In Java, how would I take an inputed IP address and convert it into its specific bits

Question

Basically what I am needing is a way to take the input from the user of their ip address, then output the subsequent 8bits for each section as to which bits are on and off.

I currently have the user inputting their ip address in each section like so:

System.out.print("Please input the first set in your IP Address: ");
strHolder = kb.next();
first = Integer.parseInt(strHolder);

System.out.print("Please input the second set in your IP Address: ");
strHolder = kb.next();
second = Integer.parseInt(strHolder);

System.out.print("Please input the third set in your IP Address: ");
strHolder = kb.next();
third = Integer.parseInt(strHolder);

System.out.print("Please input the fourth set in your IP Address: ");
strHolder = kb.next();
fourth = Integer.parseInt(strHolder);

I'm not very experienced at Java. I've taken one class in college, and most of this I had to google how to do, but I've found nothing in regards to the bit process.

http://stackoverflow.com/questions/21856626/java-integer-to-binary-string (for how to make it look pretty) , http://stackoverflow.com/questions/2406432/converting-an-int-to-a-binary-string-representation-in-java?rq=1, http://stackoverflow.com/questions/23020633/method-for-string-input-to-binary-string, etc (for how to do the conversion) — user2864740, Jan 28 '15 at 19:27
possible duplicate of [Converting an int to a binary string representation in Java?](http://stackoverflow.com/questions/2406432/converting-an-int-to-a-binary-string-representation-in-java) — user2864740, Jan 28 '15 at 19:29

score 3 · Accepted Answer · answered Jan 28 '15 at 19:24

3

String s = Integer.toBinaryString(first);

answered Jan 28 '15 at 19:24

bhspencer

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I did this, but it won't put the complete bits unless the 128 bit is turned on. I tested it by doing 192.168.2.2 and it output: 11000000.10101000.10.10 Nevermind! I found out that I can do first_bit = String.format("%8s", Integer.toBinaryString(first)).replace(' ', '0'); Thank you for your help! – Ryan Jan 28 '15 at 21:09

score 0 · Answer 2 · edited Jan 28 '15 at 19:27

If the input will be like: 123.124.125.126 (one string), you can make String[] strListHolder = strHolder.split(".") and operate on that.

Integer[] holder = new Integer[strListHolder.count];
for(int i = 0; i < strListHolder.count; i++){
    holder[i] = Integer.parseInt(strListHolder[i]);
}

Is that what you wanted? @EDIT After that you can check n'th bit of each partusing:

if ((holder[i] & (1 << n)) != 0)
{
   // The bit was set
}

score 0 · Answer 3 · edited Jan 28 '15 at 19:34

The IP address(IPV4) is of the format xxx.xxx.xxx.xxx.

Suppose the input is a string format. 1st seprate the 4 parts using split function in the string class. The you can take the individual parts and & them with 255 to get the bits which are set and use the same information to find the bits which are not set.

String[] str = ip.split(".");
for (String each : str) {
    System.out.println(Integer.parseInt(each) & 255);  // this would print the set bits just convert it to binary
}

In Java, how would I take an inputed IP address and convert it into its specific bits

3 Answers3