VB.NET控制台程序：
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Module module3
Sub Main()
Dim i As Integer
Dim num As Integer = -35
Dim R As Integer = 8
Dim arr(33) As Integer
Dim len As Integer
arr = TransToR(num, R)
len = arr(0)
Console.Write(num " 转换成 " R " 进制后：" vbTab)
If (arr(1) = 1) Then
Console.Write("-")
End If
For i = len + 1 To 2 Step -1
Console.Write(arr(i))
Next
Console.WriteLine()
arr = TransToSourceCode(num)
Console.Write(num " 转换成32位二进制源码后：" vbTab)
Console.Write(arr(1))
For i = 32 To 2 Step -1
Console.Write(arr(i))
Next
Console.WriteLine()
arr = TransToReverseCode(num)
Console.Write(num " 转换成32位二进制反码后：" vbTab)
Console.Write(arr(1))
For i = 32 To 2 Step -1
Console.Write(arr(i))
Next
Console.WriteLine()
arr = TransToCompensatoryCode(num)
Console.Write(num " 转换成32位二进制补码后：" vbTab)
Console.Write(arr(1))
For i = 32 To 2 Step -1
Console.Write(arr(i))
Next
Console.WriteLine()
End Sub
'将十进制数num转换为R(2=R=9)进制数，结果存在arr数组中
'arr(0)存储长度, arr(1)存储符号位：1为负，0为正
Function TransToR(ByVal num As Integer, ByVal R As Integer) As Integer()
Dim arr(33) As Integer
arr(0) = 0 '初始长度
'设置符号位
If (num 0) Then
arr(1) = 1
Else
arr(1) = 0
End If
num = Math.Abs(num)
While num 0
arr(0) = arr(0) + 1
arr(arr(0) + 1) = num Mod R
num \= R
End While
Return arr
End Function
'将十进制数num转换为二进制原码
Function TransToSourceCode(ByVal num As Integer) As Integer()
Dim arr(33) As Integer
Dim i As Integer
arr = TransToR(num, 2)
'高位补零
For i = arr(0) + 2 To 32
arr(i) = 0
Next
arr(0) = 32
Return arr
End Function
'将十进制数num转换为二进制反码
Function TransToReverseCode(ByVal num As Integer) As Integer()
Dim arr(33) As Integer
Dim i As Integer
arr = TransToSourceCode(num)
If num = 0 Then
Return arr
End If
'除符号位外各位取反
For i = 2 To 33
arr(i) = 1 - arr(i)
Next
Return arr
End Function
'将十进制数num转换为二进制补码
Function TransToCompensatoryCode(ByVal num As Integer) As Integer()
Dim arr(33) As Integer
Dim i As Integer
Dim c As Integer = 0 '进位
arr = TransToReverseCode(num)
If num = 0 Then
Return arr
End If
'末尾加1
i = 2
arr(i) += 1
Do While True
arr(i) = c + arr(i)
c = arr(i) \ 2
arr(i) = arr(i) Mod 2
i += 1
If c 1 Then
Exit Do
End If
Loop
Return arr
End Function
End Module
Not一个数值实际上是对这个数值进行“非”运算，即转换为2进制，0变为1，1变为0，你可以这样试：
For i = 1 To 10
s = i
debug.print not (i Mod 5 )
Next
这样可以看到not (i Mod 5 )这个表达式究竟等于多少。
按照你的想法，应该这样做结果才是对的：
If Not (i Mod 5 0) Then
Private Sub Command1_Click()
x = Val(Text1.Text)
y = Val(Text2.Text)
Max = IIf(x y, x, y)
Print "Max="; Max
End Sub
