我试图将四位数转换为IP地址。例如:
0001 ---> * .192.1.01
0011 ---> * .192.11.01
0111 ---> * .192.111.01
1111 ---> * .196.87.01
3458 ---> * .205.130.01
我认为子网掩码是255.255.192.0。
我非常感谢有关在vb.net中执行此操作的最佳方法的任何建议。
其他信息:
这是一个简单的pinger程序,用户输入一个四位数字(他们想要ping的物理站点的ID).IP寻址方案很简单,第二个和第二个。第3个八位字节用作站点编号,第四个八位字节用作站点上的设备。我没有设计这个方案,因此我不确定如何让vb.net理解它。
我尝试过的事情:
我考虑过以下方式,这是非常粗糙的。然而,这只能达到* .192.255.01,因为我不知道如果在八位字节3中超过255之后如何分割两个八位字节之间的数字。
Private Sub btnStartPing(sender As Object, e As EventArgs) Handles btnStartPing.Click
Dim Octet1 As Integer = *
Dim Octet2 As Integer = 192
Dim Octet3 As Integer = txtSiteID.text
Dim Octet4 As Integer = 01
Dim CompleteIP As String = ""
CompletIP = Octet1 & "." & Octet2 & "." & Octet3 & "." & Octet4
'PING CompleteIP
end sub
解决方案:
Dim var1 As Integer = Fix(192 + (NumericUpDown1.Value / 256))
Dim var2 As Integer = Fix((NumericUpDown1.Value Mod 256))
MsgBox("Your IP address is: " & "10." & var1 & "." & var2 & "." & "200")
End Sub
答案 0 :(得分:0)
您需要将数字分成两部分。获取前半部分并将它们添加到192,然后直接使用后半部分。
Dim number As Integer
number = 1
Console.WriteLine("*.{0}.{1}.01", 192 + ((number And &HFF00) >> 8), number And &HFF) ' *.192.1.01
number = 11
Console.WriteLine("*.{0}.{1}.01", 192 + ((number And &HFF00) >> 8), number And &HFF) ' *.192.11.01
number = 111
Console.WriteLine("*.{0}.{1}.01", 192 + ((number And &HFF00) >> 8), number And &HFF) ' *.192.111.01
number = 1111
Console.WriteLine("*.{0}.{1}.01", 192 + ((number And &HFF00) >> 8), number And &HFF) ' *.196.87.01
number = 3458
Console.WriteLine("*.{0}.{1}.01", 192 + ((number And &HFF00) >> 8), number And &HFF) ' *.205.130.01
取数字并将其拆分为两个
3458 = 0x0D82
0X0D
为0x82
然后将192添加到第一部分
0x0D + 192 = 205
0x82 = 130
答案 1 :(得分:0)
这应该是它
Public Function StartPing(txtSiteId As String) As String
Dim SiteId As Integer = Integer.Parse(txtSiteId)
'backslash performs integer division (no fractionary part)
'will throw an error when SiteId results in values greater than 255
'type Byte allows only values from 0 to 255
Dim Octet2 As Byte = 192 + (SiteId \ 256)
'Mod gets rest of division
Dim Octet3 As Byte = SiteId Mod 256
Return String.Format("*.{0}.{1}.01", Octet2, Octet3)
End Function