Does it matter how I declare a pipeline? I know of three ways:
let hello name = "Hello " + name + "!"
let solution1 = hello <| "Homer"
let solution2 = "Homer" |> hello
Which would you choose? solution1 or solution2 - and why?
As mentioned the pipe-forward operator |> helps with function composition and type inference. It allows you to rearrange the parameters of a function so that you can put the last parameter of a function first. This enables a chaining of functions that is very readable (similar to LINQ in C#). Your example doesn't show the power of this - it really shines when you have a transformation "pipeline" set up for several functions in a row.
Using |> chaining you could write:
let createPerson n =
if n = 1 then "Homer" else "Someone else"
let hello name = "Hello " + name + "!"
let solution2 =
1
|> createPerson
|> hello
|> printf "%s"
The benefit of the pipe-backward operator <| is that it changes operator precedence so it can save you a lot of brackets: Function arguments are normally evaluated left to right, using <| you don't need the brackets if you want to pass the result of one function to another function - your example doesn't really take advantage of this.
These would be equivalent:
let createPerson n =
if n = 1 then "Homer" else "Someone else"
let hello name = "Hello " + name + "!"
let solution3 = hello <| createPerson 1
let solution4 = hello (createPerson 1)
F# reads from top-to-bottom, left-to-right. For this reason, the |> operator is used much more than <| as it helps out type inference.
Related
I am new to programming so this should be an easy one.
I want to write a code that asks for 3 numbers and then finds the minimum. Something like that:
let main(): Unit =
putline ("Please enter 3 numbers:")
putline ("First number: ")
let a = getline ()
putline ("Second number: ")
let b = getline ()
putline("Third number: ")
let c = getline ()
if (a<b && a<c) then putline ("Minimum:" + a)
elif (b<c && b<a) then putline ("Minimum:" + b)
else putline ("Minimum:" + c)
I am sorry if this is terrible but I am still new to this. Also I am not allowed to use the dictionary. Any advice?
You can use the F# function min, which gives you the minimum of two values.
min 1 2 // 1
To get the minimum of three values you can use it twice:
min (min a b) c
A cleaner way to write this with F# piping is:
a |> min b |> min c
Alternatively, put the items in a list and use List.min:
[ a; b; c ] |> List.min
If, for some reason, you decide to expand beyond three numbers, you could consider using Seq.reduce
let xs = [0;-5;3;4]
xs
|> Seq.reduce min
|> printfn "%d"
// prints -5 to stdout
You can use min as the reducer because it accepts 2 arguments, which is exactly what Seq.reduce expects
Firstly your putline function. I'm assuming that this is supposed to take a value and print it to the console with a newline, so the built in F# command to do this is printfn and you would use it something like this:
let a = 1
printfn "Minimum: %d" a
The %d gets replaced with the value of a as, in this case, a is an integer. You would use %f for a float, %s for a string... the details will all be in the documentation.
So we could write your putline function like this:
let putline s = printfn "%s" s
This function has the following signature, val putline : s:string -> unit, it accepts a string and return nothing. This brings us onto your next problem, you try and say putline ("Minimum:" + a). This won't work as adding a number and a string isn't allowed, so what you could do is convert a to a string and you have several ways to do this:
putline (sprintf "Minimum: %d" a)
putline ("Minimum:" + a.ToString())
sprintf is related to printfn but gives you back a string rather than printing to the console, a.ToString() converts a to a string allowing it to be concatenated with the preceding string. However just using printfn instead of putline will work here!
You also have a logic problem, you don't consider the cases where a == b == c, what's the minimum of 1,1,3? Your code would say 3. Try using <= rather than <
For reading data from the console, there is already an answer on the site for this Read from Console in F# that you can look at.
I'm reading through an F# tutorial, and ran into an example of syntax that I don't understand. The link to the page I'm reading is at the bottom. Here's the example from that page:
let rec quicksort2 = function
| [] -> []
| first::rest ->
let smaller,larger = List.partition ((>=) first) rest
List.concat [quicksort2 smaller; [first]; quicksort2 larger]
// test code
printfn "%A" (quicksort2 [1;5;23;18;9;1;3])
The part I don't understand is this: ((>=) first). What exactly is this? For contrast, this is an example from the MSDN documentation for List.partition:
let list1 = [ 1 .. 10 ]
let listEven, listOdd = List.partition (fun elem -> elem % 2 = 0) list1
printfn "Evens: %A\nOdds: %A" listEven listOdd
The first parameter (is this the right terminology?) to List.partition is obviously an anonymous function. I rewrote the line in question as this:
let smaller,larger = List.partition (fun e -> first >= e) rest
and it works the same as the example above. I just don't understand how this construct accomplishes the same thing: ((>=) first)
http://fsharpforfunandprofit.com/posts/fvsc-quicksort/
That's roughly the same thing as infix notation vs prefix notation
Operator are functions too and follow the same rule (ie. they can be partially applied)
So here (>=) first is the operator >= with first already applied as "first" operand, and gives back a function waiting for the second operand of the operator as you noticed when rewriting that line.
