Here's my code
let rec Interest a b c =
if (c=0) then b else Interest(a ((1.0+a)*b) (c-1));;
The error is:
if (c=0) then b else Interest(a ((1.0+a)*b) (c-1));;
-------------------------^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
stdin(2,26): error FS0001: Type mismatch. Expecting a
'a but given a
'a -> int -> 'a The resulting type would be infinite when unifying ''a' and ''a -> int -> 'a'
>
You've defined Interest as a function that takes three arguments but what you pass doesn't match that. The way it's written, you're only passing in a single argument to the recursive call to Interest - the parenthesized expression a ((1.0=a)*b) (c-1). Here, a is inferred to be a function of two arguments, ((1.0=a)*b) and (c-1).
Long story short, this throws off the type inference algorithm, until it gives up saying that it can't get a hang of what type to give to Interest.
What you want is this:
let rec Interest a b c =
if (c=0) then b else Interest a ((1.0=a)*b) (c-1)
You'll also have a problem with (1.0=a). This evaluates to a boolean value that you later want to multiply with b. Not sure what the intent is, but you might be looking for something like (if 1.0 = a then 1 else 0)?
Unlike C-like languages that support implicit conversions between "bools" and ints, F# requires you to make all your conversions explicit to ensure correctness (this goes for converting between numeric types as well).
Related
Is there a way to do string formatting for Exceptions with kprintf (or similar) and still get warnings about redundant arguments? So far I have:
type MyException (s:string) =
inherit System.Exception(s)
static member Raise msg =
Printf.kprintf (fun s -> raise (MyException(s))) msg
do
MyException.Raise "boom %d" 9 1 // Gives NO warning about redundant arguments
failwithf "boom %d" 9 1 // Gives warning about redundant arguments
I know I could use the new $ formatting but would like to stay compatible with older versions of F#.
The problem does not come from kprintf, but from raise, which allows the return type of the function to be any type, including a curried function. The function failwithf has this warning as a special case.
If you return, let’s say, a string or a unit, you would actually get an error. The downside is that now you cannot raise in every position anymore, as then the return argument is not generic anymore.
You could fix that by forcing a type argument, but that would make it less than ideal for the caller.
For instance, this works, but if you change the return type of f to be a string, it fails:
open System
type MyException (s:string) =
inherit System.Exception(s)
static member Raise msg =
Printf.kprintf (fun s -> raise (MyException s) |> ignore) msg
module Test =
let f() : unit = // string won’t work
MyException.Raise "boom %d" 9 10 // error, as expected
Edit: workaround
Here's some kind of a workaround. You can use generic type constraints to limit the type, and use the fact that an F# function type is not comparable or equatable, does not have a proper value null.
Most types you'll use are equatable, because of F#'s powerful structural equality support. Other options for a type constraint are struct (only structs), null (this excludes record and DU types) or a certain inheritance constraint if that serves your use case.
Example:
type MyException (s:string) =
inherit System.Exception(s)
static member Raise<'T when 'T : equality> msg =
let raiser s =
raise (MyException s)
Unchecked.defaultof<'T> // fool the compiler
Printf.kprintf raiser msg
module Test =
let f() = MyException.Raise "boom %d" 9
let g() = MyException.Raise "boom %d" 9 10 // error
Downside, apart from the type restriction, is that the error you get will be a type error, not a warning as you requested.
The check for this warning in the compiler specifically looks for a number of known F# library functions. You can see the checks in the compiler source code and here (the functions are raise, failwith, failwithf, nullArg, invalidOp and invalidArg).
The reason for this is that, there is, in principle, nothing wrong with a generic function that returns 'T and is used so that 'T is inferred to be a function. For example, it is perfectly fine to do:
let id f = f
id sin 3.14
Here, we define a function id with one argument, but if I give it a function as an argument, it also becomes valid to call it with an extra argument as id sin 3.14.
