Pattern discriminator does not match - f#

New to F# and would like to know why I am getting a compile time error when defining the binaryMathFn.
I will also take suggestions on changing the title of this question. It's not the most descriptive :)
type Stack = StackContents of float list
let push item (StackContents stack) =
match stack with
| [] -> StackContents [item]
| _ -> StackContents (item::stack)
let pop (StackContents stack) =
match stack with
| [] -> failwith "Stack is empty, nothing to pop!"
| h::t -> (h, StackContents t)
let binaryMathFn fn (StackContents stack) = // Expression type is expected to be of type Stack?
let item1, stack' = pop stack
let item2, stack'' = pop stack'
let result = fn item1 item2
push result stack''

The answer is in the types, but before I explain what's wrong, here's one way to make the binaryMathFn function compile:
let binaryMathFn fn stack =
let item1, stack' = pop stack
let item2, stack'' = pop stack'
let result = fn item1 item2
push result stack''
Your definition of Stack is a Discriminated Union with a single union case called StackContents. When you declare a function argument as (StackContents stack), you are using a shorthand notation that states:
The input is of the type Stack, but immediately match that input with the single StackContents union case, effectively unpacking the float list from the input, and assign that float list to the symbol stack.
If you, like I suggest above, change the argument declaration to simply stack, the F# compiler can automatically infer that stack is a Stack value, because the function uses the pop function, which has this function signature:
Stack -> float * Stack
which is read as a function that takes a Stack as input, and returns a tuple of float and Stack as output. Since binaryMathFn calls pop with stack, the compiler can infer stack must be a Stack value.

So the problem is:
let binaryMathFn fn (StackContents stack)
Should be
let binaryMathFn fn (stack)
What is happening is that your definition is extracting the data. In general, using this sort of pattern matching that you have used in the function definition is unidiomatic.

Related

Splitting a list at given index f#

For a class I'm following, I have to do the following exercise:
Implement a function
let splitAt (i : int) (l : List<'a>) : List<'a> * List<'a> = ...
that splits the list into two lists, the rst one containing all the elements of l from position 0 to position i
included, and the second one containing all the remaining elements. The two resulting lists are returned
in a tuple. For example:
split 3 [3;5;4;-1;2;2] = ([3;5;4;-1],[2;2])
We have to do these problems only using functional programming, and thus I'm not allowed to use pre-existing functions.
I have the following code which seems to me to (logically) be correct:
let splitAt (i:int)(l: List<'a>): List<'a> * List<'a> =
let rec loop n startlist restlist =
if n = i then
restlist * startlist
else
match startlist with
| h :: t -> loop (n+1) [t] [(restlist :: h)]
| h :: [] -> None
loop 0 l []
and below my [<EntryPoint>]
printfn "%A" (splitAt stringlist 3)
However, this gives me a couple of errors, namely:
None of the types 'a list, 'a list support the operator *
This expression was expected to have type int but here has type char list
This expression was expected to have type List<'a> but here has type int
The * operator is used for declaring a tuple type, but when you're building a tuple you use , instead. So you want restlist, startlist.
Then you'll discover that there's another type error, because one branch of your match expression returns None. That's an option type, so the value you return should be a Some. So you want Some (restlist, startlist).
And now you'll discover one more type error, which is that you've declared that the function returns a tuple but in fact it returns a tuple option (that is, either None or Some tuple). So your type declaration needs to become (List<'a> * List<'a>) option.
For more on why * is used in declaring tuple types rather than ,, https://fsharpforfunandprofit.com/posts/tuples/ is a good read.

