BUG: Omitting Year for DateValue May Give Unexpected Results

ID: Q84115

1.00 2.00 3.00 WINDOWS kbprg kbbuglist

The information in this article applies to:

- Standard and Professional Editions of Microsoft Visual Basic for

  Windows, versions 2.0 and 3.0

- Microsoft Visual Basic programming system for Windows, version 1.0

SYMPTOMS

If you omit the year portion of the DateValue function argument, DateValue uses the current year from the computer's system date. However, if you also pass an invalid day for the month, DateValue interprets the month as the year and the day will default to 1. For example, 3/30 will be interpreted as 3/30/92, but 3/44 will not produce an error message, and will be interpreted as 3/1/44.

STATUS

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

MORE INFORMATION

The DateValue function returns a serial number that represents the date of the string argument. The date string can be in various forms. For example:

   3/30/92
   3/30/1992
   March 30, 1992
   Mar. 30, 1992
   30-Mar-1992
   30 March 92

The year portion of the string argument may be omitted, in which case the current year of the computer's system date is used. For example, 3/30 will cause DateValue to return the serial number that represents 3/30/92 (if 1992 is the year of the system date).

However, if the year is omitted and the day is not a valid day for that month of the current year, the month will be interpreted as the year and the day will default to 1. So 3/44 will be interpreted as 3/1/44.

Steps to Reproduce Problem

1. Start Visual Basic or from the File menu, choose New Project (ALT,

   F, N) if Visual Basic is already running. Form1 is created by default.

2. Add the following code to the Form_Load event procedure:

   Sub Form_Load ()
      Debug.Print "3/30 ="; DateValue("3/30")
      Debug.Print "3/30/92 ="; DateValue("3/30/92")
      Debug.Print
      Debug.Print "3/44 ="; DateValue("3/44")
      Debug.Print "3/1/44 ="; DateValue("3/1/44")
   End Sub

3. Press F5 to run the program.

Notice in the Immediate window that the serial numbers returned by the DateValue function for 3/30 and 3/30/92 are the same (assuming 1992 is the current year of the system date), and the serial numbers for 3/44 and 3/1/44 are the same. Also, no error message was produced.

Additional reference words: buglist1.00 buglist2.00 buglist3.00 1.00 2.00 3.00 KBCategory: kbprg kbbuglist KBSubcategory: APrgOther

Keywords          :  kbbuglist
Version           : 1.00 2.00 3.00
Platform          : WINDOWS

Last Reviewed: March 14, 1996