I am working on a new version of the Spiral language and as one of its features, it will have structural equality and comparison similar to F#. But I am having difficulty figuring out how to compile that efficiently to F#.
The problem is that I cannot find how F# exposes tag information for its union cases.
For example, if I wrote cmp a b I'd like the Spiral compiler to generate something like the following F# code.
type T =
| A of int32
| B of float
let cmp_t = function
| A x, A x' -> cmp_i32 (x, x')
| B x, B x' -> cmp_f64 (x, x')
| x, x' -> cmp_i32 (union_tag x, union_tag x')
I've looked around and there are some proposals to expose the tag information, but nothing is out yet. I am wondering whether the F# compiler library has some functionality that I am not aware of for extracting the tag.
Otherwise for every union type, I am going to have to generate a custom tag function like...
let tag_t = function
| A _ -> 0
| B _ -> 1
But this can't be the right choice because F# has to be using some kind of tag under the hood for each of the union cases in order to differentiate them. It has to be able to know the tag for its own comparison function. Making tag functions like these would bloat the generated code and would be inefficient at runtime.
What should I do here?
You can use F# Reflection API for this.
Reflection.FSharpType.GetUnionCases can return you a full list of DU cases, each represented as a UnionCaseInfo structure, which has such useful properties as Name and Tag:
> Reflection.FSharpType.GetUnionCases(typeof<T>) |> Array.map (fun c -> c.Tag)
[|0; 1|]
And if you have a concrete value of the type (like in your cmp_t function), check out the Reflection.FSharpValue.GetUnionFields function, which, contrary to its name, will return you not only the fields, but also the UnionCaseInfo representing the case:
> let x = A 42
> let caseInfo, fields = Reflection.FSharpValue.GetUnionFields(x, typeof<T>)
> caseInfo.Tag
0
Keep in mind, however, that for F# discriminated unions the compiler will generate structural comparison automatically, so your cmpt_t function is redundant:
> A 42 < A 54
true
> A 3 > B 1.0
false
// Or more generally:
> (A 42 :> System.IComparable<T>).CompareTo(A 54)
-1
So your cmp_t can really be just this:
let cmp_t (x: 'a when 'a :> System.IComparable<'a>) y =
(x :> System.IComparable<'a>).CompareTo(y)
> cmp_t (A 42) (A 54)
-1
And behold! There is already such function in the standard library. It's called compare:
> compare (A 42) (A 54)
-1
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.
I have found a presentation by Don Syme which shows that Erlang's
fac(0) -> 1
fac(N) -> N * fac(N-1).
is equivalent to F#'s
let rec fac = function
| 0 -> 1
| n -> n * fac (n-1)
But it looks like there is no way to use pattern matching for a different arity without losing type safety. E.g. one could use a list pattern matching, but then a type must be a common base type (such as object):
let concat = function
| [x;y] -> x.ToString() + y.ToString()
| [x] -> x.ToString()
Given that F# functions in modules do not support overloads, it looks like the only way to rewrite Erlang code into F# with static typing is to use static classes with method overloading instead of modules. Is there a better way to rewrite Erlang functions with different arity in F#?
In general, is it correct to say that Erlang's argument matching is closer to .NET's (including C#) method overloading rather than to F#'s pattern matching? Or there is no direct replacement between the two, e.g. there could be a function in Erlang with different arities + a guard:
max(x) -> x.
max(x,y) when x > y -> x.
max(x,y) -> y.
max(comparer, x, y) -> if comparer(x,y) > 0 -> x; true -> y end.
In the last case the arguments are of different types. How would you rewrite it in F#?
You can achieve something close to overloading by rethinking the problem slightly. Instead of thinking about the function as the axis of variability, think of the input as the variable part. If you do that, you'll realise that you can achieve the same with a discriminated union.
Here's a more contrived example than the one in the linked article:
type MyArguments = One of int | Two of int * int
let foo = function
| One x -> string x
| Two (x, y) -> sprintf "%i%i" x y
Usage:
> foo (One 42);;
val it : string = "42"
> foo (Two (13, 37));;
val it : string = "1337"
Obviously, instead of defining such a 'stupid' type as the above MyArguments, you'd define a discriminated union that makes sense in the domain you're modelling.