This construct combines two features: operator call with prefix notation and partial function application.
First, let's look at calling operators with prefix notation.
let x = a + b
The above code calls operator + with two arguments, a and b. Since this is a functional language, everything is a function, including operators, including operator +. It's just that operators have this funny call syntax, where you put the function between the arguments instead of in front of them. But you can still treat the operator just as any other normal function. To do that, you need to enclose it on parentheses:
let x = (+) a b // same thing as a + b.
And when I say "as any other function", I totally mean it:
let f = (+)
let x = f a b // still same thing.
Next, let's look at partial function application. Consider this function:
let f x y = x + y
We can call it and get a number in return:
let a = f 5 6 // a = 11
But we can also "almost" call it by supplying only one of two arguments:
let a = f 5 // a is a function
let b = a 6 // b = 11
The result of such "almost call" (technically called "partial application") is another function that still expects the remaining arguments.
And now, let's combine the two:
let a = (+) 5 // a is a function
let b = a 6 // b = 11
In general, one can write the following equivalency:
(+) x === fun y -> x + y
Or, similarly, for your specific case:
(>=) first === fun y -> first >= y
so I have a function named evalExpr which accepts as argument a quotation <# ... #> and it returns a value.
For example if I write
let v = evalExpr <# 22 + 2 * 22 + 45 #>
then, v is equal to 111.
Now, I want to place inside the quotations a string variable instead of the expressions, but doing this, the variable is part of the quotation and so not defined.
How can I use variable values inside quotations in F#?
Unquote features an operator evalWith : Map<string,obj> -> Quotations.Expr<'a> -> 'a that allows you to evaluate synthetic quotations with unbound variables using an environment map that provides the variable values.
First, open the Swensen.Unquote namespace to make the evalWith operator available.
open Swensen.Unquote;;
Next, construct a quotation which represents a variable x of type int:
let xvar : Quotations.Expr<int> = Quotations.Expr.Var(new Quotations.Var("x", typeof<int>)) |> Quotations.Expr.Cast;;
Next, construct a quotation with the xvar spliced in:
let q = <# %xvar + 10 #>;;
Now you can evaluate your quotation q with the variable x provided like so:
evalWith (Map.ofList [("x", box 2)]) q;;
The answer is 12!
It's unclear what exactly you're asking for, but maybe something like this would help?
type Vars<'a> private () =
static let dict = System.Collections.Generic.Dictionary<string,Quotations.Var>()
static member Var(nm) =
match dict.TryGetValue nm with
| true, v -> v
| _ ->
let v = Quotations.Var(nm, typeof<'a>)
dict.[nm] <- v
v
[<GeneralizableValue>]
let x<'a> : Quotations.Expr<'a> = Quotations.Expr.Var(Vars<'a>.Var "x") |> Quotations.Expr.Cast
[<GeneralizableValue>]
let y<'a> : Quotations.Expr<'a> = Quotations.Expr.Var(Vars<'a>.Var "y") |> Quotations.Expr.Cast
let q1 = <# %x + %y * (1 + %x) #>
let q2 = <# "test" + %x #>
I've been trying to get my head round various bits of F# (I'm coming from more of a C# background), and parsers interest me, so I jumped at this blog post about F# parser combinators:
http://santialbo.com/blog/2013/03/24/introduction-to-parser-combinators
One of the samples here was this:
/// If the stream starts with c, returns Success, otherwise returns Failure
let CharParser (c: char) : Parser<char> =
let p stream =
match stream with
| x::xs when x = c -> Success(x, xs)
| _ -> Failure
in p //what does this mean?
However, one of the things that confused me about this code was the in p statement. I looked up the in keyword in the MSDN docs:
http://msdn.microsoft.com/en-us/library/dd233249.aspx
I also spotted this earlier question:
Meaning of keyword "in" in F#
Neither of those seemed to be the same usage. The only thing that seems to fit is that this is a pipelining construct.
The let x = ... in expr allows you to declare a binding for some variable x which can then be used in expr.
In this case p is a function which takes an argument stream and then returns either Success or Failure depending on the result of the match, and this function is returned by the CharParser function.
