I'm using F# v 1.9.6.2, and I've defined a very simple computation expression:
type MaybeBuilder() =
member this.Let(x, f) =
printfn "this.Let: %A" x
this.Bind(Some x, f)
member this.Bind(x, f) =
printfn "this.Bind: %A" x
match x with
| Some(x) when x >= 0 && x <= 100 -> f(x)
| _ -> None
member this.Delay(f) = f()
member this.Return(x) = Some x
let maybe = MaybeBuilder()
I've sprinkled some print statements in the code to tell me which methods are being called in a computation expression. When I execute the following statement:
maybe {
let x = 12
let! y = Some 11
let! z = Some 30
return x + y + z
}
I expect the console to print out the following:
this.Let 12
this.Bind Some 12
this.Bind Some 11
this.Bind Some 30
But my actual results are as follows:
this.Bind: Some 11
this.Bind: Some 30
In other words, F# doesn't appear to be executing the Let member. When I re-write Let to throw an exception, the code run without an exception. Additionally, when I comment out the Let member entirely, I do not get an error message stating The field, constructor or member 'Let' is not defined, and the code executes as expected.
(I've tried investigating the code with Reflector, but as is usually the case, decompiled F# is mangled beyond readability.)
It looks like the spec for computation expressions has changed. Are let bindings no longer treated as syntax sugar, and is the Let members no longer required in computation workflows?
You had the answer yourself. From the F# spec that describes how computation expressions are translated:
{| let binds in cexpr |}C = let binds in {| cexpr |}C)
So no, you don't need to define let explicitly anymore, it's translated by the compiler.
Update: this change is mentioned in the detailed release notes of the September CTP.
Correct -- you can no longer provide a binding for let :(.
See also
http://cs.hubfs.net/forums/thread/6950.aspx
Related
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
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"
I suspect that I am missing something very obvious here but this doesn't work:
let t = Array2D.create 1 1 1.0
for x in t do printfn "%f" x;;
It fails with
error FS0001: The type 'obj' is not compatible with any of the types float,float32,decimal, arising from the use of a printf-style format string
Interestingly using printf "%A" or "%O" prints the expected values which suggests to me that the problem is with the type inference
The corresponding code for a 1D array works fine
let t = Array.create 1 1.0
for x in t do printfn "%f" x;;
For reference this is on version 2.0 (both interactive and compiler) running on the latest mono
In .NET, a 1D array implicitly implements IList, which means it also implements (by inheritance) IEnumerable<T>. So, when you run:
let t = Array.create 1 1.0
for x in t do printfn "%f" x;;
the F# compiler emits code which gets an implementation of IEnumerable<T> (seq<T> in F#) from t, then iterates over it. Since it's able to get an IEnumerable<T> from the array, x will have type T.
On the other hand, multi-dimensional arrays (2d, 3d, etc.) only implement IEnumerable (not IEnumerable<T>) so the F# compiler infers the type of x as System.Object (or obj, in F#).
There are two solutions for what you want:
Cast each individual value within the loop, before printing it:
for x in t do printfn "%f" (x :?> float);;
Or, use Seq.cast to create and iterate over a strongly-typed enumerator:
for x in (Seq.cast<float> t) do printfn "%f" x;;
As Jack pointed out, this is a problem. One easy solution is:
let t = Array2D.create 2 2 1.0
t |> Array2D.iter (printfn "%f");;
And if you really like the for .. in .. do syntax:
type Array2DForLoopBuilder() =
member __.Zero() = ()
member __.For(a, f) = Array2D.iter f a
member __.Run e = e
let a2dfor = Array2DForLoopBuilder()
let t = Array2D.init 2 2 (fun a b -> float a + float b)
a2dfor { for x in t do printfn "%f" x }
Consider the following code:
let mutable a = 0.
let b = ref 0.
a <- // works
printfn "%A" a
4. + 8.
b := // does not work
printfn "%A" a
4. + 8.
b := ( // works
printfn "%A" a
4. + 8. )
Why does the ref assignment operator (:=) have a different behaviour than the mutable assignment operator (<-)?
I can only give a partial answer.
:= is defined in terms of <- in FSharp.Core\prim-types.fs:
let (:=) x y = x.contents <- y
In your example
b := // does not work
printfn "%A" a
4. + 8.
printfn "%A" a seems to be interpreted as y, which cannot be assigned to the int ref cell (wrong type). By grouping the whole expression with ( ... ), y now also contains 4. + 8.. Maybe the two operators behave differently, because <- seems to be an intrinsic operator (i.e. part of the language, not the library).
Building on the other answers...
More elaborate expressions are allowed within assignments, so long as the final expression is one of several allowed forms. See section 6.4.9 of the spec. This allows complex assignments such as:
let x =
let rec gcd a = function
| 0 -> a
| b -> gcd b (a % b)
gcd 10 25
The compiler moves gcd to a private member, but nesting it within the assignment allows for tighter scoping. Function arguments, on the other hand, are more restricted. They don't create a new scope (that I'm aware of) and you can't define functions, for example, as part of the expression.
