在Java中，我将如何获取输入的IP地址并将其转换为特定位

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.

Answer 1

String s = Integer.toBinaryString(first);

Answer 2

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
}

Answer 3

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
}