F# Function with parameter constraints - f#

I ran into an error similar to this forum post on hubfs, which solved my problem but spawned off some questions about the code in that thread.
let test x = printfn "n"
let finall x = x : 'a -> unit
let i x = finall test x
Can someone explain to me what line 2 is accomplishing?
I see that the type of finall is
finall: ('a -> unit) -> ('a -> unit)
So its just a function that takes in a function and returns that function.
What would be the reason to do lines 2 and 3?
Can you later define a function body to finall?
it appears that you don't have to parenthesize the call on the third line, is that a result of line 2?

Yeah, the code on that thread does not make much sense. 'finall' is basically the identity function (let id x = x) except that it constrains its argument to be a function-returning-unit rather than some arbitrary value. Basically it doesn't do anything useful, you could just as easily write
let test x = printfn "n"
let i x = test x
I expect that this snippet maybe came from someone who started with an error message they didn't understand, and tried to strip it down to a tiny sample repro.
(Regarding function calls, you never need parens to call a let-bound function in F#:
f x
is a call, and function application works in the normal way to support currying, which means
f x y
means
(f x) y
which is what is happening on the 'finall test x' line.)

Related

Function that returns another function executes the body of outer function on every call of the returned function

It is actually pretty unxpected to me but consider this snippet in F#:
let f x =
printfn $"{x}"
fun x' -> x'
let y<'t> = f 1 //> val y<'t> : (obj -> obj)
y 2
//>
//1
//val it: obj = 2
what I would expect is that it will print "1" only when you bind f 1 to "y" (and that would tell me that "f" body only executes once) but seem like it executes "f" body on the every call of "y". Is it unavoidable effect related to auto curring or I'm missing something and there is a way to bypass outer function body execution on the every call of the returned function?
The hint as to what's going on here is the fact that 't has been constrained to obj and the signature of y is (obj -> obj). That's the F# compiler effectively say, "I give up, these have no real types, it is whatever it is" and emitting something that can execute at runtime but without any real type safety.
A side effect of this is that because it can't "pin down" y to a known signature, it cannot evaluate f, so it just emits y as a direct call to f, since you've effectively told the compiler that this is fine by parameterizing it with 't (which ends up just being obj, or "whatever").
Why is this happening? Value restriction!
I suspect you've evaluated this in F# Interactive block-by-block. The line of code that defines let y = f 1 is not possible to compile with more information. You can do so in two ways:
Use y with a real type that will pin its signature to the type you're using it as.
Give it an explicit signature like let y: int -> int = f 1 so that it's pinned down to a concrete type.
That's why if you execute this entire snippet in FSI or run it as a program, things work exactly like you'd expect:
let f x =
printfn $"{x}"
fun x' -> x'
let y = f 1
y 2
y 3
This is because y is generic.
Every time you refer to y, you choose a particular 't to go with that. For example:
let a = y<int>
let b = y<string>
a and b cannot be the same value, because they have been obtained from different instantiations of y. They have to be two different values. And this in turn means that y itself cannot be a single value. It has to be a function.
And that's what it is under the hood: it's compiled as a function, and every time you refer to it, the function is instantiated with the generic parameter you chose, and the body of the function is executed to obtain the result.
If you remove the generic parameter and give y a concrete type, the issue should go away:
let y = f 1 : obj -> obj

How to pass a "Function" type in FunScript?

I am encountering this Function type that I need to pass to a JQueryAnimationOptions object. I usually would pass a lambda to callbacks but these seem to be incompatible. I looked up every sample I could find in the FunScript repo. and couldn't find any workaround.
It also said the the Function is actually an interface (for what?) when used as a return statement Error: Invalid use of interface type.
So how to pass a callback argument with this Function type?
the code:
[<FunScript.JS>]
module Main
open FunScript
open FunScript.TypeScript
let sayHelloFrom (name:string) =
Globals.window.alert("Hello, " + name)
let jQuery (selector:string) = Globals.jQuery.Invoke selector
let main() =
let options = createEmpty<JQueryAnimationOptions>()
options.duration <- 3000
options.complete <- (fun _ -> sayHelloFrom("F#"))
let properties = createEmpty<Object>()
properties.Item("opacity") <- 1
let mainContent = jQuery "#mainContent"
mainContent.animate(properties, options) |> ignore
mainContent.click(fun e -> sayHelloFrom("F#") :> obj)
This works more or less as you would expect when passing lambdas between F# and C#. In F#, functions can be curried, while in C# (and JavaScript) cannot. So when you need to send a lambda from F# to C# you need to convert it first. In F# this is done by wrapping the lambda like this:
open System.Linq
open System.Collections.Generic
let ar = [|1;2;3|]
let f = fun (x: int) (y: int) -> x + y
let acc = ar.Aggregate( System.Func<int,int,int>(f) )
Actually, the F# compiler can deduce the types most of the times, so you only need to write: System.Func<_,_,_>(f). Furthermore, when passing a F# lambda to a method expecting a C# lambda, the compiler makes the wrapping automatically for you. Then the previous example becomes:
let ar = [|1;2;3|]
let acc = ar.Aggregate( fun x y -> x + y )
(Of course, in this case it would be better to use the idiomatic Array.reduce. This is just a contrived example.)
This works exactly the same when interacting with JS using FunScript. The only thing you need to be aware of is how F# lambdas get translated to JS. To allow currying, a lambda with two or more parameters like fun x y -> x + y becomes:
function (x) {
return function (y) {
return x + y;
}
}
Which may be a problem because the native JS will expect the following signature: function (x, y). In that case, you would have to wrap the lambda with System.Func<_,_,_>() as when interacting with C# (remember this is done automatically if you pass the lambda to a method).
However, lambdas with just one parameter don't suppose any problem: fun x -> x*x becomes function (x) { return x*x; }. In that case you don't need to wrap them (it doesn't hurt to do it anyway) and it's enough just to use unbox to appease the F# compiler when necessary. Just please be aware the FunScript compiler ignores unbox in the final JS code so there'll be no type check at all at runtime.
I hope the explanation is clear. Please add a comment if it isn't and I'll edit the answer.
Nevermind , I found the solution, I had to unbox the lambda:
options.complete <- unbox<Function> (fun _ -> sayHelloFrom("F#"))

