Why are parentheses needed on this F# function? - f#

Why are parentheses needed on read_rest_of_csv below?
let read_rest_of_csv() =
csv_data.Add(csv_fileH.ReadFields()) |> ignore
not csv_fileH.EndOfData
while read_rest_of_csv() do ignore None
Without the parentheses, the loop will not terminate.
open System
open System.Threading
open System.Collections.Generic
open System.Linq
open System.Text
open System.Threading.Tasks
open System.IO
open Microsoft.VisualBasic.FileIO
[<EntryPoint>]
let main argv =
let csv_fileH = new TextFieldParser("test1.csv")
csv_fileH.TextFieldType = FieldType.Delimited |> ignore
let x = csv_fileH.SetDelimiters(",")
let csv_data = new List<string[]>()
let eod = csv_fileH.EndOfData
if not eod then
let column_headings = csv_fileH.ReadFields()
csv_data.Add(column_headings) |> ignore
let read_rest_of_csv =
csv_data.Add(csv_fileH.ReadFields()) |> ignore
not csv_fileH.EndOfData
while read_rest_of_csv do ignore None
0
I apologize that I cannot remember where I saw this. I think it was in SO. It's a nice example.
Could this be that without parens I'm dealing with a function object of sorts?
I am indeed coming from not only a C, C++, and C# background, but also an intermediate Clojure background as well. In my case with F# syntax, reading my Haskell manual in a little more detail might have helped, because the syntaxes seem similar.

It seems that people coming from C-family languages (C#, Java, C, C++, JavaScript) are having problems understanding the use of brackets in F#. I certainly had, and it took me some years learning how things work.
In a nutshell, the most basic building block in F# is a value. Values can be let-bound:
let foo = bar
This means that foo is a value, which happens to be equal to bar.
Functions are also values:
// 'a -> 'a * 'a
let f = fun x -> x, x
Here, f is a function that takes some value (x) and returns a tuple with x as both the first and the second element.
That's a bit cumbersome to write, so there's a shorthand for that:
// 'a -> 'a * 'a
let f x = x, x
Notice that there are no brackets in these expressions.
Sometimes you need to adjust the precedence of operators. Just like in maths, 1 + 2 * 3 (which is equivalent to 1 + (2 * 3)) isn't the same as (1 + 2) * 3. In F#, you also use brackets to override precedence. Thus
// 'a -> string * 'a
let f x = someOtherFunction x, x
isn't the same as
// x:'a -> string
let f x = someOtherFunction (x, x)
(in this case, someOtherFunction is a function that returns a string.)
Notice that the brackets don't denote a function call; they're only there to control order of evaluation.
Sometimes, you want to define a function that doesn't take any input. You can't, however, define it like this:
let f = whatever
because that would make it a value that's immediately let-bound to whatever. Instead, you can let the function take a value of the built-in type unit. This type only has a single value, which is written ():
let f () = whatever
This means that f is a function that pattern matches its input against the only known value of unit.
Whenever you invoke f with (), the expression whatever is evaluated and returned.

Without the parentheses, the content executes once and never again. read_rest_of_csv has a type of bool: You are basically saying while true do ignore None.
The parentheses indicate that read_rest_of_csv has type unit -> bool, so every time you invoke it, it reads a row and moves the cursor. Otherwise, it will only do this once.

The answer to your question is that:
let read_rest_of_csv =
csv_data.Add(csv_fileH.ReadFields()) |> ignore
not csv_fileH.EndOfData
is not a function at all. This is no different from:
> let i = 1;;
val i : int = 1
This declares a binding with an integer value. If you want to declare a binding with a function value which takes no parameters, that looks like this:
> let i () = 1;;
val i : unit -> int
The exact same reasoning applies to read_rest_of_csv. Without the parenthesis, you are declaring a binding with type bool. With the parenthesis, you are declaring a binding with type unit->bool i.e. a binding with a function value where the function takes no inputs and returns a bool value.

Related

What is missing using interfaces compared to true type-classes?

