With most operators in F# I can use prefix or infix notation, for example:
let x = a + b
is equivalent to
let x = (+) a b
This does not work for the exponentation operator ** however, because the parenthesised version is treated as a comment. That is, (*this is a comment*) is F# syntax for a comment, so (**) is treated as an empty comment.
let x = a ** b // a raised to b
let x = (**) a b // empty comment, followed by function a applied to b
Is there an escape character I can use or is this simply a strange quirk of the language?
Try using spaces between the parentheses, as pointed by kvb in the comments:
let x = ( ** ) a b
Related
How to use a three parametered infix operator?
Eg.: base function is
let orElse labelFunc p1 p2 = {...} and operator let ( <|> ) = orElse
Now, for non-infix version this works nicely:List.reduce ((<|>) labelFunc) parserList.
Can I use it somehow still "infixing"? eg.: (p1 (<|> labelFunc) p1) does not work nor any other combination, other than using the non-infix version here as well.
First of all, I think it's best to restrict the number of custom operators you're using in your code, because custom operators make F# code hard to read. F# lets you define custom operators, but it's not particularly designed to make this a great experience - it makes sense for some small domain-specific languages (like parser combinators), but not much else.
So, while I do not recommend using this, there is a weird trick that you can use to write something like p1 (<op> l) p2, which is to make <op> infix and replace the parentheses with two more custom operators:
let (</) a b = a, b
let (/>) c d = c, d
let (!) f = f
1 </ !10 /> 2
This sample just produces a tuple with all three arguments, but if you implement your logic in the </ operator, it will actually do something like what you want. But as I said, I would not do this :-).
I don't believe there is any good way to achieve that. Once you have a parenthesized expression it won't be parsed as an operator - even 1 (+) 1 doesn't work.
Put an F# infix operator in brackets, and it behaves like a function,
let foo = (*) 3 2 // foo = 6
let factorial n = [2..n] |> List.fold (*) 1 // n!
However, this doesn't work with the :: operator (cons operator),
let ls = (::) 1 [2..5] // Error: Unexpected symbol '::' in binding.
What's the reason for this?
You can use the static method:
let ls = List.Cons (1, [2..5])
or the operator's verbose name:
let ls = op_ColonColon (1, [2..5])
(checked with F# 3.0; older versions may behave differently. For instance, MSDN suggests op_Cons)
In both cases, there's no way to curry the arguments here. Numeric operators are defined like this:
let inline ( * ) (x:int) (y:int) = ...
The list concatenation, however, requires a tuple, and this also answers your question,
What's the reason for this?
In fact, (::) is not an usual operator (a standalone function or a type member), but a union case. Here's how the List<'T> is defined in F# sources:
type List<'T> =
| ([]) : 'T list
| (::) : Head: 'T * Tail: 'T list -> 'T list
So, if your purpose is partial application of arguments, the only nice solution would be writing a wrapper function as #pad has suggested.
Because (::) (and [] for that matter) is a symbolic keyword, you can't expect to use it as an infix operator. See F# specification, section 3.6 Symbolic keywords.
In this case, you have to define an extra function e.g.
let cons x xs = x :: xs
let ls = cons 1 [2..5]
Is there a way to write an infix function not using symbols? Something like this:
let mod x y = x % y
x mod y
Maybe a keyword before "mod" or something.
The existing answer is correct - you cannot define an infix function in F# (just a custom infix operator). Aside from the trick with pipe operators, you can also use extension members:
// Define an extension member 'modulo' that
// can be called on any Int32 value
type System.Int32 with
member x.modulo n = x % n
// To use it, you can write something like this:
10 .modulo 3
Note that the space before . is needed, because otherwise the compiler tries to interpret 10.m as a numeric literal (like 10.0f).
I find this a bit more elegant than using pipeline trick, because F# supports both functional style and object-oriented style and extension methods are - in some sense - close equivalent to implicit operators from functional style. The pipeline trick looks like a slight misuse of the operators (and it may look confusing at first - perhaps more confusing than a method invocation).
That said, I have seen people using other operators instead of pipeline - perhaps the most interesting version is this one (which also uses the fact that you can omit spaces around operators):
// Define custom operators to make the syntax prettier
let (</) a b = a |> b
let (/>) a b = a <| b
let modulo a b = a % b
// Then you can turn any function into infix using:
10 </modulo/> 3
But even this is not really an established idiom in the F# world, so I would probably still prefer extension members.
Not that I know of, but you can use the left and right pipe operators. For example
let modulo x y = x % y
let FourMod3 = 4 |> modulo <| 3
To add a little bit to the answers above, you can create a custom infix operators but the vocabulary is limited to:
!, $, %, &, *, +, -, ., /, <, =, >, ?, #, ^, |, and ~
Meaning you can build your infix operator using combining these symbols.
Please check the full documentation on MS Docs.
https://learn.microsoft.com/en-us/dotnet/fsharp/language-reference/operator-overloading
In this code:
let f(a,b,c) = a * b + c - (d())
let g(a,b,c) = a * b + c -(d())
f is (int*int*int) -> int, and g is (int*int*(int*int)) -> int.
