BUG: vbKeySeparator Constant Does Not Work (188550)



The information in this article applies to:

  • Microsoft Visual Basic Learning Edition for Windows 5.0
  • Microsoft Visual Basic Learning Edition for Windows 6.0
  • Microsoft Visual Basic Professional Edition for Windows 5.0
  • Microsoft Visual Basic Professional Edition for Windows 6.0
  • Microsoft Visual Basic Enterprise Edition for Windows 5.0
  • Microsoft Visual Basic Enterprise Edition for Windows 6.0
  • Microsoft Visual Basic Standard Edition for Windows 4.0
  • Microsoft Visual Basic Professional Edition for Windows 4.0
  • Microsoft Visual Basic Enterprise Edition for Windows 4.0

This article was previously published under Q188550

SYMPTOMS

When using the constant vbKeySeparator to evaluate keystrokes in the KeyUp, KeyDown, or KeyPress event, the comparison always returns False.

CAUSE

The Visual Basic documentation lists a keycode constant of vbKeySeparator. The value of this constant is 108 (0x6C) and that value corresponds to the ENTER key on the Numeric Keypad. It also lists the constant vbKeyReturn that has the value 13 (0xD) and corresponds to the ENTER key on the Keyboard. In the KeyDown, KeyPress, or KeyUp events, the keycode returned is 13 regardless of which ENTER key is pressed. A test to see if the keycode is equal to 108 will always return False because 13 is obviously not equal to 108. The result is that there is no way to determine which ENTER key was depressed.

RESOLUTION

It is possible to differentiate between the two ENTER keys using the Win32 API PeekMessage. A bitwise comparison can then be made to determine which of the ENTER keys was pressed.

WORKAROUND

  1. Start a new Standard EXE project in Visual Basic. Form1 is created by default.
  2. Add the following code to Form1:
          Private Declare Function PeekMessage Lib "user32" Alias _
            "PeekMessageA" (lpMsg As MSG, ByVal hwnd As Long, _
            ByVal wMsgFilterMin As Long, ByVal wMsgFilterMax As Long, _
            ByVal wRemoveMsg As Long) As Long
    
          Private Type POINTAPI
             x As Long
             y As Long
          End Type
    
          Private Type MSG
             hwnd As Long
             message As Long
             wParam As Long
             lParam As Long
             time As Long
             pt As POINTAPI
          End Type
    
          Const PM_NOREMOVE = &H0
          Const WM_KEYDOWN = &H100
          Const WM_KEYUP = &H101
          Const VK_RETURN = &HD
    
          Private Sub Form_KeyDown(KeyCode As Integer, Shift As Integer)
          Dim MyMsg As MSG, RetVal As Long
    
          ' pass:
          '   MSG structure to receive message information
          '   my window handle
          '   low and high filter of 0, 0 to trap all messages
          '   PM_NOREMOVE to leave the keystroke in the message queue
          '   use PM_REMOVE (1) to remove it
          RetVal = PeekMessage(MyMsg, Me.hwnd, 0, 0, PM_NOREMOVE)
    
          ' now, per Q77550, you should look for a MSG.wParam of VK_RETURN
          ' if this was the keystroke, then test bit 24 of the lparam - if ON,
          ' then keypad was used, otherwise, keyboard was used
          If RetVal <> 0 Then
             If MyMsg.wParam = VK_RETURN Then
                If MyMsg.lParam And &H1000000 Then
                   MsgBox "Enter from Keypad pressed"
                Else
                  MsgBox "Enter from Keyboard pressed"
                End If
          End If
          Else
          MsgBox "No message waiting, or possible problems calling PeekMessage"
          End If
          End Sub
    						
  3. Run the project and press on both ENTER keys. You will get a message box correctly identifying which ENTER key was pressed.

STATUS

Microsoft has confirmed this to be a bug in the Microsoft products listed at the beginning of this article. We are researching this bug and will post new information here in the Microsoft Knowledge Base as it becomes available.

MORE INFORMATION

Steps to Reproduce Behavior

  1. Start a new Standard EXE project in Visual basic. Form1 is created by default.
  2. Add the following code to Form1.
          Private Sub Form_KeyDown(KeyCode As Integer, Shift As Integer)
          Select Case KeyCode
    
             Case vbKeyReturn
                MsgBox "KeyCode 13. Keyboard Enter Key Pressed."
    
             Case vbKeySeparator
                MsgBox "KeyCode 108. NumPad Enter Key Pressed."
    
          End Select
          End Sub
    						
  3. Run the project and press both ENTER keys. Note that the KeyCode is 13 regardless of which ENTER key is pressed.

REFERENCES

For a more detailed explanation of this bitwise comparison, please see the following article in the Microsoft Knowledge Base:

77550 Differentiating Between the Two ENTER Keys


Modification Type:MinorLast Reviewed:7/13/2004
Keywords:kbbug KB188550