The special functions are special in that they look like they return a value of a generic type, but they never actually return, because they raise an execption. The compiler cannot, in general, know whether that is the case for any custom function or method you yourself write - and so it only checks this for known special functions. (It would have to do more sophisticated analysis, possibly marking the return type of these functions as the bottom type, but that is beyond what F# does...).
I was attempting to convert this to F# but I can't figure out what I'm doing wrong as the error message (in title) is too broad of an error to search for, so I found no resolutions.
Here is the code:
let getIP : string =
let host = Dns.GetHostEntry(Dns.GetHostName())
for ip in host.AddressList do
if ip.AddressFamily = AddressFamily.InterNetwork then
ip.ToString() // big fat red underline here
"?"
A for loop in F# is for running imperative-style code, where the code inside the for loop does not produce a result but instead runs some kind of side-effect. Therefore, the expression block in an F# for loop is expected to produce the type unit, which is what side-effect functions should return. (E.g., printfn "Something" returns the unit type). Also, there's no way to exit a for loop early in F#; this is by design, and is another reason why a for loop isn't the best approach to do what you're trying to do.
What you're trying to do is go through a list one item at a time, find the first item that matches some condition, and return that item (and, if the item is not found, return some default value). F# has a specialized function for that: Seq.find (or List.find if host.AddressList is an F# list, or Array.find if host.AddressList is an array. Those three functions take different input types but all work the same way conceptually, so from now on I'll focus on Seq.find, which takes any IEnumerable as input so is most likely to be what you need here).
If you look at the Seq.find function's type signature in the F# docs, you'll see that it is:
('T -> bool) -> seq<'T> -> 'T
This means that the function takes two parameters, a 'T -> bool and seq<'T> and returns a 'T. The 'T syntax means "this is a generic type called T": in F#, the apostrophe means that what follows is the name of a generic type. The type 'T -> bool means a function that takes a 'T and returns a Boolean; i.e., a predicate that says "Yes, this matches what I'm looking for" or "No, keep looking". The second argument to Seq.find is a seq<'T>; seq is F#'s shorter name for an IEnumerable, so you can read this as IEnumerable<'T>. And the result is an item of type 'T.
Just from that function signature and name alone, you can guess what this does: it goes through the sequence of items and calls the predicate for each one; the first item for which the predicate returns true will be returned as the result of Seq.find.
But wait! What if the item you're looking for isn't in the sequence at all? Then Seq.find will throw an exception, which may not be the behavior you're looking for. Which is why the Seq.tryFind function exists: its function signature looks just like Seq.find, except for the return value: it returns 'T option rather than 'T. That means that you'll either get a result like Some "ip address" or None. In your case, you intend to return "?" if the item isn't found. So you want to convert a value that's either Some "ip address or None to either "ip address" (without the Some) or "?". That is what the defaultArg function is for: it takes a 'T option, and a 'T representing the default value to return if your value is None, and it returns a plain 'T.
So to sum up:
Seq.tryFind takes a predicate function and a sequence, and returns a 'T option. In your case, this will be a string option
defaultArg takes a 'T option and a default value, and returns a normal 'T (in your case, a string).
With these two pieces, plus a predicate function you can write yourself, we can do what you're looking for.
One more note before I show you the code: you wrote let getIP : string = (code). It seems like you intended for getIP to be a function, but you didn't give it any parameters. Writing let something = (code block) will create a value by running the code block immediately (and just once) and then assigning its result to the name something. Whereas writing let something() = (code block) will create a function. It will not run the code block immediately, but it will instead run the code block every time the function is called. So I think you should have written let getIP() : string = (code).
Okay, so having explained all that, let's put this together to give you a getIP function that actually works:
let getIP() = // No need to declare the return type, since F# can infer it
let isInternet ip = // This is the predicate function
// Note that this function isn't accessible outside of getIP()
ip.AddressFamily = AddressFamily.InterNetwork
let host = Dns.GetHostEntry(Dns.GetHostName())
let maybeIP = Seq.tryFind isInternet host.AddressList
defaultArg maybeIP "?"
