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Record type implementing an interface
f#
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type IFoo =

    abstract Bar: int

    

type Foo (value: int) =

    member this.Bar = value

    interface IFoo with

        member this.Bar = this.Bar + 1

        

type FooRec =

    { Bar: int }

    interface IFoo with

        member this.Bar = this.Bar + 1

Could anyone explain why Foo is fine but FooRec isn't? They basically compile down to the same thing, except that FooRec implements some extra interfaces inherently.

.

Just change it to

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type FooRec = { bar: int } 

   with

   interface IFoo 

      with member this.Bar = this.bar + 1

And you will be fine - just mind the "bar" VS "Bar" issue and the "with" keyword

By on 3/17/2009 10:19 PM ()

Well I understand that changing the record property name to "bar" will allow it to compile, and the "with" doesn't appear to be necessary. The issue is that I don't see why a record property can't have the same name as a member of an implemented interface. Is this a bug, or is it intended behavior, and if so, what's the rationale?

By on 3/17/2009 10:34 PM ()

This is a bug (logged as 4666); thanks.

By on 3/18/2009 11:13 AM ()

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