Enumerating a discriminated union - f#

I have read the answers from How to enumerate a discriminated union in F#?
And I like the solution suggested: solution
However, I am not sure how to write a function where I would pass the discriminated union as argument?
let testDisUnion a =
SimpleUnionCaseInfoReflection.AllCases<a>
|> Seq.iter (fun (_, instance) -> printfn "name: %s" instance)
Thank you

This is how to use a type argument as dumetrulo suggested:
let testDisUnion<'a> =
SimpleUnionCaseInfoReflection.AllCases<'a>
|> Seq.iter (fun (_, instance) -> printfn "name: %A" instance)
testDisUnion<MyType>
//name: A
//name: B
type MyType = A | B
The type argument <'a> is being passed from your "function" to the AllCases "function". I write function in quotes because although there are no proper parameters, the type argument is a type of input to the function, which means that the value is only evaluated when you "call" it with a type argument.

Related

Writing unit tests against PrintfFormat

I have a type that I'm trying to understand by writing unit tests against it, however I can't reason what to do with PrintfFormat
type ValueFormat<'p,'st,'rd,'rl,'t,'a> = {
format: PrintfFormat<'p,'st,'rd,'rl,'t>
paramNames: (string list) option
handler: 't -> 'a
}
with
static member inline construct (this: ValueFormat<_,_,_,_,_,_>) =
let parser s =
s |> tryKsscanf this.format this.handler
|> function Ok x -> Some x | _ -> None
let defaultNames =
this.format.GetFormatterNames()
|> List.map (String.replace ' ' '_' >> String.toUpperInvariant)
|> List.map (sprintf "%s_VALUE")
let names = (this.paramNames ?| defaultNames) |> List.map (sprintf "<%s>")
let formatTokens = this.format.PrettyTokenize names
(parser, formatTokens)
I feel confident that I can figure everything out but PrintfFormat is throwing me with all those generics.
The file I'm looking at for the code I want to unit test is here for the FSharp.Commandline framework.
My question is, what is PrintfFormat and how should it be used?
A link to the printf.fs file is here. It contains the definition of PrintfFormat
The PrintfFormat<'Printer,'State,'Residue,'Result,'Tuple> type, as defined in the F# source code, has four type parameters:
'Result is the type that your formatting/parsing function produces. This is string for sprintf
'Printer is a type of a function generated based on the format string, e.g. "%d and %s" will give you a function type int -> string -> 'Result
'Tuple is a tuple type generated based on the format string, e.g. "%d and %s" will give you a tuple type int * string.
'State and 'Residue are type parameters that are used when you have a custom formatter using %a, but I'll ignore that for now for simplicity (it's never needed unless you have %a format string)
There are two ways of using the type. Either for formatting, in which case you'll want to write a function that returns 'Printer as the result. The hard thing about this is that you need to construct the return function using reflection. Here is an example that works only with one format string:
open Microsoft.FSharp.Reflection
let myformat (fmt:PrintfFormat<'Printer,obj,obj,string,'Tuple>) : 'Printer =
unbox <| FSharpValue.MakeFunction(typeof<'Printer>, fun o ->
box (o.ToString()) )
myformat "%d" 1
myformat "%s" "Yo"
This simply returns the parameter passed as a value for %d or %s. To make this work for multiple arguments, you'd need to construct the function recursively (so that it's not just e.g. int -> string but also int -> (int -> string))
In the other use, you define a function that returns 'Tuple and it needs to create a tuple containing values according to the specified formatting string. Here is a small sample that only handles %s and %d format strings:
open FSharp.Reflection
let myscan (fmt:PrintfFormat<'Printer,obj,obj,string,'Tuple>) : 'Tuple =
let args =
fmt.Value
|> Seq.pairwise
|> Seq.choose (function
| '%', 'd' -> Some(box 123)
| '%', 's' -> Some(box "yo")
| _ -> None)
unbox <| FSharpValue.MakeTuple(Seq.toArray args, typeof<'Tuple>)
myscan "%d %s %d"

How to do null-conditional using Option.map? And how can F# get the property of an null object? Bug?

