This compiles
let a (t2: Type) = obj()
let c: Func<Type,obj> = a
but this
let a (t2: Type) = obj()
let b (t:Type) = a
let c: Func<Type,(Type->obj)> = b
says This expression was expected to have type
'Func<Type,(Type -> obj)>' but here has type
''a -> 'b -> obj'
I know it is something simple, but my brain has thrown the towel in.
The normal delegate conversion would be the following:
let c : System.Func<Type, Type, obj> = b
If you want to interpret b as a Func returning another function, you have to be explicit about it only having one input argument:
let a (t2: Type) = obj()
let b (t:Type) = a
let c = Func<_, _>(b)
The F# spec has this to say about delegate conversions:
[14.4]: If a formal parameter has delegate type D, an actual argument farg has known type
ty1 -> ... -> tyn -> rty, and the number of arguments of the Invoke method of delegate type
D is precisely n, interpret the formal parameter in the same way as the following:
new D(fun arg1 ... argn -> farg arg1 ... argn).
So in other words, the implicit conversion to the Func delegate of your b function is the same as
let c = System.Func<_, _, _>(fun a1 a2 -> b a1 a2)
I think the reason for this is for interop with other dotnet languages and to be consistent with how curried functions are compiled:
[8.13.4]: The compiled representation of a curried method member is a .NET method in which the arguments are concatenated into a single argument group.
I can't understand what is wrong with following bit of code:
let toClass (problem:Problem<'a>) (classID:int) (items:'a list) =
let newFreqTable = (problem.FreqTables.[classID]).count items
{ problem with FreqTables = newFreqTable :: (problem.FreqTables |> List.filter (fun i -> i.ClassID <> classID)) }
type Problem<'a> when 'a : equality with member this.toClass (classID:int) (items:list<'a>) = toClass this classID items
I have a Problem type which is nothing but a way to group up any number of FreqTables - short for "Frequency tables". So toClass method just takes appropriate freqTable (by classID argument) and returns a new one - with calculated given items.
let typeIndependentCall = toClass p 0 ["word"; "word"; "s"] // this works perfectly
let typeDependentCall = typeIndependentCall.toClass 1 ["word"; "s"]
// gives an error: "One or more of the overloads of this method has
// curried arguments. Consider redesigning these members to take
// arguments in tupled form".
I am pretty new to F# and functional programming. What is the right way to attach behavior to my type?
In F# there are 2 main ways of passing arguments to a function: curried and tupled. The curried form is what you are using in your code above, and has a few key benefits, the first and foremost being partial application.
For example, instead of thinking of
fun add a b = a + b
as a function that takes in 2 arguments and returns a value, we can think of it as a function of one argument that returns a function that with one argument. This is why the type signature of our function is
Int -> Int -> Int
or, more clearly,
Int -> (Int -> Int)
However, when overloading methods, we can only use the tupled argument form
(Int, Int) -> Int
The reason for this is for optimization, as is discussed here
To get your code to work, use
type Problem<'a> when 'a : equality with member this.toClass (classID:int, items:list<'a>) = toClass this classID items
and call it like such:
let typeDependentCall = typeIndependentCall.toClass(1, ["word"; "s"])
While refactoring some F# code, I came across a problem which I fail to understand or resolve. I have a class Problem, with 2 constructors, one default for F# consumption, one for C# convenience, which takes in Funcs and "converts" them into F# functions:
open System
type Problem<'d, 's> (data: 'd, generate: 'd -> Random -> 's, mutate: 'd -> Random -> 's -> 's, evaluate: 's -> float) =
member this.Data = data
member this.Generate = generate this.Data
member this.Mutate = mutate this.Data
member this.Evaluate = evaluate
new (data: 'd, generator: Func<'d, Random, 's> , mutator: Func<'d, 's, Random, 's> , evaluator: Func<'s, float>) =
let generate (data: 'd) rng = generator.Invoke(data, rng)
let mutate (data: 'd) (rng: Random) (solution: 's) = mutator.Invoke(data, solution, rng)
let evaluate (solution: 's) = evaluator.Invoke(solution)
Problem(data, generate, mutate, evaluate)
As far as I can tell, this builds and works as expected.
