Option Explicit
Sub test()
Dim values(5) As Byte
values(0) = 25
values(1) = 50
values(2) = 100
values(3) = 150
values(4) = 200
values(5) = 255
Debug.Print get_Bitsequenz(values(), 2, 3, 15)
End Sub
Public Function get_Bitsequenz(ByRef values() As Byte, startByte As Integer, startBit As Byte, anzahlBits As Byte) As String
If startByte > UBound(values()) Then
MsgBox "'startByte' in Array nicht vorhanden.", vbExclamation
Exit Function
End If
Dim sValues() As String
Dim i As Integer
Dim s As String
ReDim sValues(UBound(values()))
For i = 0 To UBound(values())
sValues(i) = ByteToBit(values(i))
Next i
If Len(sValues(startByte)) < startBit Then
MsgBox "'startBit' ist größer als die Bitlänge.", vbExclamation
Exit Function
End If
i = startByte
Do While Len(s) < anzahlBits And i <= UBound(sValues)
If i = startByte Then
s = s & Mid(sValues(i), startBit + 1, Len(sValues(i)) - startBit)
Else
s = s & sValues(i)
End If
i = i + 1
Loop
If Len(s) < anzahlBits Then
MsgBox "Zu wenig Bits vorhanden.", vbExclamation
Exit Function
End If
get_Bitsequenz = Left(s, anzahlBits)
End Function
Private Function ByteToBit(ByVal v As Byte) As String
Dim s As String, e As Byte, i As Integer
e = v
Do While Not e = 0
s = s & CStr(e Mod 2)
e = CByte(Int(e / 2))
Loop
For i = 1 To 8 - Len(s)
s = s & "0"
Next i
ByteToBit = StrReverse(s)
End Function