Immutable data structures in F#

I am trying to implement a very trivial stack in F# using its built-in generic type. Coming from the imperative paradigm it can sometimes be hard to imagine how to avoid mutability.
What I have so far is a simple data structure with a push and pop operator:
type Stack<'a> =
| Empty
| S of 'a list
with
member L.Push x =
match L with
| Empty | S ([]) -> S ([x])
| S (V) -> S (x :: V)
member L.Pop =
match L with
| Empty | S ([]) -> failwith "Error: Stack is empty"
| S (v::_) -> v
end
My idea was to make the Stack hold an S of 'a list where we modify the list with the cons :: operator to not mutate the list S, but to replace it with S'. As of now, the stack can at most have one element, and it doesn't grow when pushing elements to it – likewise doesn't shrink when popped.
Can anyone give me a hint on how rewrite the structure/think about it differently?
Thank you!
You could do this in a more functional way by simply having a list act as the stack. Instead of methods you can make push and pop be functions instead.
// Returns the new stack
let push item stack = item :: stack
// Returns (item, newStack) tuple or throws if stack is empty
let pop stack =
match stack with
| [] -> failwith "Stack is empty"
| item :: newStack -> item, newStack
// Example usage
let stack = []
let populatedStack = push "hello" stack
// populatedStack = ["hello"]
let item, emptiedStack = pop populatedStack
// item = "hello"
// emptiedStack = []
The built-in immutable list is already a valid Stack structure. The Push Operation is :: and you get the last item with Pattern Matching like let (first,rest) = list
If you want to create a Stack from Scratch. Here are some implementations.
1)
a) a stack is either empty
b) or a value and a reference to the previous Stack.
type Stack<'a> =
| Empty
| Value of 'a * Stack<'a>
module Stack =
let isEmpty = function
| Empty -> true
| Value _ -> false
let empty = Empty
let push x stack = Value (x, stack)
let pop (Value (x, stack)) = x, stack
let stk =
Stack.empty
|> Stack.push 1
|> Stack.push 2
|> Stack.push 3
let rec loop stack =
if Stack.isEmpty stack
then ()
else
let first, rest = Stack.pop stack
printfn "First: %d" first
loop rest
loop stk
You also could choose a record as the underlying data-structure.
type Stack<'a> = {
Value: 'a option
Next: Stack<'a> option
}
This way, a Stack with three elements looks like.
{Value=Some 3; Next=
{Value=Some 2; Next=
{Value=Some 1; Next=
{Value=None; Next=None}}}}
You also could choose a class with a value and next field and use null for the last element.
The important thing is how to work with those structures. The way to work with immutable data-structures is by using recursive functions instead of looping.
Creating a tail-recursive function that visits each element and executes a function on every element is the fold function, comparable to a forEach.