F# does not (currently) support type-classes. However, F# does support the OOP aspects of C#.
I was wondering, what is lost doing this approach compared to true type-classes?
// A concrete type
type Foo =
{
Foo : int
}
// "Trait" for things that can be shown
type IShowable =
abstract member Show : unit -> string
module Showable =
let show (showable : IShowable) =
showable.Show()
// "Witness" of IShowable for Foo
module Foo =
let asShowable (foo : Foo) =
{
new IShowable with
member this.Show() = string foo.Foo
}
// Slightly awkward usage
{ Foo = 123 }
|> Foo.asShowable
|> Showable.show
|> printfn "%s"
Your suggestion works for simple typeclasses that operate on a single value of a type, like Show. However, what happens when you need a typeclass that isn't so object-oriented? For example, when we want to add two numbers, neither one corresponds to OO's this object:
// not real F#
typeclass Numeric<'a> = // e.g. Numeric<int> or Numeric<float>
abstract member (+) : 'a -> 'a -> 'a // e.g. 2 + 3 = 5 or 2.0 + 3.0 = 5.0
...
Also, keep in mind that many useful typeclasses require higher-kinded types. For example, consider the monad typeclass, which would look something like this:
// not real F#
typeclass Monad<'m<_>> = // e.g. Monad<Option<_>> or Monad<Async<_>>
abstract member Return<'a> : 'a -> 'm<'a>
abstract member Bind<'a, 'b> : 'm<'a> -> ('a -> 'm<'b>) -> 'm<'b>
There's no good way to do this with .NET interfaces.
Higher-kinded type classes are indeed impossible to model with interfaces, but that's just because F# does not support higher-kindedness, not because of type classes themselves.
The deeper thing to note is that your encoding isn't actually correct. Sure, if you just need to call show directly, you can do asShowable like that, but that's just the simplest case. Imagine you needed to pass the value to another function that wanted to show it later? And then imagine it was a list of values, not a single one:
let needsToShow (showable: IShowable) (xs: 'a list) =
xs |> List.iter (fun x -> ??? how do I show `x` ???)
No, this wouldn't do of course. The key is that Show should be a function 'a -> string, not unit -> string. And this means that IShowable itself should be generic:
// Haskell: class Showable a where show :: a -> String
type IShowable<'a> with
abstract member Show : 'a -> string
// Haskell: instance Showable Foo where show (Foo i) = show i
module Foo =
let showable = { new IShowable<Foo> with member _.Show foo = string foo.Foo }
// Haskell: needsToShow :: Show a => [a] -> IO ()
let needsToShow (showable: IShowable<'a>) (xs: 'a list) =
xs |> List.iter (fun x -> printfn "%s" (showable.Show x))
// Haskell: needsToShow [Foo 1, Foo 42]
needsToShow Foo.showable [ { Foo: 1 }; { Foo: 42 } ]
And this is, essentially, what type classes are: they're indeed merely dictionaries of functions that are passed everywhere as extra parameters. Every type has such dictionary either available right away (like Foo above) or constructable from other such dictionaries, e.g.:
type Bar<'a> = Bar of 'a
// Haskell: instance Show a => Show (Bar a) where show (Bar a) = "Bar: " <> show a
module Bar =
let showable (showA: IShowable<'a>) =
{ new IShowable<Bar<'a>> with member _.Show (Bar a) = "Bar: " + showA.Show a }
This is completely equivalent to type classes. And in fact, this is exactly how they're implemented in languages like Haskell or PureScript in the first place: like dictionaries of functions being passed as extra parameters. It's not a coincidence that constraints on function type signatures even kinda look like parameters - just with a fat arrow instead of a thin one.
The only real difference is that in F# you have to do that yourself, while in Haskell the compiler figures out all the instances and passes them for you.
And this difference turns out to be kind of important in practice. I mean, sure, for such a simple example as Show for the immediate parameter, you can just pass the damn instance yourself. And even if it's more complicated, I guess you could suck it up and pass a dozen extra parameters.
But where this gets really inconvenient is operators. Operators are functions too, but with operators there is nowhere to stick an extra parameter (or dozen). Check this out:
x = getY >>= \y -> getZ y <&> \z -> y + 42 > z
Here I used four operators from four different classes:
>>= comes from Monad
<&> from Functor
+ from Num
> from Ord
An equivalent in F# with passing instances manually might look something like:
let x =
bind Foo.monad getY <| fun y ->
map Bar.functor (getZ y) <| fun z ->
gt Int.ord (add Int.num y 42) z
Having to do that everywhere is quite unreasonable, you have to agree.
And this is why many F# operators either use SRTPs (e.g. +) or rely on "known" interfaces (e.g. <) - all so you don't have to pass instances manually.

