This code
let rec readNLines n list =
if n = 0 then
list
else
readNLines(n-1,readInt()::list)
ends with
Type mismatch. Expecting a 'a but given a 'a -> 'a
The resulting type would be infinite when unifying ''a' and
''a -> 'a' (using built-in F# compiler)
but runs ok when last line is changed to
readNLines(n-1,(readInt()::list))
or
readNLines(n-1)(readInt()::list)
Question is: Why? :|
Only the last version can work, because readNLines takes two arguments, but
readNLines (n - 1, readInt() :: list)
passes only one argument (which is a tuple consisting of an int and the list).
readNLines (n - 1) (readInt() :: list)
passes them as two separate arguments - the difference here is using the comma (tuple) and space (two arguments).
By the way, that becomes much clearer when you use more whitespace (as I did), because the individual elements are easier to identify.
Take a look at these two functions:
> let f1 a b = a + b
val f1 : a:int -> b:int -> int
> let f2 (a, b) = a + b
val f2 : a:int * b:int -> int
As you can see, they have slightly different types. In function f1 you partially apply the arguments (you'll see the term 'curried function' used here), in function f2 you pass in a tuple of arguments in one "go", or you can think of it as only ever having a single argument (an 'uncurried' function).
What you're doing is defining a function f1 style, but later calling it f2 style, which confuses the compiler.
Related
Upon covering the predefined datatypes in f# (i.e lists) and how to sum elements of a list or a sequence, I'm trying to learn how I can work with user defined datatypes. Say I create a data type, call it list1:
type list1 =
A
| B of int * list1
Where:
A stands for an empty list
B builds a new list by adding an int in front of another list
so 1,2,3,4, will be represented with the list1 value:
B(1, B(2, B(3, B(4, A))))
From the wikibook I learned that with a list I can sum the elements by doing:
let List.sum [1; 2; 3; 4]
But how do I go about summing the elements of a user defined datatype? Any hints would be greatly appreciated.
Edit: I'm able to take advantage of the match operator:
let rec sumit (l: ilist) : int =
match l with
| (B(x1, A)) -> x1
| (B(x1, B(x2, A))) -> (x1+x2)
sumit (B(3, B(4, A)))
I get:
val it : int = 7
How can I make it so that if I have more than 2 ints it still sums the elemets (i.e. (B(3, B(4, B(5, A)))) gets 12?
One good general approach to questions like this is to write out your algorithm in word form or pseudocode form, then once you've figured out your algorithm, convert it to F#. In this case where you want to sum the lists, that would look like this:
The first step in figuring out an algorithm is to carefully define the specifications of the problem. I want an algorithm to sum my custom list type. What exactly does that mean? Or, to be more specific, what exactly does that mean for the two different kinds of values (A and B) that my custom list type can have? Well, let's look at them one at a time. If a list is of type A, then that represents an empty list, so I need to decide what the sum of an empty list should be. The most sensible value for the sum of an empty list is 0, so the rule is "I the list is of type A, then the sum is 0". Now, if the list is of type B, then what does the sum of that list mean? Well, the sum of a list of type B would be its int value, plus the sum of the sublist.
So now we have a "sum" rule for each of the two types that list1 can have. If A, the sum is 0. If B, the sum is (value + sum of sublist). And that rule translates almost verbatim into F# code!
let rec sum (lst : list1) =
match lst with
| A -> 0
| B (value, sublist) -> value + sum sublist
A couple things I want to note about this code. First, one thing you may or may not have seen before (since you seem to be an F# beginner) is the rec keyword. This is required when you're writing a recursive function: due to internal details in how the F# parser is implemented, if a function is going to call itself, you have to declare that ahead of time when you declare the function's name and parameters. Second, this is not the best way to write a sum function, because it is not actually tail-recursive, which means that it might throw a StackOverflowException if you try to sum a really, really long list. At this point in your learning F# you maybe shouldn't worry about that just yet, but eventually you will learn a useful technique for turning a non-tail-recursive function into a tail-recursive one. It involves adding an extra parameter usually called an "accumulator" (and sometimes spelled acc for short), and a properly tail-recursive version of the above sum function would have looked like this:
let sum (lst : list1) =
let rec tailRecursiveSum (acc : int) (lst : list1) =
match lst with
| A -> acc
| B (value, sublist) -> tailRecursiveSum (acc + value) sublist
tailRecursiveSum 0 lst
If you're already at the point where you can understand this, great! If you're not at that point yet, bookmark this answer and come back to it once you've studied tail recursion, because this technique (turning a non-tail-recursive function into a tail-recursive one with the use of an inner function and an accumulator parameter) is a very valuable one that has all sorts of applications in F# programming.