The F# light syntax automatically nests let .. in bindings, so for example
let x = 1
let y = x + 2
y * z
is the same as
let x = 1 in
let y = x + 2 in
y * z
Therefore, the in is not needed here and the function could have been written simply as
let CharParser (c: char) : Parser<char> =
let p stream =
match stream with
| x::xs when x = c -> Success(x, xs)
| _ -> Failure
p
The answer from Lee explains the problem. In F#, the in keyword is heritage from earlier functional languages that inspired F# and required it - namely from ML and OCaml.
It might be worth adding that there is just one situation in F# where you still need in - that is, when you want to write let followed by an expression on a single line. For example:
let a = 10
if (let x = a * a in x = 100) then printfn "Ok"
This is a bit funky coding style and I would not normally use it, but you do need in if you want to write it like this. You can always split that to multiple lines though:
let a = 10
if ( let x = a * a
x = 100 ) then printfn "Ok"
Hi everbody I am doing a project with F# but I get this error when ı use let num= line for the following code . I'm new at F# so I can not solve the problem. My code should do this things. User enter a number and calculate the fibonacci but if user enter not a number throw exception
open System
let rec fib n =
match n with
|0->0
|1->1
|2->1
|n->fib(n-1)+fib(n-2);;
let printFibonacci list =
for i=0 to (List.length list)-1 do
printf "%d " (list.Item(i));;
let control = true
while control do
try
printfn "Enter a Number:"
let num:int = Convert.ToInt32(stdin.ReadLine())
with
| :? System.FormatException->printfn "Number Format Exception";
let listFibonacci = [for i in 0 .. num-1->fib(i)]
printFibonacci(listFibonacci)
printfn "\n%A"(listFibonacci)
control<-false
Console.ReadKey(true)
exit 0;;
I'm not an F# expert but I can see 3 problems with the code you posted.
1) As Lasse V Karlsen commented - f# uses the 'offside' rule so your 'fib' expression needs the body indented in. If you are running this in the Visual Studio Shell it should warn you of this by putting a blue squiggly line under the appropriate code.
2) Both 'control' and 'num' are mutable values so need to be declared explicitly as such.
f# is a functional language so by default any expressions are immutable i.e they are not allowed to change state after they have been declared.
In f#, saying 'let n = expr' does not mean 'assign the value of expr to n' like you would in say c# or c++. Instead it means 'n fundamentally is expr' and will be forever much like a mathematical equation.
So if you want to update the value of a variable you use the special '<-' notation which is the equivalent of 'assign the value on rhs to the lhs' and you need to declare that variable as mutable i.e 'this value can be changed later'
So I think both num and control need to be declared at the top of the loop as
let mutable control = false
let mutable num = 0 // or whatever you want the initial value of num to be
As a side note you don't have to explicitly declare num as an int ( you can if you want ) but f# will infer the type for you
If I understand your code correctly, you want to keep asking for input number n until a valid number is given and print fibonacci numbers up to n. In this case, you'd better move the calculation and printing inside the try block. Here's an updated version with formatting.
open System
let rec fib n =
match n with
|0->0
|1->1
|2->1
|n->fib(n-1)+fib(n-2);;
let printFibonacci list =
for i=0 to (List.length list)-1 do
printf "%d " (list.Item(i))
let mutable control = true //you forgot to add the 'mutable' keyword
while control do
try
printfn "Enter a Number:"
let num:int = Convert.ToInt32(stdin.ReadLine())
let listFibonacci = [for i in 0 .. num-1 -> fib(i)]
printFibonacci(listFibonacci)
printfn "\n%A"(listFibonacci)
control <- false
with
| :? System.FormatException -> printfn "Number Format Exception"
//add the ignore statement to drop the resulting ConsoleKeyInfo struct
//or the compiler will complain about an unused value floating around.
Console.ReadKey(true) |> ignore
// exit 0 (* Exit isn't necessary *)
Instead of using an imperative style number entry routine and relying on exceptions for control flow, here's a recursive getNumberFromConsole function you could use as well:
open System
let rec fib n =
match n with
| 0 -> 0
| 1 | 2 -> 1
| n -> fib(n-1) + fib(n-2);;
let printFibonacci list =
for i=0 to (List.length list)-1 do
printf "%d " (list.Item(i))
//alternative number input, using recursion
let rec getNumberFromConsole() =
match Int32.TryParse(stdin.ReadLine()) with
| (true, value) -> value
| (false, _) -> printfn "Please enter a valid number"
getNumberFromConsole()
printfn "Enter a Number:"
let num = getNumberFromConsole()
let listFibonacci = [for i in 0 .. num-1 -> fib(i)]
printFibonacci(listFibonacci)
printfn "\n%A"(listFibonacci)
Console.ReadKey(true) |> ignore
P.S. Thanks for showing me stdin. I never knew it existed. Now I can write some interactive scripts.