vbObjectError Not Working as Expected

[Paul_Hossler]

Did you have any error handling specified (like in examples below)?


Post #2 at

https://www.vbforums.com/showthread....what-is-it-for

has good writeup. Old but still valid as far as I know

It seems the intent by MS was to not let things get confusing

Because Microsoft has reserved the error numbers from 1 to 2147221504 for their use and the reason why you add 2147221504=vbObjectError to what ever error number you want to raise is because things don't get confusing. Lets say you have X number of error traps within your code and you have numbered them from 1 to X without the addition of the vbObjectError constant. Then you die for some reason or are no longer working at the place you designed this code and your customer recieves error number 11 with your description but when your customer looks this error number up on MS's web site it will come back as a division by zero error.

Option Explicit


'https://docs.microsoft.com/en-us/office/vba/language/reference/user-interface-help/number-property-visual-basic-for-applications


Sub ForceError()
    Dim x As Long, y As Long
    Dim o As clsDivide
    
    Set o = New clsDivide
    
    x = Application.InputBox("Enter X as numerator")
    y = Application.InputBox("Enter Y as denominator")
    
    On Error GoTo ErrHandler
    
    MsgBox o.DivideXY(x, y)
    
    On Error GoTo 0




    Exit Sub


ErrHandler:

    ' My error
    If (Err.Number - vbObjectError) = 515 Then
        MsgBox (Err.Number - vbObjectError) & " -- " & Err.Description, vbCritical, Err.Source
     
    'another error  
    Else
        MsgBox Err.Number & " -- " & Err.Description, vbCritical, Err.Source
    End If



End Sub


Sub ForceError1()
    Dim x As Long, y As Long
    
    x = Application.InputBox("Enter X as numerator")
    y = Application.InputBox("Enter Y as denominator")
    
    On Error GoTo ErrHandler
    
    If y = 0 Then Err.Raise 515, "Force Error 1", "Can't divide by zero"
    
    On Error GoTo 0




    Exit Sub


ErrHandler:
    
    MsgBox (Err.Number) & " -- " & Err.Description, vbCritical, Err.Source


End Sub

Class clsDivideXY

Option Explicit


Private x As Long, y As Long


Function DivideXY(x As Long, y As Long) As Long


    If y = 0 Then Err.Raise vbObjectError + 515, "DivideXY", "Can't divide by zero"


    DivideXY = x / y


End Function

My thoughts, and you can experiment with the attachment, but I don't think it really makes a lot of difference