I found it's very hard to search for the simple indentation guide in F#.
Basically, I am wondering what's the rule for multiple-line statement indentation.
In C#, there is no problem because whitespace doesn't count.
Although I can write F# code according to my intuition and it works, I really want to know what's the rule for breaking one statement into multiple lines.
I write as
printfn "%d"
1
It works as expected
And if I write them in the same column, something goes wrong.
>
printfn "%A%A"
1
[];;
> //nothing is returned... and no error in this case
I want to confirm the basic rule for doing this. It's a little annoying when you can't be sure what you are doing.
Thanks in advance
I just tried another case
List.iter
(printfn "%d")
[1..10];;
And it prints out 1 to 10.
Why it's not
List.iter
((printfn "%d")
[1..10]);;
As Yin points out, the rule is that arguments of a function should be indented further than the call to the function. To add more details, your first snippet is interpreted like this:
printfn "%A%A";
1;
[];
Each of these is a valid expression that returns something (function, number, empty list) and then ignores the result and continues. Because they are written in the top-level scope, F# Interactive doesn't emit a warning that you're ignoring some values. If they were in a do block or let declaration:
do
printfn "%A%A"
1
[]
The F# compiler would emit a warning when sequencing expressions (using ;) that do not return unit:
stdin(5,3): warning FS0193: This expression is a function value, i.e. is missing arguments. Its type is 'a -> 'b -> unit.
stdin(6,3): warning FS0020: This expression should have type 'unit', but has typ
e 'int'. Use 'ignore' to discard the result of the expression, or 'let' to bind
the result to a name.
stdin(5,3): warning FS0020: This expression should have type 'unit', but has typ
e ''a list'. Use 'ignore' to discard the result of the expression, or 'let' to b
ind the result to a name.
In your second example, you should indent:
>
printfn "%A%A"
1
[];;
Otherwise the three expressions are three sequential expressions, not a single expression.
You can refer F# Language Specification for firm rules, e.g. Chapter 15 in the specification.
Related
I have some code which I'm expecting to pause when it asks for user input. It only does this however, if the last expression is Seq.initInfinite.
let consoleaction (i : int) =
Console.WriteLine ("Enter Input: ")
(Console.ReadLine().Trim(), i)
Seq.initInfinite (fun i -> consoleaction i) |> Seq.map (fun f -> printfn "%A" f)
printfn "foo" // program will not pause unless this line is commented out.
Very new to F# and I've spent way too much time on this already. Would like to know what is going on :)
If you try that piece of code in F# interactive you will see different effects depending on how you execute it.
For instance if you execute it in one shot it will create values but nothing will be executed since the Seq.initInfinite instruction is 'lost' I mean, not let-bound to anything and at the same time is a lazy expression so its side effects will not be executed. If you remove the last instruction it will start prompting, that's because fsi bounds to it the last expression so in order to show you the value of it it starts evaluating the seq expression.
Things are different if you put this in a function, for example:
open System
let myProgram() =
let consoleaction ...
Now you will get a warning on the Seq.initInfinite:
warning FS0020: This expression should have type 'unit', but has type
'seq<unit>'. Use 'ignore' to discard the result of the expression, or
'let' to bind the result to a name.
Which is very clear. Additionally to ignore as the warning suggest you can change the Seq.map to Seq.iter since you are not interested in the result of the map which will be a seq of units.
But now again your program will not execute (try myProgram())unless you remove the last line, the printfn and it's clear why, this is because it returns the last expression which is not the Seq.initInfinite which is lost since it's lazy and ignored.
If you remove the printfn it will become the 'return value' of your function so it will be evaluated when calling the function.
The precedence of F#'s member selection dot (.) operator as used in
System.Console.WriteLine("test")
has a lower precedence than [space] such that the following
ignore System.Console.WriteLine("test")
must be written explicitly as
ignore (System.Console.WriteLine("test"))
though this would be the intuition from the notion of juxtaposed symbols. Having used CoffeeScript, I can appreciate how intuitive precedence can serve to de-clutter code.
Are there any efforts being made to rationalize this kerfuffle, perhaps something along the lines that incorporated the "lightweight" syntax of the early years?
==============
Upon review, the culprit is not the "." operator but the invocation operator "()", as in "f()". So, given:
type C() = class end
then the following intuitive syntax fails:
printfn "%A" C() <-- syntax error FS0597
and must be written thus (as prescribed by the documentation):
printfn "%A" (C()) <-- OK
It seems intuitive that a string of symbols unbroken by white space should implicitly represents a block. In fact, the utility of juxtaposing is to create such a block.
a b.c is parsed as a (b.c), not (a b).c. So there are no efforts to rationalize this - it simply is not true.
Thanks to all those who responded.
My particular perplexity stemmed from treating () as an invocation operator. As an eager evaluation language, F# does not have or need such a thing. In stead, this is an expression boundary, as in, (expression). In particular, () bounds the nothing expression which is the only value of the type, unit. Consequently, () is the stipulation of a value and not a direction to resolved the associated function (though that is the practical consequence when parameters are provided to functions due to F#'s eager evaluation.)