The following code returns Some "Test" stead of None. Basically, I'm trying to implement the C# code of cObj?.B.A.P.
// Setup
[<AllowNullLiteral>]
type A() =
member x.P = "Test"
[<AllowNullLiteral>]
type B(a:A) =
member x.A = a
[<AllowNullLiteral>]
type C(b:B) =
member x.B = b
// Test
let aObj: A = null
let cObj = new C(new B(aObj))
let r =
cObj |> Option.ofObj
|> Option.map(fun c -> c.B)
|> Option.map(fun b -> b.A)
|> Option.map(fun a -> a.P) // Expect return None since a is null
// printfn "%A" a; will print <null>.
// How can F# got property of null object?
r
It seems F# doesn't treat null as None in the Option.map. Is there a simple fix to make it return None as soon as a null is found?
F#, unlike C#, tries to be explicit everywhere. In the long run this leads to more maintainable and correct programs.
In particular, null has absolutely nothing to do with Option. null is not the same as None. None is a value of type Option, while null is this very vague concept - a value that can be of any type.
If you would like to return None when the argument is null and Some otherwise, what you need is Option.bind, not Option.map. Option.bind takes a function that takes a value (extracted from a previous Option) and returns another Option. Something like this:
let maybeC = Option.ofObj cObj
let maybeB = maybeC |> Option.bind (c -> Option.ofObj c.B)
let maybeA = maybeB |> Option.bind (b -> Option.ofObj b.A)
let maybeP = maybeA |> Option.bind (a -> Option.ofObj a.P)
Or in one go:
let maybeP =
Option.ofObj cObj
|> Option.bind (c -> Option.ofObj c.B)
|> Option.bind (b -> Option.ofObj b.A)
|> Option.bind (a -> Option.ofObj a.P)
If you do this sort of thing very frequently, you can combine Option.bind and Option.ofObj calls and encode that as a separate function:
let maybeNull f = Option.bind (x -> Option.ofObj (f x))
let maybeP =
Option.ofObj cObj
|> maybeNull (c -> c.B)
|> maybeNull (b -> b.A)
|> maybeNull (a -> a.P)
However, if you find yourself buried in nulls like that, I would suggest that perhaps your domain design is not thought through very well. Nulls are not a good modeling tool, they should be avoided as much as possible. I encourage you to rethink your design.
The first argument to Option.map is a function 'T -> 'U. Its parameter is of type 'T, not 'T option. So, in your last lambda fun a -> a.P, a null argument denotes a null of type A, not a null of type A option.
Because the member P of type A just returns the string "Test," the call succeeds and returns even though the receiver is null. If you try to use the self identifier in the body of P, you will instead get a null reference exception.

Type definitions and type inference in F#

I have two example types, defined in this order:
type Quote = {QuoteNum: decimal option; ShipToID: decimal option}
type Sales = {SalesNum: decimal option; ShipToID: decimal option}
and I'm trying to write a function that can accept a both types:
let fx (y: Map<decimal option, _ list>) =
y
|> Map.map (fun key list -> list
|> List.map (fun x -> x.ShipToID))
When I try to pass a Map<decimal option, Quote list> to the function, I get an error:
Type mismatch. Expecting a
Map<decimal option,Sales list>
but given a
Map<decimal option,Quote list>
I would have thought that I'd be able to pass a map of both types to the function, but the compiler seems to be inferring that only a Map<decimal option, Sales list> is acceptable to the function. I suspect that the compiler "sees" the Sales type most recently and assumes that that's what the function needs. I thought I had made the function relatively generic by including the _ in the type annotation.
How can I make the function accept both types? Can I do this without redefining the types themselves?
Your suspicions are correct. F# needs to infer a single concrete type there, so will use the most recent matching type it finds.
You can make it work by defining an interface with ShipToID field and have both types implement it (not sure if that counts as redefining the types for you), or by providing a getter function for extracting the field value (which would let you keep the function generic):
let fx (shipToIdGetter: 'a -> decimal option) (y: Map<decimal option, 'a list>) =
y
|> Map.map (fun key list ->
list
|> List.map (fun x -> shipToIdGetter x))
Your code would work as written if F# supported row polymorphism (which happens to be the case for OCaml).
While I believe that scrwtp's answer is the way to go, you also have the choice to use inline and member constraints (or static duck typing...if thats a valid term). For the sake of completeness:
type Quote = {QuoteNum: decimal option; ShipToID: decimal option}
type Sales = {SalesNum: decimal option; ShipToID: decimal option}
let inline fx (y: Map<decimal option, 'a list>) =
let inline getId x = ( ^a : (member ShipToID : decimal option) x)
y |> Map.map (fun key list -> list |> List.map getId)
let q : Quote = { QuoteNum = None; ShipToID = None}
let s : Sales = { SalesNum = None; ShipToID = None}
fx <| Map.ofList [ None, [q]]
fx <| Map.ofList [ None, [s]]
The getId function uses the rather obscure syntax to call the ShipToID member based on ^a's expected structure.

