VB Forum / General / July 2004

Not sure if this will help or not.. But here goes some air code...

Private Type tMyUDT
   Avalible As Boolean 'Is this position Availble?
   Name As String '[Last, First (if any)]
   Age As String 'Somewhere between Birth and Death
   Sex As String '(M)ale, (F)emale, (O)ther
   Parents As Long 'Number of legal parents
   'etc...
   'etc...
End Type

Private taMyUDT() As tMyUDT

Private Sub GetInputFromWhereEver()
Dim element As tMyUDT
Dim lRtn As Long

   With element
       .Name = Text1(0).Text
       .Age = Text1(1).Text
       .Sex = Text1(2).Text
       .Parents = Text1(3).Text
       'etc...
       'etc...
   End With
   'See if On File
   lRtn = FindIndexOf(element)
   If lRtn = -1 Then 'not on file
       'Add it taMyUDT() here
       'Find First element.Availble = True
       'Add Info and set .Avaible = False
   Else
       'Duplicate Entry at Index of lRtn
       'Do what you gota do here...
   End If
   
End Sub

Private Function FindIndexOf(element As tMyUDT) As Long
Dim lLp As Long
   
   For lLp = LBound(taMyUDT) To UBound(taMyUDT)
       If Not taMyUDT(lLp).Avalible Then
           'check to see if a match occures
           With element
               If .Name = taMyUDT(lLp).Name And _
                   .Age = taMyUDT(lLp).Age And _
                   .Sex = taMyUDT(lLp).Sex And _
                   .Parents = taMyUDT(lLp).Parents Then
                   'Found a match
                   Exit For
               End If
           End With
       End If
   Next lLp
   If lLp <= UBound(taMyUDT) Then
       FindIndexOf = lLp
   Else
       FindIndexOf = -1
   End If
End Function

Have a good day...

Don

I think the answer is Is

Something, conceptually, like this:

 for each tempMy in MyArray
     if tempMy Is element then

    Bob

To check whether one UDT is actually the /same/ as another you can use
the undocumented VarPtr function that returns the memory address of a
variable

If VarPtr( UDT1 ) = VarPtr( UDT2 ) Then

I can't think of a case where I have ever needed anything like this,
but you obviously have found one.