Stackless trampoline Monad/Computation Expression

I am working on a functional programming language of my own design and I stumbled on a problem that is beyond my skills to solve. I would like to know if anyone has any advice on how to solve it or a reason for why it is impossible.
The code below is an overview of a solution that is not the ideal but a compromise.
This problem is at the heart of the runtime system I am currently using. Instead of relying on the .Net stack I am using a monad to perform operations on a trampoline. This should help with step through debugging and allow for users to not have to worry about stack space. Here is a simplified version of the monad I am currently using.
type 't StackFree =
|Return of 't //Return a value
|StackPush of ('t->'t StackFree)*'t StackFree //Pushes a return handler onto the "Stack"
|Continuation of (unit->'t StackFree) //Perform a simple opperation
type StackFreeMonad() =
member this.Delay(fn) =
Continuation(fn)
member this.Bind(expr,fn) =
StackPush(fn,expr)
member this.Return(value) =
Return(value)
member this.ReturnFrom(x) =x
let stackfree = StackFreeMonad()
This was not the original design but it was the best I could get to work with F# computation expressions in an ideal world the above computation expression would work on this type.
type 't Running =
|Result of 't
|Step of (unit->'t Running)
So in order to convert a StackFree into a Running type I have to use this conversion function
//this method loops through the StackFree structure finding the next computation and managing a pseudo stack with a list.
let prepareStackFree<'t> :'t StackFree->'t Running =
let rec inner stack stackFree =
Step(fun ()->
match stackFree with
//takes the return values and passes it to the next function on the "Stack"
|Return(value)->
match stack with
|[]->Result(value)
|x::xs -> inner xs (x value)
//pushes a new value on the the "Stack"
|StackPush(ret,next) ->
inner (ret::stack) next
//performs a single step
|Continuation(fn)->
inner stack (fn()))
inner []
Here is a brief example of the two types in action.
let run<'t> :'t StackFree->'t =
let rec inner = function
|Step(x)-> inner (x())
|Result(x)-> x
stackFreeToRunning>>inner
//silly function to recompute an intiger value using recursion
let rec recompute number = stackfree {
if number = 0 then return 0
else
let! next = recompute (number-1)
return next+1
}
let stackFreeValue = recompute 100000
let result = run stackFreeValue
do printfn "%i" result
I have spent several hours trying to get a Computation Expression that works directly on the Running type and cutting out the middleman StackFree. However I cannot figure out how to do it. At this point I am seriously considering the possibility that a solution to this problem is impossible. However I cannot figure out the reason that it is impossible.
I have gotten close a few times but the resulting solutions ended up using the stack in some confusing way.
Is it possible to have a computation expression that operates on the Running type without utilizing the .Net stack? If this is not possible why is it not possible. There must be some simple mathematical reasoning that I am missing.
NB: These are not the actual types I am using they are simplified for this questions the real ones keep track of scope and position in the script. Furthermore I am aware of the serious performance cost of this type of abstraction
Edit: Here is another way to approach the problem. This implementation is flawed because it uses the stack. Is there anyway to get the exact behavior below without using the stack?
type RunningMonad() =
member this.Delay(fn) =
Step(fun ()->fn ())
member this.Bind(m, fn) =
Step(fun ()->
match m with
|Result(value)-> fn value
//Here is the problem
|Step(next)-> this.Bind(next(),fn))
member this.Return(v) =
Result(v)
member this.ReturnFrom(x) = x
The bind implementation in the above computation expression creates a function that calls another function. So as you go deeper and call bind more and more you have to chase a bunch of function calls and then eventually you hit a stackoverflow exception.
Edit2: Clarity.
Better late than never!
This is addressed in section 4 of Stackless Scala with Free Monads. Bjarnason tackles the problem by adding a new constructor to the Trampoline datatype, representing a subroutine call to another trampoline. He keeps this new constructor private, in order to ensure that you can't build left-nested Binds (which would overflow the stack when executing the trampoline).
I am by no means an F#er, but I'll muddle through. In WishF#ul, an imaginary dialect of F# which I just made up, you can express the new existentially quantified constructor directly:
type Tram<'a> =
| Done of 'a
| Step of (unit -> Tram<'a>)
| Call<'x> of Tram<'x> * ('x -> Tram<'a>) // don't export this
type TramMonad() =
member this.Return(x) = Done(x)
member this.Bind(ma, f) = match ma with
| Call(mx, k) -> Call(mx, fun x -> this.Bind(k(x), f))
| _ -> Call(ma, f)
// i confess to not quite understanding what your Delay and ReturnFrom methods are for
let tram = new TramMonad()
let rec runTram t =
let next mx f = match mx with
| Done(x) -> f x
| Step(k) -> Step(fun () -> tram.Bind(k(), f))
| Call(my, g) -> tram.Bind(my, fun x -> tram.Bind(g x, f))
match t with
| Done(x) -> x
| Step(k) -> runTram(k())
| Call(mx, f) -> runTram(next mx f)
Note that all of the recursive calls to runTram are in tail position. It takes a bit of puzzling, but you can convince yourself that Bind won't construct a deeply-nested continuation, so runT will always operate in O(1) stack space.
Sadly we're working in F#, not WishF#ul, so we have to resort to an object-oriented encoding of the existential type in the Call constructor. Here goes...
module rec Trampoline =
type Call<'a> =
abstract member Rebind<'b> : ('a -> Tram<'b>) -> Tram<'b>
abstract member Next : unit -> Tram<'a>
type Tram<'a> =
| Done of 'a
| Step of (unit -> Tram<'a>)
| Call of Call<'a>
type TramMonad() =
member this.Return(x) = Done(x)
member this.Bind(ma, f) =
match ma with
| Call(aCall) -> aCall.Rebind(f)
| _ -> call ma f
let tram = new TramMonad()
let rec call<'a, 'x>(mx : Tram<'x>) (f : 'x -> Tram<'a>) : Tram<'a> = Call {
new Call<'a> with
member this.Rebind<'b>(g : 'a -> Tram<'b>) : Tram<'b> =
call<'b, 'x> mx (fun x -> tram.Bind(f x, g) : Tram<'b>)
member this.Next() =
match mx with
| Done(x) -> f x
| Step(k) -> Step(fun () -> tram.Bind(k(), f))
| Call(aCall) -> aCall.Rebind(f)
}
let rec runTram t =
match t with
| Done(x) -> x
| Step(k) -> runTram(k())
| Call(aCall) -> runTram(aCall.Next())
I recommend reading the whole paper, which goes on to generalise this stackless construction to any free monad, not just trampolines (which are Free (Unit -> _)). Phil Freeman's Stack Safety for Free builds on this work, generalising the trampoline paper's free monad to a free monad transformer.