I'm trying to write some function that handle errors by returning double options instead of doubles. Many of these functions call eachother, and so take double options as inputs to output other double options. The problem is, I can't do with double options what I can do with doubles--something simple like add them using '+'.
For example, a function that divides two doubles, and returns a double option with none for divide by zero error. Then another function calls the first function and adds another double option to it.
Please tell me if there is a way to do this, or if I have completely misunderstood the meaning of F# option types.
This is called lifting - you can write function to lift another function over two options:
let liftOpt f o1 o2 =
match (o1, o2) with
| (Some(v1), Some(v2)) -> Some(f v1 v2)
| _ -> None
then you can supply the function to apply e.g.:
let inline addOpt o1 o2 = liftOpt (+) o1 o2
liftA2 as mentioned above will provide a general way to 'lift' any function that works on the double arguments to a function that can work on the double option arguments.
However, in your case, you may have to write special functions yourself to handle the edge cases you mention
let (<+>) a b =
match (a, b) with
| (Some x, Some y) -> Some (x + y)
| (Some x, None) -> Some (x)
| (None, Some x) -> Some (x)
| (None, None) -> None
Note that liftA2 will not put the cases where you want to add None to Some(x) in automatically.
The liftA2 method for divide also needs some special handling, but its structure is generally what we would write ourselves
let (</>) a b =
match (a, b) with
| (Some x, Some y) when y <> 0.0d -> Some (x/y)
| _ -> None
You can use these functions like
Some(2.0) <+> Some(3.0) // will give Some(5.0)
Some(1.0) </> Some(0.0) // will give None
Also, strictly speaking, lift is defined as a "higher order function" - something that takes a function and returns another function.
So it would look something like this:
let liftOpt2 f =
(function a b ->
match (a, b) with
| (Some (a), Some (b)) -> f a b |> Some
| _ -> None)
In the end, I realized what I was really looking for was the Option.get function, which simply takes a 'a option and returns an 'a. That way, I can pattern match, and return the values I want.
In this case you might want to consider Nullables over Options, for two reasons:
Nullables are value types, while Options are reference types. If you have large collections of these doubles, using Nullables will keep the numbers on the stack instead of putting them on the heap, potentially improving your performance.
Microsoft provides a bunch of built-in Nullable Operators that do let you directly perform math on nullables, exactly as you're trying to do with options.
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.
In almost all examples, a y-combinator in ML-type languages is written like this:
let rec y f x = f (y f) x
let factorial = y (fun f -> function 0 -> 1 | n -> n * f(n - 1))
This works as expected, but it feels like cheating to define the y-combinator using let rec ....
I want to define this combinator without using recursion, using the standard definition:
Y = λf·(λx·f (x x)) (λx·f (x x))
A direct translation is as follows:
let y = fun f -> (fun x -> f (x x)) (fun x -> f (x x));;
However, F# complains that it can't figure out the types:
let y = fun f -> (fun x -> f (x x)) (fun x -> f (x x));;
--------------------------------^
C:\Users\Juliet\AppData\Local\Temp\stdin(6,33): error FS0001: Type mismatch. Expecting a
'a
but given a
'a -> 'b
The resulting type would be infinite when unifying ''a' and ''a -> 'b'
How do I write the y-combinator in F# without using let rec ...?
As the compiler points out, there is no type that can be assigned to x so that the expression (x x) is well-typed (this isn't strictly true; you can explicitly type x as obj->_ - see my last paragraph). You can work around this issue by declaring a recursive type so that a very similar expression will work:
type 'a Rec = Rec of ('a Rec -> 'a)
Now the Y-combinator can be written as:
let y f =
let f' (Rec x as rx) = f (x rx)
f' (Rec f')
Unfortunately, you'll find that this isn't very useful because F# is a strict language,
so any function that you try to define using this combinator will cause a stack overflow.