Being somewhat obsessive-compulsive, I noted that the mutator Func in the C#-friendly constructor had a different order for the arguments, so I proceeded and rewrote it this way (first part unchanged):
new (data: 'd, generator: Func<'d, Random, 's> , mutator: Func<'d, Random, 's, 's> , evaluator: Func<'s, float>) =
let generate = fun (data: 'd) rng -> generator.Invoke(data, rng)
let mutate = fun (data: 'd) (rng: Random) (solution: 's) -> mutator.Invoke(data, rng, solution)
let evaluate = fun (solution: 's) -> evaluator.Invoke(solution)
Problem(data, generate, mutate, evaluate)
While the 3 functions seem to have the right signature, the last line fails and displays a red squiggly, telling me that "A unique overload for method 'Problem`2' could not be determined based on type information prior to this program point. The available overloads are shown below (or in the Error List window). A type annotation may be needed."
Can anyone help me see what I am missing? I tried to type-annotate the 4 arguments in the last line, to no avail - and I have no idea how to fix this. What also strikes me as odd is that the previous version worked, simply by having 2 arguments reversed in a Func.
Reading the spec (8.13.6) gives:
The first type-directed conversion converts anonymous function expressions and other function-valued arguments to delegate types.
Given:
A formal parameter of delegate type D
· An actual argument farg of known type ty1 -> ... -> tyn -> rty
· Precisely n arguments to the Invoke method of delegate type D
Then:
· The parameter is interpreted as if it were written:
new D(fun arg1 ... argn -> farg arg1 ... argn)
As these conditions apply, the Func delegates are converted to curried form, so an ambiguity exists.
I need to cast a function with the signature Thing -> Thing -> int to a Comparison<Thing>.
For example, when I have:
Array.Sort(values, mySort)
I get an error stating "This expression was expected to have type Comparison but here has type Thing -> Thing -> int"
When I try this:
Array.Sort(values, (fun (a, b) -> mySort a b))
(actually, this is by way of a wrapper object)
I get an error stating "Type constraint mismatch. The type ''a -> 'b' is not compatible with the type 'Comparison'"
How am I supposed to provide a Comparison<T>?
FYI: Comparison<T> is a delegate with the signature 'T * 'T -> int
This works for me:
let values = [|1;2;3|]
let mySort a b = 0
Array.Sort(values, mySort)
But have you tried:
Array.Sort(values, Comparison(mySort))
Oddly, despite the fact that delegate signatures are expressed in tupled form (e.g. 'T * 'T -> int), you need to provide your function in curried form (e.g. fun a b -> mySort a b). See section 6.4.2 of the spec.
I wonder what this means in F#.
“a function taking an integer, which returns a function which takes an integer and returns an integer.”
But I don't understand this well.
Can anyone explain this so clear ?
[Update]:
> let f1 x y = x+y ;;
val f1 : int -> int -> int
What this mean ?
F# types
Let's begin from the beginning.
F# uses the colon (:) notation to indicate types of things. Let's say you define a value of type int:
let myNumber = 5
F# Interactive will understand that myNumber is an integer, and will tell you this by:
myNumber : int
which is read as
myNumber is of type int
F# functional types
So far so good. Let's introduce something else, functional types. A functional type is simply the type of a function. F# uses -> to denote a functional type. This arrow symbolizes that what is written on its left-hand side is transformed into what is written into its right-hand side.
Let's consider a simple function, that takes one argument and transforms it into one output. An example of such a function would be:
isEven : int -> bool
This introduces the name of the function (on the left of the :), and its type. This line can be read in English as:
isEven is of type function that transforms an int into a bool.
Note that to correctly interpret what is being said, you should make a short pause just after the part "is of type", and then read the rest of the sentence at once, without pausing.
In F# functions are values
In F#, functions are (almost) no more special than ordinary types. They are things that you can pass around to functions, return from functions, just like bools, ints or strings.
So if you have:
myNumber : int
isEven : int -> bool
You should consider int and int -> bool as two entities of the same kind: types. Here, myNumber is a value of type int, and isEven is a value of type int -> bool (this is what I'm trying to symbolize when I talk about the short pause above).
