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Interfaces and recursive types
f#

Hi!

I have a compilation error. Here is a minimalist example which reproduces the problem:

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open StructuredFormat


type 'a foo when 'a :> IFormattable =

 | Item of 'a * 'a foo

 | Empty

with

  interface IFormattable with

    member x.GetLayout(env) = ...

   end

end

and exp =

 | Foo of exp foo  // compilation error is here

 | Bar

with

  interface IFormattable with

    member x.GetLayout(env) = ...

  end

end

The error message is "The type exp is not compatible with the type IFormattable". I didn't have the problem when I didn't use the 'a parameter in foo. If I write "exp foo" after the type declarations, it's allowed. So, it looks like exp is not considered as an IFormattable until type declaration is completed.

Does anyone have a solution?

.

I am no F# wiz but I believe only classes (and thus object expressions) can subscribe to interfaces. Type exp is simply an algebraic data type, not a class. A solution would depend much on what you exactly mean by your code, which is not clear to me.

Cheers,

Jorge.

By on 3/7/2007 6:10 AM ()

No, it works with algebraic data types. I've used several times, and my example works when there's not the type parameter (and the recursion).

This code was minimalist. I have a quite big AST that implements the IFormattable interface (for pretty-printing). It's working fine, but I'd like to add a type parameter to make it more generic. When I added the type parameter (with the type constraint), I got the message "The type exp is not compatible with the type IFormattable".

So my current solution is to duplicate type definitions, but it's not pretty.

Laurent.

By on 3/7/2007 7:08 AM ()

Hi LLB,

This is indeed a bug. We will of course fix it for the next release.

Kind regards

Don

By on 3/9/2007 7:13 PM ()

Hi,

Thank you Don.
I have another problem with the interface. I might be the same bug, but I prefer to be sure. I can't call recursively GetLayout (error: "member 'GetLayout' is not defined"). As a workaround, I defined a get_layout function, but it's not elegant. Minimalist example:

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let get_layout (env: #IEnvironment) (e: #IFormattable) = e.GetLayout(env)


type foo = Item of int * foo | Empty

with

   interface IFormattable with

      member x.GetLayout(env) =

          match x with

//           | Item (_,a) -> a.GetLayout(env) // not allowed

            | Item (_,a) -> get_layout env a   // ok

            | Empty -> emptyL

    end

end

Laurent.

By on 3/10/2007 3:04 PM ()

The issue here is that the "this" value i.e. "a" has type "foo". Methods associated with intefaces are not immediately accessible from a value of type "foo" - you have to cast to the interface type first. Try the following

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type foo = Item of int * foo | Empty
with
  interface IFormattable with
   member x.GetLayout(env) =
     match x with
     | Item (_,a) -> (a :> IFormattable).GetLayout(env) 
     | Empty -> emptyL
  end
end

Don

By on 3/10/2007 4:05 PM ()

OK, thank you.
I hoped the compiler could know it's a IFormattable. Actually, I will keep my solution, because it's shorter (I would need to add a lot of casts).

Laurent.

By on 3/10/2007 6:21 PM ()

Hi Laurent,

Just to let you know that this bug has been fixed and the fix will be in the 1.9.1 release of F#, due out "real soon now".

Thanks

don

By on 3/15/2007 10:55 AM ()

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