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Break Binary String into Blocks

Posted on 2001-07-26
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Last Modified: 2008-03-03
VB6 Pro.......Win98 2nd Ed.

 I have a binary string i.e.
11010101111001
 Would like to break up into block of 4 i.e.
11-0101-0111-1001

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Question by:et1dkn
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11 Comments
 
LVL 7

Expert Comment

by:Z_Beeblebrox
ID: 6324503
Here you go.

Zaphod.


Dim intPosition as integer
Dim strNewBinary as string
intPosition = len(strBinary)
If len(strBinary) mod 4 <> 0 then
   strNewBinary = left(strBinary, len(strBinary) mod 4) & "-"
End if
For intPosition = len(strBinary) mod 4 to len(strBinary)
    strNewBinary = strNewBinary & mid(strBinary, intPosition,4) & "-"
Next
strBinary = left(strBinary,len(strBinary)-1)
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LVL 7

Expert Comment

by:Z_Beeblebrox
ID: 6324509
Oh, strBinary contains the original binary string, and this will fail if the string is empty, so you might want to add a check at the start.

Zaphod.
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LVL 7

Expert Comment

by:Z_Beeblebrox
ID: 6324519
Doh, small error:

Dim intPosition as integer
Dim strNewBinary as string
intPosition = len(strBinary)
If len(strBinary) mod 4 <> 0 then
  strNewBinary = left(strBinary, len(strBinary) mod 4) & "-"
End if
For intPosition = (len(strBinary) mod 4)+1 to len(strBinary)
   strNewBinary = strNewBinary & mid(strBinary, intPosition,4) & "-"
Next
strBinary = left(strBinary,len(strBinary)-1)
0
 

Author Comment

by:et1dkn
ID: 6324741
Z B..........
Your routine returns the original string...
that is no "-" inserted in string.  Believe you have
some conflict with strBinary and strNewBinary.
0
 
LVL 22

Expert Comment

by:rspahitz
ID: 6324979
I was going to suggest using a masked edit box with the mask = "####-####-####-####" but I can't figure out how to right-justify without putting in a lot of code, in which case, just do it by hand:

Dim binstring As String
Dim z As Integer
binstring = "11010101111001"
For z = Len(binstring) - 3 To 1 Step -4
  binstring = Left$(binstring, z - 1) & "-" & Mid$(binstring, z)
Next z
MsgBox binstring
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LVL 22

Expert Comment

by:rspahitz
ID: 6324988
I guess this is similar to the Z_Beeblebrox comment, but by working backwords, you don't have to worry about the expanding size of the result and can therefore keep it in the same variable, if that's what you want.
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LVL 7

Accepted Solution

by:
Z_Beeblebrox earned 100 total points
ID: 6325032
Dim intPosition As Integer
Dim strNewBinary As String

intPosition = Len(strBinary)
If Len(strBinary) Mod 4 <> 0 Then
 strNewBinary = Left(strBinary, Len(strBinary) Mod 4) & "-"
End If
For intPosition = (Len(strBinary) Mod 4) + 1 To Len(strBinary) Step 4
  strNewBinary = strNewBinary & Mid(strBinary, intPosition, 4) & "-"
Next
strNewBinary = Left(strNewBinary, Len(strNewBinary) - 1)

I made a couple mistakes, I guess this just isn't my day. Anyways, that should do it, I actually tested it this time.

Zaphod.
0
 
LVL 1

Expert Comment

by:ramani_gr
ID: 6325033
Try this.

    Dim intPosition As Integer
    Dim strNewBinary As String
    Dim strbinary  As String
   
    strbinary = "11010101111001"
    For intPosition = Len(strbinary) To 0 Step -4
       If intPosition - 3 > 0 Then
            strNewBinary = Mid(strbinary, intPosition - 3, 4) & IIf(strNewBinary = "", "", "-") & strNewBinary
        Else
            strNewBinary = Mid(strbinary, 1, intPosition) & IIf(strNewBinary = "", "", "-") & strNewBinary
        End If
    Next
    MsgBox strNewBinary
    '11-0101-0111-1001

this could slove ur problem.

0
 
LVL 27

Expert Comment

by:Ark
ID: 6325090
  s = "11010101111001"
   Debug.Print Format(s, "&&&&-&&&&-&&&&-&&&&")
Cheers
0
 
LVL 27

Expert Comment

by:Ark
ID: 6325104
PS. If you want to split output string:

s = "11010101111001"
sBin = Format(s, "&&&&-&&&&-&&&&-&&&&")
and then use Split function.

or another solution

Private Type FullString
   str As String * 16
End Type

Private Type PartString
   First As String * 4
   Second As String * 4
   Third As String * 4
   Fourth As String * 4
End Type

Private Sub Command1_Click()
   Dim fs As FullString
   Dim ps As PartString
   RSet fs.str = "11010101111001"
   LSet ps = fs
   Debug.Print ps.First, ps.Second, ps.Third, ps.Fourth
End Sub



0
 

Author Comment

by:et1dkn
ID: 6325401
Z B
Thanks much for the prompt reply.........
Dan
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