Inc() and Dec() functions

[Eric_Williams]

Actually, this is easy to do in Xojo with exactly the syntax you’re looking for.

Language Enhancements

You can’t name a method using a character like “+”. Best you can do is: Module IntegerExtensions Public Sub Increment(byref extends i as integer) i=i+1 End Sub Public Sub Add(byref extends i as integer, value as integer) i=i+value End Sub End i.Increment i.Add(3)

[Frank_Bicknell]

I hope you have gotten your answer too.

[Arnaud_N]

@Frank_Bicknell

Function INC(byref value as integer) As Integer
  value = value + 1
  return value
End Function

Here, you’re doing the work twice. Using byref, there’s no need to also return the value.

In other words, you can simplify to two versions with your code above.

Version 1 (using a function):

Function INC(value as integer) As Integer
  return value+1
End Function

Called like this:
Value=INC(value)
(note: in this version, the value isn’t passed as ByRef, so it’s not modifiable from inside the function, hence why I’m straight returning value+1 instead of having the “value=value+1” step)

Version 2 (using ByRef):

Sub INC(byref value as integer)
  value=value+1
End Sub

Called like this:
INC(value)

The main difference, and is the purpose of ByRef, is that in version 2, you can modify “value” directly, without needing to return anything. The variable is modified “in place”.
As it looks obvious, version 2 is cleaner.

[Tim_Hare]

@Arnaud_N

Here, you’re doing the work twice.

Yes, because he is setting 2 separate variables at the same time.

[Frank_Bicknell]

Yes I have it sorted out. What I was doing is assigning the value to a pi-dog dataview cell. Ex. cell = Inc(RowNo). Now I am just calling Inc(RowNo) the using cell = RowNo. Thanks.

[Antonio_Rinaldi]

you can define both, with return if you need the result assigned to something else than the origin otherwise without return if you need only to increment (or whatever)

[Arnaud_N]

Indeed, but both ways in the same method is rarely useful.