Besides tail-recursion, generic programming may be a concept of importance for the functional learner. Why go to the trouble of creating a custom data type, if it only can hold integer values?
The sum of all elements of a list can be abstracted as the repeated application of the addition operator to all elements of the list and an accumulator primed with an initial state. This can be generalized as a functional fold:
type 'a list1 = A | B of 'a * 'a list1
let fold folder (state : 'State) list =
let rec loop s = function
| A -> s
| B(x : 'T, xs) -> loop (folder s x) xs
loop state list
// val fold :
// folder:('State -> 'T -> 'State) -> state:'State -> list:'T list1 -> 'State
B(1, B(2, B(3, B(4, A))))
|> fold (+) 0
// val it : int = 10
Making also the sum function generic needs a little black magic called statically resolved type parameters. The signature isn't pretty, it essentially tells you that it expects the (+) operator on a type to successfully compile.
let inline sum xs = fold (+) Unchecked.defaultof<_> xs
// val inline sum :
// xs: ^a list1 -> ^b
// when ( ^b or ^a) : (static member ( + ) : ^b * ^a -> ^b)
B(1, B(2, B(3, B(4, A))))
|> sum
// val it : int = 10
I have found a presentation by Don Syme which shows that Erlang's
fac(0) -> 1
fac(N) -> N * fac(N-1).
is equivalent to F#'s
let rec fac = function
| 0 -> 1
| n -> n * fac (n-1)
But it looks like there is no way to use pattern matching for a different arity without losing type safety. E.g. one could use a list pattern matching, but then a type must be a common base type (such as object):
let concat = function
| [x;y] -> x.ToString() + y.ToString()
| [x] -> x.ToString()
Given that F# functions in modules do not support overloads, it looks like the only way to rewrite Erlang code into F# with static typing is to use static classes with method overloading instead of modules. Is there a better way to rewrite Erlang functions with different arity in F#?
In general, is it correct to say that Erlang's argument matching is closer to .NET's (including C#) method overloading rather than to F#'s pattern matching? Or there is no direct replacement between the two, e.g. there could be a function in Erlang with different arities + a guard:
max(x) -> x.
max(x,y) when x > y -> x.
max(x,y) -> y.
max(comparer, x, y) -> if comparer(x,y) > 0 -> x; true -> y end.
In the last case the arguments are of different types. How would you rewrite it in F#?
You can achieve something close to overloading by rethinking the problem slightly. Instead of thinking about the function as the axis of variability, think of the input as the variable part. If you do that, you'll realise that you can achieve the same with a discriminated union.
Here's a more contrived example than the one in the linked article:
type MyArguments = One of int | Two of int * int
let foo = function
| One x -> string x
| Two (x, y) -> sprintf "%i%i" x y
Usage:
> foo (One 42);;
val it : string = "42"
> foo (Two (13, 37));;
val it : string = "1337"
Obviously, instead of defining such a 'stupid' type as the above MyArguments, you'd define a discriminated union that makes sense in the domain you're modelling.
let mapTuple f (a,b) = (f a, f b)
I'm trying to create a function that applies a function f to both items in a tuple and returns the result as a tuple. F# type inference says that mapTuple returns a 'b*'b tuple. It also assumes that a and b are of the same type.
I want to be able to pass two different types as parameters. You would think that wouldn't work because they both have to be passed as parameters to f. So I thought if they inherited from the same base class, it might work.
Here is a less generic function for what I am trying to accomplish.
let mapTuple (f:Map<_,_> -> Map<'a,'b>) (a:Map<int,double>,b:Map<double, int>) = (f a, f b)
However, it gives a type mismatch error.
How do I do it? Is what I am trying to accomplish even possible in F#?
Gustavo is mostly right; what you're asking for requires higher-rank types. However,
.NET (and by extension F#) does support (an encoding of) higher-rank types.
Even in Haskell, which supports a "nice" way of expressing such types (once you've enabled the right extension), they wouldn't be inferred for your example.
