Check valid IPv4 Address in Java

Question

I am using the sun.net.util.IPAddressUtil package to check whether the string contains a valid IPv4 and IPv6 address or not.

Code Snippet is:-

String ipv4addr="200";

    if(IPAddressUtil.isIPv4LiteralAddress(ipv4addr))
    {
        System.out.println("valid ipv4 address");
    }
    else
    {
        System.out.println("not valid");

    }

But for addresses such as 200 and 300 it is still saying it is a valid IPv4 address, which it isn't. When I used the same package and checked for IPV6 address using :-

String ipv6addr="200";

    if(IPAddressUtil.isIPv6LiteralAddress(ipv6addr))
    {
        System.out.println("valid ipv6 address");
    }
    else
    {
        System.out.println("not valid");

    }

I get the correct result. However, IPv4 does not seem to be working or may be I am using it incorrectly. Please guide me. I don't want to use regex for IPv4 validation...

Answer 1

There's a reason you're getting a "valid" result: 200 is a valid IPv4 address.

See, to the computer, an IPv4 address is just a 32-bit number. The dots are entirely for our convenience, because we humans suck at memorizing big precise numbers. But they don't have to be there; there are rules about how an address gets parsed depending on how many parts it has.

When an address consists of one number, it's considered a 32-bit number, and each byte is 8 bits of that number. If you were to parse "200" as an IP address, it would be equivalent to 0.0.0.200. Likewise, "2130706433" would be equivalent to 127.0.0.1.

There are also standards for when an address has two parts like 0.200 (first part is the first byte, and the second part is a 24-bit number representing the other 3 bytes), and even 0.0.200 (first two numbers are bytes, the last part is 16 bits and takes up the other 2 bytes). The "unusual" formats are leftovers from the days of IP address classes, but almost all software that has to parse addresses will understand them. (If you pop open your browser and go to http://1249739112^* or even http://74.125.33128^*, for example, Google's home page will come up.)

^{* See the comments for clickable links. Thanks, "link validator". :P}

See http://download.oracle.com/javase/6/docs/api/java/net/Inet4Address.html or http://www.perlmonks.org/?node_id=221512, or http://en.wikipedia.org/wiki/IPv4#Address_representations, for some more details.

Java understands these formats as well (as does .net, as well as any decent OS), and parses the address correctly whether it contains 1, 2, 3, or 4 parts.

If you want to check that a would-be address actually looks like "xxx.xxx.xxx.xxx", then you'll probably want to explicitly check that using a pattern, or using a validation library that considers 32-bit numbers as invalid addresses (even though they are valid). I wouldn't bother, though -- if you use the lookup functions provided, you can accept an address in any standard format and it will work.

(All this mess changes with IPv6; there's a much stricter format, and you can't just type in some 36-digit number and expect it to work. But the platform still knows how to parse an address, and you should trust it to do so.)

Answer 2

Check out Guava's InetAddresses class which contains static utility methods for working with IP addresses. (As I understand it uses the sun.net.util.IPAddressUtil class behind the scenes.)

System.out.println(InetAddresses.isInetAddress("400")); // false