Solved

Simple VB Question

Posted on 2003-10-29
3
164 Views
Last Modified: 2010-05-01
I am not a VB expert and I am trying to make a simple function that will count the number of decimal places there is in a number.  This is what I've done:

Public Function countDecimalPlaces(number As Double) As Integer
Do While number > Math.Round(number)
number = number * 10
countDecimalPlaces = countDecimalPlaces + 1
Loop
End Function

The problem is that it only works for up to 5 digits because in the comparison line for the loop (Do While number > Math.Round(number)), the Math.Round(number) part makes an integer, thus the the 5 digits the function will work until.  I have tried a few other things for the right side of the inequation (ie: multiplying and dividing by powers of 10, using Int(number), etc.)  but got the same results.

I can't seem to cast the right side of the inequation into a double properly.

Your help is much appreciated.

0
Comment
Question by:djschick
3 Comments
 
LVL 3

Accepted Solution

by:
kenspencer earned 50 total points
ID: 9645271
Hi,
Why not something like this:

Public Function countDecimalPlaces(number as Double) As Integer
Dim sNum As String
dim x        As Integer

    sNum = Cstr(number)
    x = Instr(sNum,".")
    If x > 0 then x = Len(sNum) - x
    countDecimalPlaces = x
Exit Function

Ken
0
 
LVL 17

Expert Comment

by:zzzzzooc
ID: 9645544
I came up with the top method after realizing Ken basically did the bottom one. Darn skimming. :P


Private Sub Form_Load()
    MsgBox CountDecimal(1.12345678)
End Sub
Private Function CountDecimal(ByVal Number As Double) As Integer
    CountDecimal = Len(CStr(Number)) - Round(Number, 0) - 1
End Function



or...



Private Sub Form_Load()
    MsgBox CountDecimal(1.123456789)
End Sub
Private Function CountDecimal(ByVal Number As Double) As Integer
    Dim iPos As Integer
    iPos = InStr(1, Number, ".")
    If iPos > 0 Then
        CountDecimal = Len(CStr(Number)) - iPos
    Else
        CountDecimal = 0
    End If
End Function
0
 

Expert Comment

by:clangl
ID: 9649546
Public Function NumberOfCharInString(strSearchIn as string, strSerarchFor as string) as integer
dim intCnt as Integer, intSavePlace as Intger
intCnt=0
intSavePlace=1
Do while intSavePlace>0
  intSavePlace=  Instr(intsavePlace, strSeachIn, strSearchFor)
  If intSacePlace>0 then
     intCnt=intCnt+1
  end if
loop
NumberOfCharInString=intCnt
end function

Then you can count what ever you want in a string
by calling NumberOfCharInString ("LOOK.FOR.DECIMALS",".")
0

Featured Post

Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

The debugging module of the VB 6 IDE can be accessed by way of the Debug menu item. That menu item can normally be found in the IDE's main menu line as shown in this picture.   There is also a companion Debug Toolbar that looks like the followin…
If you need to start windows update installation remotely or as a scheduled task you will find this very helpful.
As developers, we are not limited to the functions provided by the VBA language. In addition, we can call the functions that are part of the Windows operating system. These functions are part of the Windows API (Application Programming Interface). U…
Show developers how to use a criteria form to limit the data that appears on an Access report. It is a common requirement that users can specify the criteria for a report at runtime. The easiest way to accomplish this is using a criteria form that a…

863 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

20 Experts available now in Live!

Get 1:1 Help Now