Why this F# function runs only once? I call twice and it runs only once

I wrote the following code to test some MonteCarlo code in F#.
My problem is I only see the random numbers and the "oi" once in my console. I call two times the oneRun function, but it looks that it only runs once.
Here is the code:
let genRandomNumbers count =
let rnd = System.Random()
printf "oi "
List.init count (fun _ -> rnd.NextDouble ())
let oneRun =
let numberofClicks = 0
let randomNumber = genRandomNumbers 50
let action numberofClicks random = if random <= 0.10
then numberofClicks+1
else numberofClicks
randomNumber |> Seq.iter (printf "%f ")
randomNumber |> List.fold action numberofClicks
[<EntryPoint>]
let main argv =
let a = oneRun
printf "%d " a
let b = oneRun
printf "%d " b
let key_info = Console.ReadKey()
0 //
Any hints? Ideas?
To expand a little on Mankarse's correct comment, the F# syntax for defining values and functions looks very similar, so it's easy to get confused between them.
This is a value:
let sum = 42
This is a function:
let addThree x = x + 3
Both values and functions can have blocks following them, not just single lines:
let sumWithSideEffects =
// This will only be evaluated once
printfn "Side effect happens here"
42
let addThree x =
// This will run every time you call the function
let result = x + 3
printfn "Added three to %d and got %d" x result
result
A let declaration that just declares a name is a value. Values are only evaluated once, so any side effects in the value will happen just once. Exactly when they happen is not defined precisely by the language spec, so you can't count on when the side effects will happen. Functions, on the other hand, are evaluated every time the function is called.
Now, when you have a function that takes no parameters, how do you declare it? Well, you declare it by giving it a parameter, but a parameter that doesn't matter. Specifically, you declare that it takes a parameter of type unit. The unit type is a special type in F#. It basically corresponds to an empty tuple, and is written as ().
Think about the empty-tuple type for a minute. If you have a tuple of two bool values, how many possible values can this tuple have? Four: it could be (false, false), or (false, true), or (true, false), or (true, true). If you have a tuple of just one bool, it could have two values: (true) or (false). If you have a tuple of zero values (of whatever type: bool, int, string, doesn't matter), then there's only one possible value it could have: (), the empty tuple. And since that's a type with only one possible value, that's why it's called the unit type.
So if you want a function rather than a value, but that function doesn't need to take any meaningful parameters, you define it like this:
let myFunction () =
printfn "I'm being called purely for the side effects"
Note how I put a space between the function name and the unit parameter. You don't actually have to have that space there — it's perfectly legal to write let myFunction() = ... — but I want you to see that the () is not just function-declaration syntax, it's an actual value of an actual type. This distinction becomes important when you start doing advanced things with functions, so I want you to be clear about it now.
BTW, normally you'd have a parameter name in your function declaration rather than a value, but the unit type is treated specially: since there's only one possible value of unit, you already know what value your function will be called with, so you don't really need to assign that to a name anyway. So F# lets you declare a function whose input type is unit by just having a () in the parameter list, instead of making you choose a name that you'd never actually use in the function body.
I hope this clears things up for you.