Function application
Values of types that contain -> happens to be also called functions, and have special powers: you can apply a function to a value. So, for example,
isEven myNumber
means that you are applying the function called isEven to the value myNumber. As you can expect by inspecting the type of isEven, it will return a boolean value. If you have correctly implemented isEven, it would obviously return false.
A function that returns a value of a functional type
Let's define a generic function to determine is an integer is multiple of some other integer. We can imagine that our function's type will be (the parenthesis are here to help you understand, they might or might not be present, they have a special meaning):
isMultipleOf : int -> (int -> bool)
As you can guess, this is read as:
isMultipleOf is of type (PAUSE) function that transforms an int into (PAUSE) function that transforms an int into a bool.
(here the (PAUSE) denote the pauses when reading out loud).
We will define this function later. Before that, let's see how we can use it:
let isEven = isMultipleOf 2
F# interactive would answer:
isEven : int -> bool
which is read as
isEven is of type int -> bool
Here, isEven has type int -> bool, since we have just given the value 2 (int) to isMultipleOf, which, as we have already seen, transforms an int into an int -> bool.
We can view this function isMultipleOf as a sort of function creator.
Definition of isMultipleOf
So now let's define this mystical function-creating function.
let isMultipleOf n x =
(x % n) = 0
Easy, huh?
If you type this into F# Interactive, it will answer:
isMultipleOf : int -> int -> bool
Where are the parenthesis?
Note that there are no parenthesis. This is not particularly important for you now. Just remember that the arrows are right associative. That is, if you have
a -> b -> c
you should interpret it as
a -> (b -> c)
The right in right associative means that you should interpret as if there were parenthesis around the rightmost operator. So:
a -> b -> c -> d
should be interpreted as
a -> (b -> (c -> d))
Usages of isMultipleOf
So, as you have seen, we can use isMultipleOf to create new functions:
let isEven = isMultipleOf 2
let isOdd = not << isEven
let isMultipleOfThree = isMultipleOf 3
let endsWithZero = isMultipleOf 10
F# Interactive would respond:
isEven : int -> bool
isOdd : int -> bool
isMultipleOfThree : int -> bool
endsWithZero : int -> bool
But you can use it differently. If you don't want to (or need to) create a new function, you can use it as follows:
isMultipleOf 10 150
This would return true, as 150 is multiple of 10. This is exactly the same as create the function endsWithZero and then applying it to the value 150.
Actually, function application is left associative, which means that the line above should be interpreted as:
(isMultipleOf 10) 150
That is, you put the parenthesis around the leftmost function application.
Now, if you can understand all this, your example (which is the canonical CreateAdder) should be trivial!
Sometime ago someone asked this question which deals with exactly the same concept, but in Javascript. In my answer I give two canonical examples (CreateAdder, CreateMultiplier) inf Javascript, that are somewhat more explicit about returning functions.
I hope this helps.
The canonical example of this is probably an "adder creator" - a function which, given a number (e.g. 3) returns another function which takes an integer and adds the first number to it.
So, for example, in pseudo-code
x = CreateAdder(3)
x(5) // returns 8
x(10) // returns 13
CreateAdder(20)(30) // returns 50
I'm not quite comfortable enough in F# to try to write it without checking it, but the C# would be something like:
public static Func<int, int> CreateAdder(int amountToAdd)
{
return x => x + amountToAdd;
}
Does that help?
EDIT: As Bruno noted, the example you've given in your question is exactly the example I've given C# code for, so the above pseudocode would become:
let x = f1 3
x 5 // Result: 8
x 10 // Result: 13
f1 20 30 // Result: 50
It's a function that takes an integer and returns a function that takes an integer and returns an integer.
This is functionally equivalent to a function that takes two integers and returns an integer. This way of treating functions that take multiple parameters is common in functional languages and makes it easy to partially apply a function on a value.
For example, assume there's an add function that takes two integers and adds them together:
let add x y = x + y
You have a list and you want to add 10 to each item. You'd partially apply add function to the value 10. It would bind one of the parameters to 10 and leaves the other argument unbound.
let list = [1;2;3;4]
let listPlusTen = List.map (add 10)
This trick makes composing functions very easy and makes them very reusable. As you can see, you don't need to write another function that adds 10 to the list items to pass it to map. You have just reused the add function.