I hope that's clear enough; if there's anything you don't understand, let me know and I'll try to explain further.
Edit: The above has one possible flaw: the fact that F# may not be able to infer the type of the ip argument in isInternet without an explicit type declaration. It's clear from the code that it needs to be some class with an .AddressFamily property, but the F# compiler can't know (at this point in the code) which class you intend to pass to this predicate function. That's because the F# compiler is a single-pass compiler, that works its way through the code in a top-down, left-to-right order. To be able to infer the type of the ip parameter, you might need to rearrange the code a little, as follows:
let getIP() = // No need to declare the return type, since F# can infer it
let host = Dns.GetHostEntry(Dns.GetHostName())
let maybeIP = host.AddressList |> Seq.tryFind (fun ip -> ip.AddressFamily = AddressFamily.InterNetwork)
defaultArg maybeIP "?"
This is actually more idiomatic F# anyway. When you have a predicate function being passed to Seq.tryFind or other similar functions, the most common style in F# is to declare that predicate as an anonymous function using the fun keyword; this works just like lambdas in C# (in C# that predicate would be ip => ip.AddressFamily == AddressFamily.InterNetwork). And the other thing that's common is to use the |> operator with things like Seq.tryFind and others that take predicates. The |> operator basically* takes the value that's before the |> operator and passes it as the last parameter of the function that's after the operator. So foo |> Seq.tryFind (fun x -> xyz) is just like writing Seq.tryFind (fun x -> xyz) foo, except that foo is the first thing you read in that line. And since foo is the sequence that you're looking in, and fun x -> xyz is how you're looking, that feels more natural: in English, you'd say "Please look in my closet for a green shirt", so the concept "closet" comes up before "green shirt". And in idiomatic F#, you'd write closet |> Seq.find (fun shirt -> shirt.Color = "green"): again, the concept "closet" comes up before "green shirt".
With this version of the function, F# will encounter host.AddressList before it encounters fun ip -> ..., so it will know that the name ip refers to one item in host.AddressList. And since it knows the type of host.AddressList, it will be able to infer the type of ip.
* There's a lot more going on behind the scenes with the |> operator, involving currying and partial application. But at a beginner level, just think of it as "puts a value at the end of a function's parameter list" and you'll have the right idea.
In F# any if/else/then-statement must evaluate to the same type of value for all branches. Since you've omitted the else-branch of the expression, the compiler will infer it to return a value of type unit, effectively turning your if-expression into this:
if ip.AddressFamily = AddressFamily.InterNetwork then
ip.ToString() // value of type string
else
() // value of type unit
Scott Wlaschin explains this better than me on the excellent F# for fun and profit.
This should fix the current error, but still won't compile. You can solve this either by translating the C#-code more directly (using a mutable variable for the localIP value, and doing localIP <- ip.ToString() in your if-clause, or you could look into a more idiomatic approach using something like Seq.tryFind.
Why does this statement give me a type mismatch error,
let x = List.rev [] in (3::x, true::x);;
while this statement does not?
let x = [] in (3::x, true::x);;
I'm assuming it is because x is given a function call in the first statement, while it is only give an empty list in the second statement. But, I am not sure exactly why the second works and the first does not? Any help will be much appreciated. Thanks!
Try the following:
let x = [] ;;
Result: val x : 'a list. F# knows that x is a list of as-yet unknown type. If it had any contents, its type would be known. This works perfectly well.
But the following will not work:
let x = List.rev [] ;;
Result:
error FS0030: Value restriction. The value 'x' has been inferred to have generic type
val x : '_a list
Either define 'x' as a simple data term, make it a function with explicit arguments or, if you do not intend for it to be generic, add a type annotation.
The "value restriction" error in F# can be hard to understand — why is [] allowed when List.rev [] is not? — but this article goes into some details. Essentially, F# can always feel safe making functions generic, but it can only feel safe making values generic if the following two conditions apply:
The expression on the right-hand side of the let is a pure value, e.g. it has no side-effects,
and
The expression on the right-hand side of the let is immutable.
