -> 'b. The resulting type
would be infinite when unifying ''b'
and ''a -> 'b'.
Of course, these are intentionally simple samples.
I was wondering if I am somehow mistaken. Perhaps I missed some type of necessary annotation?
You can also do something like
type 'a RecType = RecType of ('a -> 'a RecType)
to create a named type through which to perform the recursion. Now this works:
let rec specialF = RecType (fun _ -> specialF)
type d<'T> = delegate of 'T -> d<'T> //'
let del : d<int> = null
let anotherDel = del.Invoke(1).Invoke(2).Invoke(3)
I think you need a named type that is representable directly in CLI to break the recursion, so in F# this means you need an actual delegate type as well.
Recursive record types should work as well.
type A = { A : A }
let rec a : A = { A = a }
I'd be interested in a practical application. Or even an impractical one :)