:= is a function (try (:=);; in FSI) which has a type : 'a ref -> 'a -> unit
So
b := // does not work
printfn "%A" a
4. + 8.
is being parsed as because of the infix call parsing rule:
(:=) b (printfn "%A" a)
4. + 8.
Which is invalid as par the (:=) function type.
Other example:
let c = 10 +
11
12
c would be 12 here
Looks like a discrepancy in the indentation-sensitive parser rather than anything specifically to do with those operators.
I want to be able to write a computation expression in F# that will be able to retry an operation if it throws an exception. Right now my code looks like:
let x = retry (fun() -> GetResourceX())
let y = retry (fun() -> GetResourceY())
let z = retry (fun() -> DoThis(x, y))
etc. (this is obviously an astract representation of the actual code)
I need to be able to retry each of the functions a set number of times, which I have defined elswhere.
I was thinking a computation expression could help me here, but I don't see how it could help me remove explicitly wrapping each right hand side to a Retryable<'T>
I could see the computation expression looking something like:
let! x = Retryable( fun() -> GetResourceX())
etc.
I understand that Monads, in a crude fashion, are wrapper types, but I was hoping a way around this. I know I can overload an operator and have a very succinct syntax for converting an operation into a Retryable<'T>, but to me that's just making the repetition/wrapping more succinct; it's still there. I could wrap each function to be a Retryable<'T>, but once again, I don't see the value over doing what's done at the top of the post (calling retry on each operation. At least it's very explicit).
Maybe computation expressions are the wrong abstraction here, I'm not sure. Any ideas on what could be done here?
Computation expressions have a few extensions (in addition to the standard monadic features), that give you a nice way to do this.
As you said, the monads are essentially wrappers (creating e.g. Retryable<'T>) that have some additional behavior. However, F# computation expression can also define Run member which automatically unwraps the value, so the result of retry { return 1 } can have just a type int.
Here is an example (the builder is below):
let rnd = new System.Random()
// The right-hand side evaluates to 'int' and automatically
// retries the specified number of times
let n = retry {
let n = rnd.Next(10)
printfn "got %d" n
if n < 5 then failwith "!" // Throw exception in some cases
else return n }
// Your original examples would look like this:
let x = retry { return GetResourceX() }
let y = retry { return GetResourceY() }
let z = retry { return DoThis(x, y) }
Here is the definition of the retry builder. It is not really a monad, because it doesn't define let! (when you use computation created using retry in another retry block, it will just retry the inner one X-times and the outer one Y-times as needed).
type RetryBuilder(max) =
member x.Return(a) = a // Enable 'return'
member x.Delay(f) = f // Gets wrapped body and returns it (as it is)
// so that the body is passed to 'Run'
member x.Zero() = failwith "Zero" // Support if .. then
member x.Run(f) = // Gets function created by 'Delay'
let rec loop(n) =
if n = 0 then failwith "Failed" // Number of retries exceeded
else try f() with _ -> loop(n-1)
loop max
let retry = RetryBuilder(4)
A simple function could work.
let rec retry times fn =
if times > 1 then
try
fn()
with
| _ -> retry (times - 1) fn
else
fn()
Test code.
let rnd = System.Random()
let GetResourceX() =
if rnd.Next 40 > 1 then
"x greater than 1"
else
failwith "x never greater than 1"
let GetResourceY() =
if rnd.Next 40 > 1 then
"y greater than 1"
else
failwith "y never greater than 1"
let DoThis(x, y) =
if rnd.Next 40 > 1 then
x + y
else
failwith "DoThis fails"
let x = retry 3 (fun() -> GetResourceX())
let y = retry 4 (fun() -> GetResourceY())
let z = retry 1 (fun() -> DoThis(x, y))
Here is a first try at doing this in a single computation expression. But beware that it's only a first try; I have not thoroughly tested it. Also, it's a little bit ugly when re-setting the number of tries within the computation expression. I think the syntax could be cleaned-up a good bit within this basic framework.
let rand = System.Random()
let tryIt tag =
printfn "Trying: %s" tag
match rand.Next(2)>rand.Next(2) with
| true -> failwith tag
| _ -> printfn "Success: %s" tag
type Tries = Tries of int
type Retry (tries) =
let rec tryLoop n f =
match n<=0 with
| true ->
printfn "Epic fail."
false
| _ ->
try f()
with | _ -> tryLoop (n-1) f
member this.Bind (_:unit,f) = tryLoop tries f
member this.Bind (Tries(t):Tries,f) = tryLoop t f
member this.Return (_) = true
let result = Retry(1) {
do! Tries 8
do! tryIt "A"
do! Tries 5
do! tryIt "B"
do! tryIt "C" // Implied: do! Tries 1
do! Tries 2
do! tryIt "D"
do! Tries 2
do! tryIt "E"
}
printfn "Your breakpoint here."
p.s. But I like both Tomas's and gradbot's versions better. I just wanted to see what this type of solution might look like.