Re: nested _next for same object class, confusion

[Brian G <brian@xxxxxxxxxxxxxxxx>]

On 5/30/24 13:46, Benoît Minisini wrote:
> Le 30/05/2024 à 20:15, Brian G a écrit :
> > I have a question, how to implement nested key value for a class with 
> > a _next function where the key is not equal to the enum.index value.
> >
> > I include a simple script to demonstrate the issue. Which happened to 
> > me by accident.
> >
> > After the nested _next, the key is of course equal to the value of 
> > the last nested cycle of the inner most next.
> >
> > Here a script(extremely simplified) that shows the issue I am having. 
> > Maybe I am not seeing the simple answer.
>
> 'Enum.Index' is a Variant, so you can put whatever you want in it. In 
> your case, you must put both the index and the key.
>
> Another solution is making the 'Key' property compute the key from the 
> current index, and not store it in a global variable.
>
> Regards,
I Tried that

The enum.index access caused the error 'Script Error > Not an enumeration' when not accessing it inside the _next function
here is the updated script doing just that.

To be clear I am trying to print the key from the first for each after the inner one completes

-------------------------------------------------------------------------------------------------------------
#!/usr/bin/env gbs3
' Gambas Script File Created 05/30/2024 08:36:16

Public Sub main()
Dim MyHendrix As New MyNext(6)
Print "begin"
For Each v As Variant In MyHendrix
   Print "Outer Value="; v, "Key="; MyHendrix.key
   If v = MyHendrix.count / 2 Then
     For Each vv As Variant In MyHendrix
         Print "   Inner Value="; vv, "Key="; MyHendrix.key
     Next
      Print "   Outer Value="; v, "Key="; MyHendrix.key
   Endif
Next
Print "end"
Quit 0
Catch
Error "Script Error >";; error.text & "\n" & error.where
End

Class SetKey
  Public Key As Integer
  Public index As Integer
End Class

Class Mynext
'Private $key As Variant
Private hendrix As New Variant[]
Property Read key As Variant
Property Read Count As Integer

Private Function key_read() As Variant
  Return enum.index.key                     ' doing this outside the actual _next causes error
End

Public Sub _new(count As Integer)
  For i As Integer = 0 To count - 1
     hendrix.push(i)
  Next
End

Private Function count_read() As Integer
  Return hendrix.count
End

Public Sub _next() As Variant
  Dim DynKey As SetKey
  If hendrix.count = 0 Then
      enum.stop
      Return
  Endif
  If IsNull(enum.index) Then
      DynKey = New SetKey
      enum.index = DynKey
      enum.index.key = 1
      enum.index.index = 0
      Return hendrix[enum.index.index]
  Else
      Inc enum.index.index
      If enum.index.index > hendrix.max Then
         enum.stop
         Return
      Else
         enum.index.key += 1
      Endif
  Endif
  
  Return hendrix[enum.index.index]
End
End Class

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--
~~~~ Brian

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