How to interpret a function signature

I am still confused on how to read function signatures.
The Option.map signature is the following:
/// map f inp evaluates to match inp with None -> None | Some x -> Some (f x).
/// mapping: A function to apply to the option value.
/// option: The input option.
val map : mapping:('T -> 'U) -> option:'T option -> 'U option
However, I have no clue what that signature means.
I read it as the following:
There's a function called map that takes a function as an input that we'll call "mapping" and it will yield a result that is also a function that we'll call "option".
Mapping Parameter:
mapping:('T -> 'U)
The function that we pass in as input takes Titanium (i.e. 'T) as the input and yields Uranium (i.e. 'U) as output.
The Option returned
option:'T option -> 'U option
We'll call the output of the map function "option".
Thus, this "option" that is returned from executing the map function is also a function as referenced above. It's takes a Titanium option and yields a Uranium option.
Example:
type String20 = String20 of string
type Name = { First:String20
Last:String20
Suffix:String20 option }
let tryCreateName (first:string) (last:string) (suffix:string option) =
let isValid = [first; last]
|> List.forall (fun x -> x.Length > 2 && x.Length <= 20)
if isValid then
Some { First = String20(first);
Last = String20(last);
Suffix = Option.map String20 suffix }
else None
How does the following expression map:
Option.map String20 suffix
Based on the expression above, where is the "returned function" of Titanium option -> Uranium option?
First off take a look at Option.map<'T,'U> Function (F#) and notice
The expression map f inp evaluates to match inp with None -> None |
Some x -> Some (f x).
So lets convert this comment to working code. First map is a method of the type Option, but to make it easier we will make it a function outside of a type and to avoid conflicts with other map functions we will give it the name OptionMap.
let OptionMap = f inp =
match inp with
| None -> None
| Some x -> Some (f x)
Now to get the signature of this function just send it to F# Interactive
val OptionMap : f:('a -> 'b) -> inp:'a option -> 'b option
and to make the types obvious we will type the parameters of the function.
let optionMap (f : ('a -> 'b)) (inp : 'a option) : 'b option =
match inp with
| None -> None
| Some x -> Some (f x)
Now to test this we can use
let (test : String20 option) = optionMap String20 (Some("something"))
printfn "test: %A" test
// test: Some (String20 "something")
So what happened in OptionMap that allowed this to work
If we add some print statements let sees what happens
let optionMap (f : ('a -> 'b)) (inp : 'a option) : 'b option =
printfn "f: %A" f
printfn "inp: %A" inp
match inp with
| None -> None
| Some x ->
let result = Some (f x)
printfn "result: %A" result
result
we get
f: <fun:test#63>
inp: Some "something"
result: Some (String20 "something")
we see that f is a function which is String20
So how can String20 be a function?
If we send String20 to F# Interactive it gives.
> String20;;
val it : arg0:string -> String20 = <fun:clo#4>
So String20 is a function that takes a string and returns a type of String20. That smells of a constructor.
Let's test that out in F# interactive.
> let test1 = String20 "something";;
val test1 : String20 = String20 "something"
Sure enough, String20 is a constructor, but we didn't specifically create a constructor as is done in the Object-Oriented world.
You have to think of a type, even a discriminated union as having constructors. The constructors are not specifically written but do exist. So String20 is a constructor that takes one value, a string, which is a function with the correct type signature for the Option.map function.
I gave the answer a lot more detail so that one can learn a process on how to break down problems and look at the inner workings as a tool to solving these kinds of problems.
To learn more about the lower level details of how functional programming works one needs to understand lambda calculus. The best way I know to learn it is to read An Introduction To Functional Programming Through Lambda Calculus by Greg Michaelson and look at the info in the lambda calculus tag.
Also if you like the book, you should buy a copy instead of using the free version.
There are two ways of looking at it.
let f x y = x + y
// val f: x:int -> y:int -> int
One way is to say that function f takes two parameters, x of type int and y of type int, and returns an int. So I can supply two arguments and get the result:
let a = f 4 5
// val a: int = 9
The other way is that the function takes one parameter, x of type int, and returns another function, which takes one parameter, y of type int, and returns an int. So I can supply one argument and get a function as result:
let b = f 4
// val b: int -> int
Mathematically, it's always the second way - all functions are one-parameter functions. This notion is very convenient for programming with higher-order functions.
But the first way is usually more understandable to humans, so you will often see functions discussed as if they take multiple parameters.
String20 is the case constructor for the String20 Discriminated Union case. It's a function with the type string -> String20. So string takes the place of 'T, and String20 takes the place of 'U in the mapping you supply to Option.map.
Generally, if you have a function T1 -> T2 -> ... and you apply it to one parameter, you get a function T2 -> .... In the case of Option.map, T1 is itself a function, but that is of no consequence to the way the arguments are applied.
I find it confusing to call a function "option", a string "Titanium" and a type called String20 "Uranium", so I'll stick with type names.
You ask where the "returned function" that maps a string option to a String20 option is in the expression Option.map String20 suffix. It is simply
Option.map String20
Since String20 constructs a String20 from a string, Option.map String20 is a function that maps a string to a String20 if it is present, and to None otherwise.
If you write Option.map String20 suffix, this function is applied to suffix, which would be a string option. The result of this expression is a String20 option.

