Artem's blog

Mainly .NET (C#, ASP.NET) and my projects

Archives for VB.NET

Investigating mod

Almost all languages have operators. Example of these are, “+” – addition, “-” – subtraction, “*” – multiplication, “/” division. There are also more advanced operators such as “Mod”, “Exponent”, and “And”. Let us investigate the mod.

I have seen that there are much misunderstanding about mod. Some claims that it performs a modulo operation, some claims that it is the reminder. Actually it is a reminder, and not a modulo operation.

In this investigation, I chosen Visual Basic as the default language. So, let us try to use mod operation as a modulo operation.

'default mod, "reminder"
MsgBox(3 - 4 Mod 6) ' result is -1
MsgBox(-1 + 4 Mod 6) ' result is 3

I would like to add that there is no problem to get the same answer back, i.e., 3 -4 mod 6 is -1, and then by adding 4 and again mod, will result 3. However, that is not the modulo. To display the modulo, use the function below:

   Function mod_r(ByVal x As Integer, ByVal y As Integer) As Integer
        'the "real" modulo operation
        Return x - y * Int(x / y)
    End Function

In arrays, tables, indexes, where the index should be positive, we can use the function above.

        MsgBox(mod_r(3 - 4, 6)) ' result is 5
        MsgBox(mod_r(5 + 4, 6)) ' result is 3

Sample of Bubble sort

The MsgBox is there to show how it actually goes through the integer array. You can remove it as well!

    ' A sample of Bubble sort
    ' Have fun!

        Dim i() As Integer = New Integer(4) {2, 5, 3, 1, 4}

        Dim t As Boolean = True
        While (t)
            t = False
            For j As Integer = 0 To 3

                If i(j) > i(j + 1) Then

                    Dim _vec As Integer = i(j)
                    i(j) = i(j + 1)
                    i(j + 1) = _vec
                    t = True

                    'this is only to check the value of i
                    Dim _result As String = ""
                    For y As Integer = 0 To 4
                        _result &= i(y) & ","
                End If
        End While

CRC32 Method [VB.NET]

Please take a look over the code I’ve founded at
The author of this code is Tim Hartwig. If you want to read more about the code/algorithm, please go here!

' This code have to be included in the header of file
Imports System.IO
Public Function GetCRC32(ByVal sFileName As String) As String
        Dim FS As FileStream = New FileStream(sFileName, FileMode.Open, FileAccess.Read, FileShare.Read, 8192)
        Dim CRC32Result As Integer = &HFFFFFFFF
        Dim Buffer(4096) As Byte
        Dim ReadSize As Integer = 4096
        Dim Count As Integer = FS.Read(Buffer, 0, ReadSize)
        Dim CRC32Table(256) As Integer
        Dim DWPolynomial As Integer = &HEDB88320
        Dim DWCRC As Integer
        Dim i As Integer, j As Integer, n As Integer

        'Create CRC32 Table
        For i = 0 To 255
            DWCRC = i
            For j = 8 To 1 Step -1
                If (DWCRC And 1) Then
                    DWCRC = ((DWCRC And &HFFFFFFFE)  2&) And &H7FFFFFFF
                    DWCRC = DWCRC Xor DWPolynomial
                    DWCRC = ((DWCRC And &HFFFFFFFE)  2&) And &H7FFFFFFF
                End If
            Next j
            CRC32Table(i) = DWCRC
        Next i

        'Calcualting CRC32 Hash
        Do While (Count > 0)
            For i = 0 To Count - 1
                n = (CRC32Result And &HFF) Xor Buffer(i)
                CRC32Result = ((CRC32Result And &HFFFFFF00)  &H100) And &HFFFFFF
                CRC32Result = CRC32Result Xor CRC32Table(n)
            Next i
            Count = FS.Read(Buffer, 0, ReadSize)
        Return Hex(Not (CRC32Result))
    Catch ex As Exception
        Return ""
    End Try
End Function