"int -> int -> int" What does this mean in F# ?

Question

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 ?

Answer 1

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.

Answer 2

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

Answer 3

You usually interpret this as a function that takes two integers and returns an integer. You should read about currying.

Answer 4

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

Answer 5

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;
    }
}

Answer 6

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#)

Answer 7

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.

Answer 8

You may want to read

F# function types: fun with tuples and currying

Answer 9

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.

Answer 10

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);
}

Answer 11

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.