Ip address format

Question

Hi Everyone,
One of methods in my COM object is
public void setIP(Variant ip)
with Variant data type supposes to be 4 byte int. I don't know how to convert a string like "192.167.1.108" to an int so I can pass in the above function. Would anybody please help me out!
Thanks in advance
Hung

Answer 1

In java, there is no such thing as an unsigned int. How to put this data into an int is going to depend on how he reads it out. Can you find this out? Does he strictly read each 8 bits as an unsigned byte? If this is the case, the previous examples (mine included) will have problems because signed data types use 2's complement for negative numbers (see http://forum.java.sun.com/thread.jsp?forum=54&thread=143155 ). I've modified the method for this case:
    static int ipToInt(String ip)
     StringTokenizer st = new StringTokenizer(ip, ".");
     int numParts = st.countTokens();
     if (numParts == 0)
          throw new IllegalArgumentException("Invalid IP address: empty argument");
     if (numParts > 4)
          throw new IllegalArgumentException("Invalid IP address: too many tokens");
     long ipLong = 0;
     for (int i = 1; i < numParts; i++)
     /* For all but the last token in an IP address, the nth token corresponds to the
         nth byte from the left (or 4 - n from the right)  */
          String token = st.nextToken();
          int tokenVal = Integer.parseInt(token);
          // Above will throw NumberFormatException if the token is not an integer
          int processedVal = tokenVal << (8 * (4 - i));
          // Moves the byte to the correct portion of the int
          ipLong = ipLong + processedVal;
     /* The right most token of an IP address is always added as its given value
        (never shifted) */
     String token = st.nextToken();
     int tokenVal = Integer.parseInt(token);
     // Above will throw NumberFormatException if the token is not an integer
     ipLong = ipLong + tokenVal;
     if (ipLong > Integer.MAX_VALUE)
     /* This will ensure the proper bits are set using 2's complement when
        converting back to an int */
          ipLong = Integer.MIN_VALUE + (ipLong - Integer.MAX_VALUE - 1);
     int ipInt = new Long(ipLong).intValue();
     return ipInt;
    }If you do the following, you can see each 8 bit section matches:
     String ipString = "192.167.1.108";
     int ipInt = ipToInt(ipString);
     System.out.println("IP:   \t" + ipString);
     System.out.println("ipInt:\t" + ipInt);
     System.out.println("      \t" + Integer.toBinaryString(ipInt)); The results are:
IP:     192.167.1.108
ipInt:  -1062796948
        11000000101001110000000101101100Note the value of this bitwise representation is negative.