Digging into point 2 may be valuable: given map f a b = (f a, f b), why doesn't Haskell infer a more general type than map :: (t1 -> t) -> t1 -> t1 -> (t, t)? The reason is that once you include higher-rank types, it's not typically possible to infer a single "most general" type for a given expression. Indeed, there are many possible higher-rank signatures for map given its simple definition above:
map :: (forall t. t -> t) -> x -> y -> (x, y)
map :: (forall t. t -> z) -> x -> y -> (z, z)
map :: (forall t. t -> [t]) -> x -> y -> ([x], [y])
(plus infinitely many more). But note that these are all incompatible with each other (none is more general than another). Given the first one you can call map id 1 'c', given the second one you can call map (\_ -> 1) 1 'c', and given the third one you can call map (\x -> [x]) 1 'c', but those arguments are only valid with each of those types, not with the other ones.
So even in Haskell you need to specify the particular polymorphic signature you want to use - this may be a bit of a surprise if you're coming from a more dynamic language. In Haskell, this is relatively clean (the syntax is what I've used above). However, in F# you'll have to jump through an additional hoop: there's no clean syntax for a "forall" type, so you'll have to create an additional nominal type instead. For example, to encode the first type above in F# I'd write something like this:
type Mapping = abstract Apply : 'a -> 'a
let map (m:Mapping) (a, b) = m.Apply a, m.Apply b
let x, y = map { new Mapping with member this.Apply x = x } (1, "test")
Note that in contrast to Gustavo's suggestion, you can define the first argument to map as an expression (rather than forcing it to be a member of some separate type). On the other hand, there's clearly a lot more boilerplate than would be ideal...
This problem has to do with rank-n types which are supported in Haskell (through extensions) but not in .NET type system.
One way I found to workaround this limitation is to pass a type with a single method instead of a function and then define an inline map function with static constraints, for example let's suppose I have some generic functions: toString and toOption and I want to be able to map them to a tuple of different types:
type ToString = ToString with static member inline ($) (ToString, x) = string x
type ToOption = ToOption with static member ($) (ToOption, x) = Some x
let inline mapTuple f (x, y) = (f $ x, f $ y)
let tuple1 = mapTuple ToString (true, 42)
let tuple2 = mapTuple ToOption (true, 42)
// val tuple1 : string * string = ("True", "42")
// val tuple2 : bool option * int option = (Some true, Some 42)
ToString will return the same type but operating with arbitrary types. ToOption will return two Generics of different types.
By using a binary operator type inference creates the static constraints for you and I use $ because in Haskell it means apply so a nice detail is that for haskellers f $ x reads already apply x to f.
At the risk of stating the obvious, a good enough solution might be to have a mapTuple that takes two functions instead of one:
let mapTuple fa fb (a, b) = (fa a, fb b)
If your original f is generic, passing it as fa and fb will give you two concrete instantiations of the function with the types you're looking for. At worst, you just need to pass the same function twice when a and b are of the same type.
It's not a practically important issue, but I'd like to see an example of tacit programming in F# where my point-free functions can have multiple arguments (not in form of a list or tuple).
And secondly, how such functions can manipulate a complex data structure. I'm trying it out in F# Interactive, but have no success yet.
I tried, for instance:
> (fun _ -> (fun _ -> (+))) 333 222 111 555
Is that right way?
And:
> (fun _ -> (fun _ -> (+))) "a" "b" "c" "d";;
val it : string = "cd"
F# doesn't contain some of the basic functions that are available in Haskell (mainly because F# programmers usually prefer the explicit style of programming and use pointfree style only in the most obvious cases, where it doesn't hurt readability).
However you can define a few basic combinators like this:
// turns curried function into non-curried function and back
let curry f (a, b) = f a b
let uncurry f a b = f (a, b)
// applies the function to the first/second element of a tuple
let first f (a, b) = (f a, b)
let second f (a, b) = (a, f b)
Now you can implement the function to add lengths of two strings using combinators as follows:
let addLengths =
uncurry (( (first String.length) >> (second String.length) ) >> (curry (+)))
This constructs two functions that apply String.length to first/second element of a tuple, then composes them and then adds the elements of the tuple using +. The whole thing is wrapped in uncurry, so you get a function of type string -> string -> int.