Removing the brackets around d() in g causes the "Successive arguments should be separated by spaces or tupled" error.
What's going on?
#bytebuster is quite correct in his comment, but to put it into layman's terms ;-] one is parsed as the binary subtraction operator and the other is parsed as the unary negation operator – you're simply fighting operator precedence here.
I'm trying to learn F# by going through some of the Euler problems and I found an issue I haven't been able to figure out. This is my naive solution.
let compute =
let mutable f = false
let mutable nr = 0
while f = false do
nr <- nr + 20
f = checkMod nr
nr
When i do this I get the error message warning FS0020: This expression should have type 'unit', but has type 'bool' on the expression "nr <- nr +20". I've tried rewriting and moving the expressions around and I always get that error on the line below the while statement.
I'm writing this using VS2010 Beta.
Since I can imagine this weg page becoming the 'canonical' place to look up information about warning FS0020, here's my quick summary of the three commonest cases in which you get the warning, and how to fix them.
Intentionally discarding the result of a function that is called only for its side-effects:
// you are calling a function for its side-effects, intend to ignore result
let Example1Orig() =
let sb = new System.Text.StringBuilder()
sb.Append("hi") // warning FS0020
sb.Append(" there") // warning FS0020
sb.ToString()
let Example1Fixed() =
let sb = new System.Text.StringBuilder()
sb.Append("hi") |> ignore
sb.Append(" there") |> ignore
sb.ToString()
Warning is useful, pointing out an error (function has no effects):
// the warning is telling you useful info
// (e.g. function does not have an effect, rather returns a value)
let Example2Orig() =
let l = [1;2;3]
List.map (fun x -> x * 2) l // warning FS0020
printfn "doubled list is %A" l
let Example2Fixed() =
let l = [1;2;3]
let result = List.map (fun x -> x * 2) l
printfn "doubled list is %A" result
Confusing assignment operator and equality comparison operator:
// '=' versus '<-'
let Example3Orig() =
let mutable x = 3
x = x + 1 // warning FS0020
printfn "%d" x
let Example3Fixed() =
let mutable x = 3
x <- x + 1
printfn "%d" x
The following line:
f = checkMod nr
is an equality check, not an assignment as I believe you are intending. Change it to:
f <- checkMod nr
and all should work fine. I'm not sure why you've used the correct syntax on the previous line and not that line...
Also, the line while f = false do should really be simplified to while not f do; equality checks on booleans are rather convoluted.
As I side note, I feel a need to point out that you are effectively trying to use F# as an imperative language. Use of mutable variables and while loops are strongly discouraged in functional languages (including F#), especially when a purely functional (and simpler) solution exists, as in this situation. I recommend you read up a bit on programming in the functional style. Of course, just getting to grips with the syntax is a useful thing in itself.
If you're trying to adopt the functional style, try to avoid mutable values.
For example like this:
let nr =
let rec compute nr =
if checkMod nr then nr else compute (nr + 20)
compute 0
while expressions in F# take a little getting used to if you're coming from an imperative language. Each line in a while expression must evaluate to unit (think void from C++/C#). The overall expression then also evaluates to unit.
In the example:
nr <- nr + 20
evaluates to unit whereas
f = checkMod nr
evaluates to a bool as Noldorin noted. This results in a warning message being reported. You can actually turn the warning off if you so desire. Just put the following at the top of your file:
#nowarn "0020"
I've been programming in an imperative style for a long time, so getting used to the functional programming mindset took a while.
In your example, you're trying to find the first multiple of 20 that passes your checkMod test. That's the what part. For the functional how part, I recommend browsing through the methods available to sequences. What you need is the first element of a sequence (multiples of 20) passing your test, like this:
let multi20 = Seq.initInfinite (fun i -> i*20)
let compute = multi20 |> Seq.find checkMod
The first let generates an infinite list of twentyples (I made that one up). The second let finds the first number in said list that passes your test. Your task is to make sure that there actually is a number that will pass the test, but that's of course also true for the imperative code.
If you want to condense the two above lines into one, you can also write
let computeCryptic = Seq.initInfinite ((*) 20) |> Seq.find checkMod
but I find that pulling stunts like that in code can lead to headaches when trying to read it a few weeks later.
In the same spirit as Brian's post, here is another way to get warning FS0020: In a nutshell, I accidentally tupled the function arguments.
Being an F# newbie, I had a difficult time debugging the code below, which for the second line (let gdp...) gave the warning FS0020: This expression should have type 'unit', but has type '(string -> ^a -> unit) * string * float'. It turns out that line was not the problem at all; instead, it was the printfn line that was messed up. Removing the comma separators from the argument list fixed it.
for country in wb.Regions.``Arab World``.Countries do
let gdp = country.Indicators.``GDP per capita (current US$)``.[2010]
let gdpThous = gdp / 1.0e3
printfn "%s, %s (%.2f)" country.Name, country.CapitalCity, gdpThous