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!).

Can't get operator overloading to work with Linq Expression Trees

I am creating Linq expression trees from F# that operates on a custom datatype I have. The type is a very simple discriminated union that has the usual arithmetic operators overloaded. But for some reason I cannot create arithmetic linq expression nodes due to the fact that it can't find the correct overload. Thing is, I swear I had this working some time ago but I can't figure out what I changed to make it break.
I'll attach a small code sample showing the problem. The datatype below has the Addition operator overloaded. Using the overloaded operator works like a charm, but when I try to create an addition expression tree node using Expression.Add(lhs, rhs) the system throws an exception complaining that it can't find the overload for the Add operation.
Does anyone have an idea of what I am doing wrong?
Thank you,
Rickard
open System.Linq.Expressions
module DataType =
exception NotImplementedYet of string
type DataCarrier =
| ScalarCarrier of float
| VectorCarrier of float array
member this.Add(other) =
match (this, other) with
| ScalarCarrier(x), ScalarCarrier(y) -> ScalarCarrier(x + y)
| VectorCarrier(u), VectorCarrier(v) ->
VectorCarrier(Array.map2 (fun x y -> x + y) u v)
| _,_ -> raise (NotImplementedYet("No go!"))
static member (+) (lhs:DataCarrier, rhs) =
lhs.Add(rhs)
module Main =
let createAddOp (lhs:DataType.DataCarrier) (rhs:DataType.DataCarrier) =
let clhs = Expression.Constant(lhs)
let crhs = Expression.Constant(rhs)
Expression.Add(clhs, crhs)
(* no problems with this one *)
printf "Testing operator overloading: %A" (DataType.ScalarCarrier(1.0)
+ DataType.ScalarCarrier(2.0))
(* this throws an exception *)
printf "Testing expr construction %A" (Main.createAddOp
(DataType.ScalarCarrier(1.0))
(DataType.ScalarCarrier(2.0)))
One solution is to explicitly type the Expression operands (giving them the static type DataType.DataCarrier instead of their runtime type DataType.DataCarrier.ScalarCarrier):
module Main =
let createAddOp (lhs:DataType.DataCarrier) (rhs:DataType.DataCarrier) =
let clhs = Expression.Constant(lhs, typeof<DataType.DataCarrier>)
let crhs = Expression.Constant(rhs, typeof<DataType.DataCarrier>)
Expression.Add(clhs, crhs)
Another option would be to explicitly pass the addition operator to use:
module Main =
let createAddOp (lhs:DataType.DataCarrier) (rhs:DataType.DataCarrier) =
let clhs = Expression.Constant(lhs)
let crhs = Expression.Constant(rhs)
Expression.Add(clhs, crhs, typeof<DataType.DataCarrier>.GetMethod("op_Addition"))
I am surprised that your original code doesn't work, though. It appears to be a limitation in how expression trees find relevant add operators (that is, it appears that Linq only looks for add operators on the runtime types of operands).

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