In F#, the arity of functions is fixed, so you're not going to be able to write both
(op) 1 2
and
(op) 1 2 3 4
for any given operator op. You will need to use a list or other data structure if that's what you want. If you're just trying to avoid named variables, you can always do "1 + 2 + 3 + 4". The most idiomatic way to add a list of numbers in F# is List.sum [1;2;3;4], which also avoids variables.
I've been trying to get into F# on and off for a while but I keep getting put off. Why?
Because no matter which 'beginners' resource I try to look at I see very simple examples that start using the operator ->.
However, nowhere have I found as yet that provides a clear simple explanation of what this operator means. It's as though it must be so obvious that it doesn't need explanation even to complete newbies.
I must therefore be really dense or perhaps it's nearly 3 decades of previous experience holding me back.
Can someone please, explain it or point to a truly accessible resource that explains it?
'->' is not an operator. It appears in the F# syntax in a number of places, and its meaning depends on how it is used as part of a larger construct.
Inside a type, '->' describes function types as people have described above. For example
let f : int -> int = ...
says that 'f' is a function that takes an int and returns an int.
Inside a lambda ("thing that starts with 'fun' keyword"), '->' is syntax that separates the arguments from the body. For example
fun x y -> x + y + 1
is an expression that defines a two argument function with the given implementation.
Inside a "match" construct, '->' is syntax that separates patterns from the code that should run if the pattern is matched. For example, in
match someList with
| [] -> 0
| h::t -> 1
the stuff to the left of each '->' are patterns, and the stuff on the right is what happens if the pattern on the left was matched.
The difficulty in understanding may be rooted in the faulty assumption that '->' is "an operator" with a single meaning. An analogy might be "." in C#, if you have never seen any code before, and try to analyze the "." operator based on looking at "obj.Method" and "3.14" and "System.Collections", you may get very confused, because the symbol has different meanings in different contexts. Once you know enough of the language to recognize these contexts, however, things become clear.
It basically means "maps to". Read it that way or as "is transformed into" or something like that.
So, from the F# in 20 minutes tutorial,
> List.map (fun x -> x % 2 = 0) [1 .. 10];;
val it : bool list
= [false; true; false; true; false; true; false; true; false; true]
The code (fun i -> i % 2 = 0) defines
an anonymous function, called a lambda
expression, that has a parameter x and
the function returns the result of "x
% 2 = 0", which is whether or not x is
even.
First question - are you familiar with lambda expressions in C#? If so the -> in F# is the same as the => in C# (I think you read it 'goes to').
The -> operator can also be found in the context of pattern matching
match x with
| 1 -> dosomething
| _ -> dosomethingelse
I'm not sure if this is also a lambda expression, or something else, but I guess the 'goes to' still holds.
Maybe what you are really referring to is the F# parser's 'cryptic' responses:
> let add a b = a + b
val add: int -> int -> int
This means (as most of the examples explain) that add is a 'val' that takes two ints and returns an int. To me this was totally opaque to start with. I mean, how do I know that add isn't a val that takes one int and returns two ints?
Well, the thing is that in a sense, it does. If I give add just one int, I get back an (int -> int):
> let inc = add 1
val inc: int -> int
This (currying) is one of the things that makes F# so sexy, for me.
For helpful info on F#, I have found that blogs are FAR more useful that any of the official 'documentation': Here are some names to check out
Dustin Campbell (that's diditwith.net, cited in another answer)
Don Symes ('the' man)
Tomasp.net (aka Tomas Petricek)
Andrew Kennedy (for units of measure)
Fsharp.it (famous for the Project Euler solutions)
http://lorgonblog.spaces.live.com/Blog (aka Brian)
Jomo Fisher
(a -> b) means "function from a to b". In type annotation, it denotes a function type. For example, f : (int -> String) means that f refers to a function that takes an integer and returns a string. It is also used as a contstructor of such values, as in
val f : (int -> int) = fun n -> n * 2
which creates a value which is a function from some number n to that same number multiplied by two.
There are plenty of great answers here already, I just want to add to the conversation another way of thinking about it.
' -> ' means function.
'a -> 'b is a function that takes an 'a and returns a 'b
('a * 'b) -> ('c * 'd) is a function that takes a tuple of type ('a, 'b) and returns a tuple of ('c, 'd). Such as int/string returns float/char.
