Visual Basic中的循環位移

Question

我已經編寫了一個解決方案，我相信它可以在Visual Basic中進行循環位移。不過，我對這門語言很陌生，我不能100％確定這是有效的或功能性的。有沒有更好的方法來做到這一點？

如果您好奇，我試圖執行ARIA cipher，我需要這個功能來做到這一點。

Private Function CircularRotationLeft(ByVal bytes As Byte(), ByVal times As Integer) As Byte() 
    Dim carry As Boolean = False 
    If times < 0 Then 
     Return Nothing 
    End If 
    While times > bytes.Length * 8 
     times -= bytes.Length * 8 
    End While 
    If times = 0 Then 
     Return bytes 
    End If 
    Array.Reverse(bytes) 
    For index As Integer = 1 To times 
     For Each bits As Byte In bytes 
      If bits > 127 Then 
       bits -= 128 
       bits *= 2 
       If carry Then 
        bits += 1 
       End If 
       carry = True 
      Else 
       bits *= 2 
       If carry Then 
        bits += 1 
       End If 
       carry = False 
      End If 
     Next 
     If carry Then 
      bytes(0) += 1 
     End If 
    Next 
    Array.Reverse(bytes) 
    Return bytes 
End Function 

Private Function CircularRotationRight(ByVal bytes As Byte(), ByVal times As Integer) As Byte() 
    Dim carry As Boolean = False 
    If times < 0 Then 
     Return Nothing 
    End If 
    While times > bytes.Length * 8 
     times -= bytes.Length * 8 
    End While 
    If times = 0 Then 
     Return bytes 
    End If 
    Array.Reverse(bytes) 
    For index As Integer = 1 To times 
     For Each bits As Byte In bytes 
      If bits Mod 2 = 0 Then 
       bits /= 2 
       If carry Then 
        bits += 128 
       End If 
       carry = False 
      Else 
       bits /= 2 
       If carry Then 
        bits += 128 
       End If 
       carry = True 
      End If 
     Next 
     If carry Then 
      bytes(0) += 128 
     End If 
    Next 
    Array.Reverse(bytes) 
    Return bytes 
End Function 

Answer 1

旋轉在Visual Basic .NET中的32位有符號整數的一種方式，不應該是很難適應你的需求：

Public Function RotateCircularLeft(n As Int32, nBits As Byte) As Int32 
    Return (n << nBits) Or ((n >> (32 - nBits)) And (Not (-1 << nBits))) 
End Function 


Public Function RotateCircularRight(n As Int32, nBits As Byte) As Int32 
    Return (n << (32 - nBits)) Or ((n >> nBits) And (Not (-1 << (32 - nBits)))) 
End Function 

還要注意，調用函數RotateCircularRight(x, n)等同於調用RotateCircularLeft(x, 32-n)，因此您可以實時刪除其中一項功能。

我還沒有基準哪個是更快的方法。