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[WinForm] C# specifies the IP segment to generate the IP address

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Posted on 12/14/2015 3:01:40 PM | | | |


  1. private void button1_Click(object sender, EventArgs e)  
  2.         {  
  3.             string StartIp = "";  
  4.             string EndIp = "";  
  5.             StartIp = Startinput.Text;  
  6.             EndIp = Endinput.Text;  
  7.             uint iStartip = ipTint(StartIp);  
  8.             uint iEndIp = ipTint(EndIp);  
  9.             //string ip_result="";  
  10.             StringBuilder ip_result=new StringBuilder();  
  11.             if (iEndIp >= iStartip)  
  12.             {  
  13.                 for (uint ip = iStartip; ip <= iEndIp; ip++)  
  14.                 {  
  15.                     ip_result.Append(intTip(ip)).Append("\r\n");  
  16.                     //ip_result = ip_result + intTip(ip)+"\r\n";  
  17.                 }  
  18.                 resultTextBox.Text = ip_result.ToString();   //RichTextBox  
  19.             }  
  20.             else  
  21.             {  
  22.                 MessageBox.Show("天啊，起始ip居然比终止ip还大");  
  23.             }  
  24.         }  
  25.         public static uint ipTint(string ipStr)  
  26.         {  
  27.             string[] ip = ipStr.Split('.');  
  28.             uint ipcode = 0xFFFFFF00 | byte.Parse(ip[3]);  
  29.             ipcode = ipcode & 0xFFFF00FF | (uint.Parse(ip[2]) << 0x8);  
  30.             ipcode = ipcode & 0xFF00FFFF | (uint.Parse(ip[1]) << 0xF);  
  31.             ipcode = ipcode & 0x00FFFFFF | (uint.Parse(ip[0]) << 0x18);  
  32.             return ipcode;  
  33.         }  
  34.         public static string intTip(uint ipcode)  
  35.         {  
  36.             byte a = (byte)((ipcode & 0xFF000000) >> 0x18);  
  37.             byte b = (byte)((ipcode & 0x00FF0000) >> 0xF);  
  38.             byte c = (byte)((ipcode & 0x0000FF00) >> 0x8);  
  39.             byte d = (byte)(ipcode & 0x000000FF);  
  40.             string ipStr = string.Format("{0}.{1}.{2}.{3}", a, b, c, d);  
  41.             return ipStr;  
  42.         }  
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Posted on 12/14/2015 3:54:44 PM |
 Landlord| Posted on 12/31/2015 4:25:19 PM |
If there is an error in the generated IP address above, use the following method



Posted on 1/2/2016 3:11:56 PM |
Learned, thank you for sharing
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