Function argument is null, even though a non-null argument is passed

F# newbie here, and sorry for the bad title, I'm not sure how else to describe it.
Very strange problem I'm having. Here's the relevant code snippet:
let calcRelTime (item :(string * string * string)) =
tSnd item
|>DateTime.Parse
|> fun x -> DateTime.Now - x
|> fun y -> (floor y.TotalMinutes).ToString()
|>makeTriple (tFst item) (tTrd item) //makeTriple switches y & z. How do I avoid having to do that?
let rec getRelativeTime f (l :(string * string * string) list) =
match l with
| [] -> f
| x :: xs -> getRelativeTime (List.append [calcRelTime x] f) xs
I step through it with Visual Studio and it clearly shows that x in getRelativeTime is a 3-tuple with a well-formed datetime string. But when I step to calcRelTime item is null. Everything ends up returning a 3-tuple that has the original datetime string, instead of one with the total minutes past. There's no other errors anywhere, until the that datetime string hits a function that expects it to be an integer string.
Any help would be appreciated! (along with any other F# style tips/suggestions for these functions).
item is null, because it hasn't been constructed yet out of its parts. The F# compiler compiles tupled parameters as separate actual (IL-level) parameters rather than one parameter of type Tuple<...>. If you look at your compiled code in ILSpy, you will see this signature (using C# syntax):
public static Tuple<string, string, string> calcRelTime(string item_0, string item_1, string item_2)
This is done for several reasons, including interoperability with other CLR languages as well as efficiency.
To be sure, the tuple itself is then constructed from these arguments (unless you have optimization turned on), but not right away. If you make one step (hit F11), item will obtain a proper non-null value.
You can also see these compiler-generated parameters if you go to Debug -> Windows -> Locals in Visual Studio.
As for why it's returning the original list instead of modified one, I can't really say: on my setup, everything works as expected:
> getRelativeTime [] [("x","05/01/2015","y")]
val it : (string * string * string) list = [("x", "y", "17305")]
Perhaps if you share your test code, I would be able to tell more.
And finally, what you're doing can be done a lot simpler: you don't need to write a recursive loop yourself, it's already done for you in the many functions in the List module, and you don't need to accept a tuple and then deconstruct it using tFst, tSnd, and tTrd, the compiler can do it for you:
let getRelativeTime lst =
let calcRelTime (x, time, y) =
let parsed = DateTime.Parse time
let since = DateTime.Now - parsed
let asStr = (floor since.TotalMinutes).ToString()
(x, asStr, y)
List.map calRelTime lst
let getRelativeTime' list =
let calc (a, b, c) = (a, c, (floor (DateTime.Now - (DateTime.Parse b)).TotalMinutes).ToString())
list |> List.map calc
Signature of the function is val getRelativeTime : list:('a * string * 'b) list -> ('a * 'b * string) list
You can deconstruct item in the function declaration to (a, b, c), then you don't have to use the functions tFst, tSnd and tTrd.
The List module has a function map that applies a function to each element in a list and returns a new list with the mapped values.

Value restriction when there are no generic parameters

I get the value restriction error on let makeElem in the following code:
let elemCreator (doc: XmlDocument) =
fun name (value: obj) ->
let elem = doc.CreateElement(name)
match value with
| :? seq<#XmlNode> as childs ->
childs |> Seq.iter (fun c -> elem.AppendChild(c) |> ignore)
elem
| _ -> elem.Value <- value.ToString(); elem
let doc = new XmlDocument()
let makeElem = elemCreator doc
Why I get the value restriction error if anonymous function returned from elemCreator hasn't any generic parameters?
The compiler states that the infered type of makeElem is (string -> 'a -> XmlNode). But why it infers second parameter as 'a if I've declared it as obj?
I believe that this may be the "expected" behavior (although unfortunate in this case), as a result of the compiler's generalization and condensation processes. Consider Tomas's example:
let foo (s:string) (a:obj) = a
If you were to define
let bar a = foo "test" a
then the compiler will infer the type bar : 'a -> obj because it generalizes the type of the first argument. In your case, you have the equivalent of
let bar = foo "test"
so bar is a value rather than a syntactic function. The compiler does essentially the same inference procedure, except now the value restriction applies. This is unfortunate in your case, since it means that you have to explicitly annotate makeElem with a type annotation (or make it a syntactic function).
This looks like an unexpected behavior to me. It can be demonstrated using a simpler function:
let foo (s:string) (a:obj) = a
let bar = foo "bar" // Value restriction
One possible explanation might be that the F# compiler allows you to call a function taking parameter of some type with an argument of any subtype. So, you can call foo "hi" (new A()) without explicitly casting A to obj (which used to be required some time ago).
This implicit casting could mean that the compiler actually interprets bar as something like this:
let bar a = foo "bar" (a :> obj)
...and so it thinks that the argument is generic. Anyway, this is just a speculation, so you could try sending this as a bug report to fsbugs at microsoft dot com.
(The following is based solely on observation.)
If you have a function obj -> 'a, calls to that function are not used to infer/solve the type of its argument. An illustration:
let writeLine (arg: obj) = System.Console.WriteLine(arg)
writeLine is obj -> unit
let square x =
writeLine x
x * x
In the above function x is inferred as int because of (*). If a type could be constrained by obj then this function would not work (x would be inferred as obj prior to the use of (*), which would cause an error along the lines of: type obj does not support operator (*)).
I think this behavior is a Good Thing. There's no need to restrict a type as obj because every type is already implicitly convertible to obj. This allows your program to be more generic and provides better interoperability with the .NET BCL.
In short, obj has no bearing on type inference (yay!).