Where it gets interesting is in the cascade case of 'a -> 'b -> 'c. This is a function that takes an 'a and returns a function ('b -> 'c), or a function that takes a 'b -> 'c.
So if you write:
let f x y z = ()
The type will be f : 'a -> 'b -> 'c -> unit, so if you only applied the first parameter, the result would be a curried function 'b -> 'c -> 'unit.
From Microsoft:
Function types are the types given to
first-class function values and are
written int -> int. They are similar
to .NET delegate types, except they
aren't given names. All F# function
identifiers can be used as first-class
function values, and anonymous
function values can be created using
the (fun ... -> ...) expression form.
Many great answers to this questions, thanks people. I'd like to put here an editable answer that brings things together.
For those familiar with C# understanding -> being the same as => lamba expression is a good first step. This usage is :-
fun x y -> x + y + 1
Can be understood as the equivalent to:-
(x, y) => x + y + 1;
However its clear that -> has a more fundemental meaning which stems from concept that a function that takes two parameters such as the above can be reduced (is that the correct term?) to a series of functions only taking one parameter.
Hence when the above is described in like this:-
Int -> Int -> Int
It really helped to know that -> is right associative hence the above can be considered:-
Int -> (Int -> Int)
Aha! We have a function that takes Int and returns (Int -> Int) (a curried function?).
The explaination that -> can also appear as part of type definiton also helped. (Int -> Int) is the type of any of function which takes an Int and returns an Int.
Also helpful is the -> appears in other syntax such as matching but there it doesn't have the same meaning? Is that correct? I'm not sure it is. I suspect it has the same meaning but I don't have the vocabulary to express that yet.
Note the purpose of this answer is not to spawn further answers but to be collaboratively edited by you people to create a more definitive answer. Utlimately it would be good that all the uncertainies and fluf (such as this paragraph) be removed and better examples added. Lets try keep this answer as accessible to the uninitiated as possible.
In the context of defining a function, it is similar to => from the lambda expression in C# 3.0.
F#: let f = fun x -> x*x
C#: Func<int, int> f = x => x * x;
The -> in F# is also used in pattern matching, where it means: if the expression matches the part between | and ->, then what comes after -> should be given back as the result:
let isOne x = match x with
| 1 -> true
| _ -> false
The nice thing about languages such as Haskell (it's very similar in F#, but I don't know the exact syntax -- this should help you understand ->, though) is that you can apply only parts of the argument, to create curried functions:
adder n x y = n + x + y
In other words: "give me three things, and I'll add them together". When you throw numbers at it, the compiler will infer the types of n x and y. Say you write
adder 1 2 3
The type of 1, 2 and 3 is Int. Therefore:
adder :: Int -> Int -> Int -> Int
That is, give me three integers, and I will become an integer, eventually, or the same thing as saying:
five :: Int
five = 5
But, here's the nice part! Try this:
add5 = adder 5
As you remember, adder takes an int, an int, an int, and gives you back an int. However, that is not the entire truth, as you'll see shortly. In fact, add5 will have this type:
add5 :: Int -> Int -> Int
It will be as if you have "peeled off" of the integers (the left-most), and glued it directly to the function. Looking closer at the function signature, we notice that the -> are right-associative, i.e.:
addder :: Int -> (Int -> (Int -> Int))
This should make it quite clear: when you give adder the first integer, it'll evaluate to whatever's to the right of the first arrow, or:
add5andtwomore :: Int -> (Int -> Int)
add5andtwomore = adder 5
Now you can use add5andtwomore instead of "adder 5". This way, you can apply another integer to get (say) "add5and7andonemore":
add5and7andonemore :: Int -> Int
add5and7andonemore = adder 5 7
As you see, add5and7andonemore wants exactly another argument, and when you give it one, it will suddenly become an integer!
> add5and7andonemore 9
=> ((add5andtwomore) 7) 9
=> ((adder 5) 7) 9)
<=> adder 5 7 9
Substituting the parameters to adder (n x y) for (5 7 9), we get:
> adder 5 7 9 = 5 + 7 + 9
=> 5 + 7 + 9
=> 21
In fact, plus is also just a function that takes an int and gives you back another int, so the above is really more like:
> 5 + 7 + 9
=> (+ 5 (+ 7 9))
=> (+ 5 16)
=> 21
There you go!