Going more idiomatic with F#

A very simple example of what I'm trying to do: I know it's possible to write:
let myFunc = anotherFunc
instead of
let myFunc = fun x -> anotherFunc x
I've got two functions fDate1, fDate2 - both of type DateTime -> bool. I need to construct a function that takes a date and verifies if any of fDate1, fDate2 returns true. For now I've invented the following expression:
let myDateFunc = fun x -> (fDate1 x) || (fDate2 x)
Is there a better way of doing these (e.g. using '>>' or high order funcions) ?
I don't think there is anything non-idiomatic with your code. In my opinion, one of the strong points about F# is that you can use it to write simple and easy-to-understand code. From that perspective, nothing could be simpler than writing just:
let myDateFunc x = fDate1 x || fDate2 x
If you had more functions than just two, then it might make sense to write something like:
let dateChecks = [ fDate1; fDate2 ]
let myDateFunc x = dateChecks |> Seq.exists (fun f -> f x)
But again, this only makes sense when you actually need to use a larger number of checks or when you are adding checks often. Unnecessary abstraction is also a bad thing.
You can define a choice combinator:
let (<|>) f g = fun x -> f x || g x
let myDateFunc = fDate1 <|> fDate2
In general, you should use explicit function arguments. The elaborated form of myDateFunc can be written as:
let myDateFunc x = fDate1 x || fDate2 x
As other answers say, your current approach is fine. What is not said is that idiomatic style often produces less readable code. So if you are working in a real project and expect other developers to understand your code, it is not recommended to go too far with unnecessary function composition.
However, for purposes of self-education, you may consider the following trick, a bit in FORTH style:
// Define helper functions
let tup x = x,x
let untup f (x,y) = f x y
let call2 f g (x,y) = f x, g y
// Use
let myFunc =
tup
>> call2 fDate1 fDate2
>> untup (||)
Here, you pass the original value x through a chain of transformations:
make a tuple of the same value;
apply each element of the tuple to corresponding function, obtaining a tuple of results;
"fold" a tuple of booleans with or operator to a single value;
There are many drawbacks with this approach, including that both of fDate1 and fDate2 will be evaluated while it may not be necessary, extra tuples created degrading performance, and more.

F# and lisp-like apply function

For starters, I'm a novice in functional programming and F#, therefore I don't know if it's possible to do such thing at all. So let's say we have this function:
let sum x y z = x + y + z
And for some reason, we want to invoke it using the elements from a list as an arguments. My first attempt was just to do it like this:
//Seq.fold (fun f arg -> f arg) sum [1;2;3]
let rec apply f args =
match args with
| h::hs -> apply (f h) hs
| [] -> f
...which doesn't compile. It seems impossible to determine type of the f with a static type system. There's identical question for Haskell and the only solution uses Data.Dynamic to outfox the type system. I think the closest analog to it in F# is Dynamitey, but I'm not sure if it fits. This code
let dynsum = Dynamitey.Dynamic.Curry(sum, System.Nullable<int>(3))
produces dynsum variable of type obj, and objects of this type cannot be invoked, furthermore sum is not a .NET Delegate.So the question is, how can this be done with/without that library in F#?
F# is a statically typed functional language and so the programming patterns that you use with F# are quite different than those that you'd use in LISP (and actually, they are also different from those you'd use in Haskell). So, working with functions in the way you suggested is not something that you'd do in normal F# programming.
If you had some scenario in mind for this function, then perhaps try asking about the original problem and someone will help you find an idiomatic F# approach!
That said, even though this is not recommended, you can implement the apply function using the powerful .NET reflection capabilities. This is slow and unsafe, but if is occasionally useful.
open Microsoft.FSharp.Reflection
let rec apply (f:obj) (args:obj list) =
let invokeFunc =
f.GetType().GetMethods()
|> Seq.find (fun m ->
m.Name = "Invoke" &&
m.GetParameters().Length = args.Length)
invokeFunc.Invoke(f, Array.ofSeq args)
The code looks at the runtime type of the function, finds Invoke method and calls it.
let sum x y z = x + y + z
let res = apply sum [1;2;3]
let resNum = int res
At the end, you need to convert the result to an int because this is not statically known.

How do I define y-combinator without "let rec"?

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 ^^

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