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Propagating optional arguments

The following code does not compile.

type A(?arg) =
  member __.Arg : string option = arg

type B(?arg) =
  inherit A(arg) //ERROR expected type string but has type 'a option

I assume this is because an instance of the underlying type of the option must be provided, and the compiler handles passing Some/None based on syntax.

Assuming my assumption has been correctly assumed, is there a workaround for this? Is it possible to propagate optional arguments?

like image 882
Daniel Avatar asked Aug 17 '11 15:08

Daniel


1 Answers

F# spec 8.13.5 Optional arguments to method members

Callers may specify values for optional arguments by using the following techniques:

  • By name, such as arg2 = 1.
  • By propagating an existing optional value by name, such as ?arg2=None or ?arg2=Some(3) or ?arg2=arg2. This can be useful when building one method that passes optional arguments on to another.
  • By using normal, unnamed arguments matched by position.

    type A(?arg) =
        member __.Arg : string option = arg
    
    type B(?arg) =
        inherit A(?arg = arg) 
    
    printfn "1. %A" (B()).Arg // None
    printfn "2. %A" (B("1")).Arg // Some "1"
    
    printfn "3. %A" (A()).Arg // None
    printfn "4. %A" (A("1")).Arg // Some "1"
    
like image 171
desco Avatar answered Oct 23 '22 21:10

desco