When the expression is [], then F# knows that it's a pure, immutable value, so it can make it generic ('a list). But when the expression is someFunction [], the F# compiler doesn't know what someFunction is going to do. Even though in this case, we know that List.rev is part of the standard library and matches those two scenarios, the F# compiler can't know that. In a completely pure language like Haskell, you can know from a function's type signature that it's pure, but F# is a more pragmatic language. So F# takes the guaranteed-safe approach and does not make the result of List.rev [] generic.
Therefore, when you write let x = [] in (3::x, true::x), the [] value is a generic empty list, so it can become either an int list or a bool list as needed. But when you write let x = List.rev [] in (3::x, true::x), F# cannot make List.rev [] generic. It can say "This is a list of a type I don't yet know", and wait for the type to become clear. Then the first expression of a specific type that uses this list, in this case 3::x, will "lock" the type of that list. I.e., F# cannot consider this list to be generic, so now it has figured out that this empty list is an empty list of ints. And then when you try to append a bool to an empty int list, that's an error.
If you flipped the tuple around so that it was true::x, 3::x then the type error would "flip" as well: it would expect a bool list and find an int list.
So the short version of this answer is: you're hitting the F# value restriction, even though that's not immediately obvious since the error you got didn't mention value restriction at all.
See Understanding F# Value Restriction Errors for a good discussion of the value restriction, including the most common place where you'd normally see it (partially-applied functions).
I thought that conversions between F# functions and System.Func had to be done manually, but there appears to be a case where the compiler (sometimes) does it for you. And when it goes wrong the error message isn't accurate:
module Foo =
let dict = new System.Collections.Generic.Dictionary<string, System.Func<obj,obj>>()
let f (x:obj) = x
do
// Question 1: why does this compile without explicit type conversion?
dict.["foo"] <- fun (x:obj) -> x
// Question 2: given that the above line compiles, why does this fail?
dict.["bar"] <- f
The last line fails to compile, and the error is:
This expression was expected to have type
System.Func<obj,obj>
but here has type
'a -> obj
Clearly the function f doesn't have a signature of 'a > obj. If the F# 3.1 compiler is happy with the first dictionary assignment, then why not the second?
The part of the spec that should explain this is 8.13.7 Type Directed Conversions at Member Invocations. In short, when invoking a member, an automatic conversion from an F# function to a delegate will be applied. Unfortunately, the spec is a bit unclear; from the wording it seems that this conversion might apply to any function expression, but in practice it only appears to apply to anonymous function expressions.
The spec is also a bit out of date; in F# 3.0 type directed conversions also enable a conversion to a System.Linq.Expressions.Expression<SomeDelegateType>.
EDIT
In looking at some past correspondence with the F# team, I think I've tracked down how a conversion could get applied to a non-syntactic function expression. I'll include it here for completeness, but it's a bit of a strange corner case, so for most purposes you should probably consider the rule to be that only syntactic functions will have the type directed conversion applied.
The exception is that overload resolution can result in converting an arbitrary expression of function type; this is partly explained by section 14.4 Method Application Resolution, although it's pretty dense and still not entirely clear. Basically, the argument expressions are only elaborated when there are multiple overloads; when there's just a single candidate method, the argument types are asserted against the unelaborated arguments (note: it's not obvious that this should actually matter in terms of whether the conversion is applicable, but it does matter empirically). Here's an example demonstrating this exception:
type T =
static member M(i:int) = "first overload"
static member M(f:System.Func<int,int>) = "second overload"
let f i = i + 1
T.M f |> printfn "%s"
EDIT: This answer explains only the mysterious promotion to 'a -> obj. #kvb points out that replacing obj with int in OPs example still doesn't work, so that promotion is in itself insufficient explanation for the observed behaviour.