How to implement delay in the maybe computation builder?

Here is what I have so far:
type Maybe<'a> = option<'a>
let succeed x = Some(x)
let fail = None
let bind rest p =
match p with
| None -> fail
| Some r -> rest r
let rec whileLoop cond body =
if cond() then
match body() with
| Some() ->
whileLoop cond body
| None ->
fail
else
succeed()
let forLoop (xs : 'T seq) f =
using (xs.GetEnumerator()) (fun it ->
whileLoop
(fun () -> it.MoveNext())
(fun () -> it.Current |> f)
)
whileLoop works fine to support for loops, but I don't see how to get while loops supported. Part of the problem is that the translation of while loops uses delay, which I could not figure out in this case. The obvious implementation below is probably wrong, as it does not delay the computation, but runs it instead!
let delay f = f()
Not having delay also hinders try...with and try...finally.
There are actually two different ways of implementing continuation builders in F#. One is to represent delayed computations using the monadic type (if it supports some way of representing delayed computations, like Async<'T> or the unit -> option<'T> type as shown by kkm.
However, you can also use the flexibility of F# computation expressions and use a different type as a return value of Delay. Then you need to modify the Combine operation accordingly and also implement Run member, but it all works out quite nicely:
type OptionBuilder() =
member x.Bind(v, f) = Option.bind f v
member x.Return(v) = Some v
member x.Zero() = Some ()
member x.Combine(v, f:unit -> _) = Option.bind f v
member x.Delay(f : unit -> 'T) = f
member x.Run(f) = f()
member x.While(cond, f) =
if cond() then x.Bind(f(), fun _ -> x.While(cond, f))
else x.Zero()
let maybe = OptionBuilder()
The trick is that F# compiler uses Delay when you have a computation that needs to be delayed - that is: 1) to wrap the whole computation, 2) when you sequentially compose computations, e.g. using if inside the computation and 3) to delay bodies of while or for.
In the above definition, the Delay member returns unit -> M<'a> instead of M<'a>, but that's perfectly fine because Combine and While take unit -> M<'a> as their second argument. Moreover, by adding Run that evaluates the function, the result of maybe { .. } block (a delayed function) is evaluated, because the whole block is passed to Run:
// As usual, the type of 'res' is 'Option<int>'
let res = maybe {
// The whole body is passed to `Delay` and then to `Run`
let! a = Some 3
let b = ref 0
while !b < 10 do
let! n = Some () // This body will be delayed & passed to While
incr b
if a = 3 then printfn "got 3"
else printfn "got something else"
// Code following `if` is delayed and passed to Combine
return a }
This is a way to define computation builder for non-delayed types that is most likely more efficient than wrapping type inside a function (as in kkm's solution) and it does not require defining a special delayed version of the type.
Note that this problem does not happen in e.g. Haskell, because that is a lazy language, so it does not need to delay computations explicitly. I think that the F# translation is quite elegant as it allows dealing with both types that are delayed (using Delay that returns M<'a>) and types that represent just an immediate result (using Delay that returns a function & Run).
According to monadic identities, your delay should always be equivalent to
let delay f = bind (return ()) f
Since
val bind : M<'T> -> ('T -> M<'R>) -> M<'R>
val return : 'T -> M<'T>
the delay has the signature of
val delay : (unit -> M<'R>) -> M<'R>
'T being type-bound to unit. Note that your bind function has its arguments reversed from the customary order bind p rest. This is technically same but does complicate reading code.
Since you are defining the monadic type as type Maybe<'a> = option<'a>, there is no delaying a computation, as the type does not wrap any computation at all, only a value. So you definition of delay as let delay f = f() is theoretically correct. But it is not adequate for a while loop: the "body" of the loop will be computed before its "test condition," really before the bind is bound. To avoid this, you redefine your monad with an extra layer of delay: instead of wrapping a value, you wrap a computation that takes a unit and computes the value.
type Maybe<'a> = unit -> option<'a>
let return x = fun () -> Some(x)
let fail = fun() -> None
let bind p rest =
match p() with
| None -> fail
| Some r -> rest r
Note that the wrapped computation is not run until inside the bind function, i. e. not run until after the arguments to bind are bound themselves.
With the above expression, delay is correctly simplified to
let delay f = fun () -> f()

Resources