Why is Type Annotation Required Here?

This function has the signature: (UnionCaseInfo -> bool) -> 'T option
let private findCase<'T> f =
match FSharpType.GetUnionCases typeof<'T> |> Array.filter f with
|[|case|] -> Some (FSharpValue.MakeUnion(case,[||]) :?> 'T)
|_ -> None
This function, which calls the above function, has the signature: int -> obj
let CreateFromId<'T> id =
match findCase (fun case -> case.Tag = id) with
| Some c -> c
| None -> failwith (sprintf "Lookup for union case by \"%d\" failed." id)
In the pattern for CreateFromId, intellisense shows that c is inferred to be of type obj, even though it shows the correct signature for findCase. Why does the type seem to have been "lost" in the pattern?
(I can workaround this by specifying the return type of CreateFromId to 'T)
Because the type parameter 'T is not referenced in the body of the function, so type inference has no way to know your intention was to name 'T the return type.
So you can either add it in the body as the return type (as you already figured it out) or remove it from the declaration:
let CreateFromId id = ...
By removing it works because F# does automatic generalization, the only different is it will use an arbitrary name for the type variable, but even if you want to name that type variable 'T what I would do is add it as a return type but no in the declaration between brackets:
let CreateFromId id : 'T = ...
The type of CreateFromId was inferred to by int -> obj because there's nothing "linking" the two 'T type arguments on your functions.
findCase is properly generic over 'T, but CreateFromId is only declared to be generic, and the generic type argument is never used.
Annotating the function with the desired type is good enough to make the types line up. You can also call findCase explicitly providing the type:
let CreateFromId<'T> id =
match findCase<'T> (fun case -> case.Tag = id) with
| Some c -> c
| None -> failwith (sprintf "Lookup for union case by \"%d\" failed." id)
Or as the other answer suggests, just drop the 'T from CreateFromId and let type inference do its thing.

Resources