To increase flexibility, the F# type elaborator may under certain conditions promote a named function from f : SomeType -> OtherType to f<'a where 'a :> SomeType> : 'a -> OtherType. This is to reduce the need for upcasts. (See spec. 14.4.2.)
Question 2 first:
dict["bar"] <- f (* Why does this fail? *)
Because f is a "named function", its type is promoted from f : obj -> obj following sec. 14.4.2 to the seemingly less restrictive f<'a where 'a :> obj> : 'a -> obj. But this type is incompatible with System.Func<obj, obj>.
Question 1:
dict["foo"] <- fun (x:obj) -> x (* Why doesn't this, then? *)
This is fine because the anonymous function is not named, and so sec. 14.4.2 does not apply. The type is never promoted from obj -> obj and so fits.
We can observe the interpreter exhibit behaviour following 14.4.2:
> let f = id : obj -> obj
val f : (obj -> obj) (* Ok, f has type obj -> obj *)
> f
val it : ('a -> obj) = <fun:it#135-31> (* f promoted when used. *)
(The interpreter doesn't output constraints to obj.)
From the book by Tomas Petricek the following code doesn't work as compiler is unable to infer the type of the dt parameter:
> Option.map (fun dt -> dt.Year) (Some(DateTime.Now));;
error FS0072: Lookup on object of indeterminate type.
And if we specify type explicitly everything works fine:
> Option.map (fun (dt:DateTime) -> dt.Year) (Some(DateTime.Now));;
val it : int option = Some(2008)
Or we can use pipelining operator to "help" compiler to infer type:
> Some(DateTime.Now) |> Option.map (fun dt -> dt.Year);;
val it : int option = Some(2008)
The question is why F# compiler can't infer the type of the dt parameter? In this particular case it looks quite easy to infer the dt's type.
The logic behind it can be the following:
the signature of Option.map is map : ('T -> 'U) -> 'T option -> 'U option
the type of the last parameter is DateTime option
so our map usage looks like map : ('T -> 'U) -> 'DateTime option -> 'U option
the compiler then can try to substitute DateTime as 'T to see if it would be correct, so we have (DateTime -> 'U) -> 'DateTime option -> 'U option
then it can infer the 'U type by looking at the body of the lambda-function, so the 'U becomes int
and we finally have (DateTime -> int) -> 'DateTime option -> 'int option
So why F# can't do this inference? Tomas mentions in his book that F# infers types by going from the first to the last argument and that's why the order of arguments matters. And that's why F# can't infer the types in the first example. But why F# can't behave like C#, i.e. try to infer types incrementally starting with what is known?
In most cases F# is much more powerful when speaking about type inference... that't why I'm confused a bit.
So why F# can't do this inference?
F# could do that as OCaml does that. The disadvantage of this kind of more sophisticated inference is the obfuscation of error messages. OCaml taught us that the result generated such incomprehensible errors that, in practice, you always resort to annotating types in order to prevent the compiler from being led down a type inference garden path. Consequently, there was little motivation to implement this in F# because OCaml had already shown that it is not very pragmatic.
For example, if you do that in OCaml but mis-spell the method name then you will get a huge error message at some later point in the code where two inferred class types mismatch and you will have to hunt through it to find the discrepancy and then search back through your code to find the actual location of the error.
IMO, Haskell's type classes also suffer from an equivalent practical problem.
F# can do everything C#'s type inference can do...and much, much more. AFAIK, the extent of C#'s type inference is auto-typing a variable based on the right-hand side of an assignment.
var x = new Dictionary<string, int>();
The equivalent F# would be:
let x = Dictionary()
or
let x = Dictionary<_,_>()
or
let x = Dictionary<string,_>()
or
let x = Dictionary<string,int>()
You can provide as much or as little type information as you want, but you would almost never declare the type of x. So, even in this simple case, F#'s type inference is obviously much more powerful. Hindley-Milner type inference types entire programs, unifying all the expressions involved. As far as I can tell, C# type inference is limited to a single expression, assignment at that.