You usually interpret this as a function that takes two integers and returns an integer.
You should read about currying.
a function taking an integer, which returns a function which takes an integer and returns an integer
The last part of that:
a function which takes an integer and returns an integer
should be rather simple, C# example:
public int Test(int takesAnInteger) { return 0; }
So we're left with
a function taking an integer, which returns (a function like the one above)
C# again:
public int Test(int takesAnInteger) { return 0; }
public int Test2(int takesAnInteger) { return 1; }
public Func<int,int> Test(int takesAnInteger) {
if(takesAnInteger == 0) {
return Test;
} else {
return Test2;
}
}
You may want to read
F# function types: fun with tuples and currying
In F# (and many other functional languages), there's a concept called curried functions. This is what you're seeing. Essentially, every function takes one argument and returns one value.
This seems a bit confusing at first, because you can write let add x y = x + y and it appears to add two arguments. But actually, the original add function only takes the argument x. When you apply it, it returns a function that takes one argument (y) and has the x value already filled in. When you then apply that function, it returns the desired integer.
This is shown in the type signature. Think of the arrow in a type signature as meaning "takes the thing on my left side and returns the thing on my right side". In the type int -> int -> int, this means that it takes an argument of type int — an integer — and returns a function of type int -> int — a function that takes an integer and returns an integer. You'll notice that this precisely matches the description of how curried functions work above.
Example:
let f b a = pown a b //f a b = a^b
is a function that takes an int (the exponent) and returns a function that raises its argument to that exponent, like
let sqr = f 2
or
let tothepowerofthree = f 3
so
sqr 5 = 25
tothepowerofthree 3 = 27
The concept is called Higher Order Function and quite common to functional programming.
Functions themselves are just another type of data. Hence you can write functions that return other functions. Of course you can still have a function that takes an int as parameter and returns something else. Combine the two and consider the following example (in python):
def mult_by(a):
def _mult_by(x):
return x*a
return mult_by
mult_by_3 = mult_by(3)
print mylt_by_3(3)
9
(sorry for using python, but i don't know f#)
There are already lots of answers here, but I'd like to offer another take. Sometimes explaining the same thing in lots of different ways helps you to 'grok' it.
I like to think of functions as "you give me something, and I'll give you something else back"
So a Func<int, string> says "you give me an int, and I'll give you a string".
I also find it easier to think in terms of 'later' : "When you give me an int, I'll give you a string". This is especially important when you see things like myfunc = x => y => x + y ("When you give curried an x, you get back something which when you give it a y will return x + y").
(By the way, I'm assuming you're familiar with C# here)
So we could express your int -> int -> int example as Func<int, Func<int, int>>.
Another way that I look at int -> int -> int is that you peel away each element from the left by providing an argument of the appropriate type. And when you have no more ->'s, you're out of 'laters' and you get a value.
(Just for fun), you can transform a function which takes all it's arguments in one go into one which takes them 'progressively' (the official term for applying them progressively is 'partial application'), this is called 'currying':
static void Main()
{
//define a simple add function
Func<int, int, int> add = (a, b) => a + b;
//curry so we can apply one parameter at a time
var curried = Curry(add);
//'build' an incrementer out of our add function
var inc = curried(1); // (var inc = Curry(add)(1) works here too)
Console.WriteLine(inc(5)); // returns 6
Console.ReadKey();
}
static Func<T, Func<T, T>> Curry<T>(Func<T, T, T> f)
{
return a => b => f(a, b);
}
Here is my 2 c. By default F# functions enable partial application or currying. This means when you define this:
let adder a b = a + b;;
You are defining a function that takes and integer and returns a function that takes an integer and returns an integer or int -> int -> int. Currying then allows you partiallly apply a function to create another function:
let twoadder = adder 2;;
//val it: int -> int
The above code predifined a to 2, so that whenever you call twoadder 3 it will simply add two to the argument.
The syntax where the function parameters are separated by space is equivalent to this lambda syntax:
let adder = fun a -> fun b -> a + b;;
Which is a more readable way to figure out that the two functions are actually chained.