Instead, you need to use the applicative-order version of the Y-combinator (\f.(\x.f(\y.(x x)y))(\x.f(\y.(x x)y))):
let y f =
let f' (Rec x as rx) = f (fun y -> x rx y)
f' (Rec f')
Another option would be to use explicit laziness to define the normal-order Y-combinator:
type 'a Rec = Rec of ('a Rec -> 'a Lazy)
let y f =
let f' (Rec x as rx) = lazy f (x rx)
(f' (Rec f')).Value
This has the disadvantage that recursive function definitions now need an explicit force of the lazy value (using the Value property):
let factorial = y (fun f -> function | 0 -> 1 | n -> n * (f.Value (n - 1)))
However, it has the advantage that you can define non-function recursive values, just as you could in a lazy language:
let ones = y (fun ones -> LazyList.consf 1 (fun () -> ones.Value))
As a final alternative, you can try to better approximate the untyped lambda calculus by using boxing and downcasting. This would give you (again using the applicative-order version of the Y-combinator):
let y f =
let f' (x:obj -> _) = f (fun y -> x x y)
f' (fun x -> f' (x :?> _))
This has the obvious disadvantage that it will cause unneeded boxing and unboxing, but at least this is entirely internal to the implementation and will never actually lead to failure at runtime.
I would say it's impossible, and asked why, I would handwave and invoke the fact that simply typed lambda calculus has the normalization property. In short, all terms of the simply typed lambda calculus terminate (consequently Y can not be defined in the simply typed lambda calculus).
F#'s type system is not exactly the type system of simply typed lambda calculus, but it's close enough. F# without let rec comes really close to the simply typed lambda calculus -- and, to reiterate, in that language you cannot define a term that does not terminate, and that excludes defining Y too.
In other words, in F#, "let rec" needs to be a language primitive at the very least because even if you were able to define it from the other primitives, you would not be able to type this definition. Having it as a primitive allows you, among other things, to give a special type to that primitive.
EDIT: kvb shows in his answer that type definitions (one of the features absent from the simply typed lambda-calculus but present in let-rec-less F#) allow to get some sort of recursion. Very clever.
Case and let statements in ML derivatives are what makes it Turing Complete, I believe they're based on System F and not simply typed but the point is the same.
System F cannot find a type for the any fixed point combinator, if it could, it wasn't strongly normalizing.
What strongly normalizing means is that any expression has exactly one normal form, where a normal form is an expression that cannot be reduced any further, this differs from untyped where every expression has at max one normal form, it can also have no normal form at all.
If typed lambda calculi could construct a fixed point operator in what ever way, it was quite possible for an expression to have no normal form.
Another famous theorem, the Halting Problem, implies that strongly normalizing languages are not Turing complete, it says that's impossible to decide (different than prove) of a turing complete language what subset of its programs will halt on what input. If a language is strongly normalizing, it's decidable if it halts, namely it always halts. Our algorithm to decide this is the program: true;.
To solve this, ML-derivatives extend System-F with case and let (rec) to overcome this. Functions can thus refer to themselves in their definitions again, making them in effect no lambda calculi at all any more, it's no longer possible to rely on anonymous functions alone for all computable functions. They can thus again enter infinite loops and regain their turing-completeness.
Short answer: You can't.
Long answer:
The simply typed lambda calculus is strongly normalizing. This means it's not Turing equivalent. The reason for this basically boils down to the fact that a Y combinator must either be primitive or defined recursively (as you've found). It simply cannot be expressed in System F (or simpler typed calculi). There's no way around this (it's been proven, after all). The Y combinator you can implement works exactly the way you want, though.
I would suggest you try scheme if you want a real Church-style Y combinator. Use the applicative version given above, as other versions won't work, unless you explicitly add laziness, or use a lazy Scheme interpreter. (Scheme technically isn't completely untyped, but it's dynamically typed, which is good enough for this.)
See this for the proof of strong normalization:
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.127.1794
After thinking some more, I'm pretty sure that adding a primitive Y combinator that behaves exactly the way the letrec defined one does makes System F Turing complete. All you need to do to simulate a Turing machine then is implement the tape as an integer (interpreted in binary) and a shift (to position the head).
Simply define a function taking its own type as a record, like in Swift (there it's a struct) :)
Here, Y (uppercase) is semantically defined as a function that can be called with its own type. In F# terms, it is defined as a record containing a function named call, so for calling a y defined as this type, you have to actually call y.call :)
type Y = { call: Y -> (int -> int) }
let fibonacci n =
let makeF f: int -> int =
fun x ->
if x = 0 then 0 else if x = 1 then 1 else f(x - 1) + f(x - 2)
let y = { call = fun y -> fun x -> (makeF (y.call y)) x }
(y.call y) n
It's not supremely elegant to read but it doesn't resort to recursion for defining a y combinator that is supposed to provide recursion all by itself ^^