Java implementation of IP address and decimal number of mutual conversion _java

First look at the example
The code is as follows

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Classip
{
Privatestaticlongiptolong (Stringstrip)
Converts an IP address in the form 127.0.0.1 to a 10-binary integer, without any error handling
{
intj=0;
inti=0;
LONG[]IP=NEWLONG[4];
Intposition1=strip.indexof (".");
Intposition2=strip.indexof (".", position1+1);
Intposition3=strip.indexof (".", position2+1);
Ip[0]=long.parselong (strip.substring (0,position1));
Ip[1]=long.parselong (strip.substring (Position1+1,position2));
Ip[2]=long.parselong (strip.substring (Position2+1,position3));
Ip[3]=long.parselong (strip.substring (position3+1));
Return (IP[0]&LT;&LT;24) + (ip[1]<<16) + (ip[2]<<8) +IP[3];//IP1*256*256*256+IP2*256*256+IP3*256+IP4
}
Privatestaticstringlongtoip (LONGLONGIP)
Converts a 10-in-integer form to an IP address in 127.0.0.1 form, entering ping3396362403l at a command prompt
{
Stringbuffersb=newstringbuffer ("");
Sb.append (string.valueof (longip>>>24))/Direct 24-bit right shift
Sb.append (".");
Sb.append (string.valueof (LONGIP&AMP;0X00FFFFFF) >>>16);//will be 8 digits high 0, then 16 digits to the right
Sb.append (".");
Sb.append (string.valueof (LONGIP&AMP;0X0000FFFF) >>>8));
Sb.append (".");
Sb.append (string.valueof (LONGIP&AMP;0X000000FF));
Sb.append (".");
Returnsb.tostring ();
}
Publicstaticvoidmain (String[]args)
{
System.out.println ("Various representations of IP address: rn");
System.out.print ("32-bit binary form:");
System.out.println (long.tobinarystring (3396362403l));
System.out.print ("decimal form:");
System.out.println (Iptolong ("202.112.96.163"));
System.out.print ("ordinary form:");
System.out.println (Longtoip (3396362403l));
}
}

Run Result:
Various representations of IP addresses:
32-bit binary form: 11001010011100000110000010100011
Decimal form: 3396362403
General form: 202.112.96.163.
Output complete (time consuming 1 seconds)-Normal termination
Let's take another step apart to analyze
Knowledge Point: A binary number, bitwise left n bit, is to multiply the value of the number by 2 of the n-th side
Binary is one of the two right shifts
1. Convert IP address to integer
Principle: IP address can be considered as a 8-bit unsigned integer that is 0-255, split each segment into a binary form, and then convert the binary number to
An unsigned 32 is an integer.
Example: An IP address for 10.0.3.193
Number of binary digits corresponding to each segment
1000001010
000000000
300000011
19311000001
The combination is: 00001010000000000000001111000001, conversion to 10 is: 167773121, that is, the IP address after the conversion of the number is it.
The code is as follows

Copy Code code as follows:

publicclassip{
Publicstaticvoidmain (String[]args) {
System.out.print (Ip2int ("10.0.3.193"));
}
Publicstaticlongip2int (STRINGIP) {
String[]items=ip.split (".");
Returnlong.valueof (Items[0]) <<24
| Long.valueof (items[1]) <<16
| Long.valueof (items[2]) <<8
| Long.valueof (Items[3]);
}
}

2. Convert integer to IP address
Principle: Converts this integer to a 32-bit binary number. From left to right, every 8 bits are split, get 4 8-bit binary numbers, convert these binary numbers to integers and add ". "That's the IP address.
Example: 167773121
The binary representation is: 00001010000000000000001111000001
Split into four segments: 00001010,00001010,00000011,11000001, converted to integers, plus ". "I got 10.0.3.193.
The code is as follows

Copy Code code as follows:

publicclassip{
Publicstaticvoidmain (String[]args) {
System.out.print (Int2ip (167773121));
}
Publicstaticstringint2ip (longipint) {
Stringbuildersb=newstringbuilder ();
Sb.append (Ipint&0xff). Append (".");
Sb.append ((ipint>>8) &0xff). Append (".");
Sb.append ((ipint>>16) &0xff). Append (".");
Sb.append ((ipint>>24) &0xff);
Returnsb.tostring ();
}
}