As a result, the following expression
ignore System.Console.WriteLine("test")
actually surfaces three distinct values,
ignore System.Console.WriteLine ("test")
which are interpreted according to the left-to-right precedence evaluation order or F# (which then permits partial function application and perhaps other things)
( ignore System.Console.WriteLine ) ("test")
...but the result of (ignore expr) will be unit, which does not expect a parameter. Hence, syntax error (strong typing, yea!). So, an expression boundary is required. In particular,
ignore ( System.Console.WriteLine ("test") )
or
ignore (System.Console.WriteLine "test")
or
ignore <| System.Console.WriteLine "test"
or
System.Console.WriteLine "test" |> ignore
I am currently reading Programming in F# 3.0 and I read that F# supports partial function application.
But if I try
List.map (+1) [1 .. 10];;
Then I get error FS0001: This expression was expected to have type 'a -> 'b
but List.map (fun x -> x + 1) [1 .. 10];; compiles fine. Any ideas why?
I suppose you come from Haskell, for expecting it to work this way :)
Unfortunately, F# doesn't have the syntactic sugar where you can partially apply an operator by parenthesizing it preceded or followed by an expression. In your code, +1 is considered a single number token, which is why it complains that it is not a function.
What F# does have, however, is the syntax where you can parenthesize the operator alone. Which leads to #Lee's solution. Be careful about this syntax though: (+) 1 is equivalent to Haskell's (1+), not (+1). Obviously for addition it doesn't matter, but for asymmetrical operations such as subtraction, it can be misleading.
You need to put brackets around the +:
List.map ((+)1) [1..10]
How can I write a no-op statement in F#?
Specifically, how can I improve the second clause of the following match statement:
match list with
| [] -> printfn "Empty!"
| _ -> ignore 0
Use unit for empty side effect:
match list with
| [] -> printfn "Empty!"
| _ -> ()
The answer from Stringer is, of course, correct. I thought it may be useful to clarify how this works, because "()" insn't really an empty statement or empty side effect...
In F#, every valid piece of code is an expression. Constructs like let and match consist of some keywords, patterns and several sub-expressions. The F# grammar for let and match looks like this:
<expr> ::= let <pattern> = <expr>
<expr>
::= match <expr> with
| <pat> -> <expr>
This means that the body of let or the body of clause of match must be some expression. It can be some function call such as ignore 0 or it can be some value - in your case it must be some expression of type unit, because printfn ".." is also of type unit.
The unit type is a type that has only one value, which is written as () (and it also means empty tuple with no elements). This is, indeed, somewhat similar to void in C# with the exception that void doesn't have any values.
BTW: The following code may look like a sequence of statements, but it is also an expression:
printf "Hello "
printf "world"
The F# compiler implicitly adds ; between the two lines and ; is a sequencing operator, which has the following structure: <expr>; <expr>. It requires that the first expression returns unit and returns the result of the second expression.
This is a bit surprising when you're coming from C# background, but it makes the langauge surprisingly elegant and consise. It doesn't limit you in any way - you can for example write:
if (a < 10 && (printfn "demo"; true)) then // ...
(This example isn't really useful - just a demonstration of the flexibility)
When quoting
<# 1 + 1 #>
I want "1 + 1"
instead of
"Call (None, Int32 op_Addition[Int32,Int32,Int32](Int32, Int32),
[Value (1), Value (1)])"
You'll have to write it yourself. See the F# quotations visualizer code as a guide for transforming the quotations abstract syntax tree.
I have implemented a quotation decompiler as part of a larger open source project Unquote. It can decompile many simple F# quoted expressions as single-line non-light syntax strings (see the project's home page for a list of decompiler features). For example,
> decompile <# (11 + 3) / 2 = String.length ("hello world".Substring(4, 5)) #>;;
val it : string =
"(11 + 3) / 2 = String.length ("hello world".Substring(4, 5))"
#Kurt Schelfthout is correct about the many challenges faced when decompiling F# Quotations into human readable form. But from my work so far, I believe that it is possible to write a quotation decompiler which can generate correct F# code. Take match expressions and computation expressions for example, the Unquote decompiler can produce correct F# code in the following simple cases:
> decompile <# match true with | true -> "hi" | _ -> "bye" #>;;
val it : string =
"let matchValue = true in if matchValue then "hi" else "bye""
> decompile <# seq {yield 1; yield 2} #>;;
val it : string =
"seq (Seq.delay (fun unitVar -> Seq.append (Seq.singleton 1) (Seq.delay (fun unitVar -> Seq.singleton 2))))"
Infix and prefix operators are not too hard (as you can see in the first example), but source structure such as new lines and indentation is an interesting topic (though not terribly difficult, I think). However, single-line non-light syntax is sufficient for Unquote's requirements.
There is none, and it's not quite that easy, except in very simple cases. One of the main problems, for example, is the match construct. It is syntactic sugar for a whole bunch of if and switch statements (try printing a quotation with a match in, you'll see). Another one of those biggies are computation expressions, but I guess you could skip those at first.
Then there is a the rabbit hole of ambiguities you'll have to resolve, with conventions like the pipe operator starts a new line, let starts a new line, indentation, infix, prefix, special cases like the (::) operator and so forth.
All in all, doable, but not trivial. Sort of like decompiling.