IPAddress: Facing problem with IPAddressString contains(IPAddressString other) method

Hello @seancfoley, I am using contains method to check whether one IPAddressString contains other,

Examples, 192.168.1.0/24 contains 192.168.1.0/12.(true) 192.168.1.0/24 contains 192.168.2.0/24.(false) 192.168.1.1-10 contains 192.168.1.5-7.(true) 192.168.1.5-7 contains 192.168.1.1-10.(false) 192.168.. contains 192.168.1..(true) 192.168.1. contains 192.168...(false) 192.168.1.0/24 contains 192.168.1.1-10.(true) 192.168.1.0/24 contains 192.168.2.1-10.(false) 192.168.1.* contains 192.168.1.0/24.(true)

All the above cases are working fine,but some of the below cases are failing,

• Case-1

IIPAddressString ipAddressStringSubnetError1 = new IPAddressString( "10.162.155.1-51" );
        IPAddressString ipAddressStringSubnetError2 = new IPAddressString( "10.162.155.1-255" );
        System.out.println( String.format( "%s contains %s %s", ipAddressStringSubnetError2,
                ipAddressStringSubnetError1, ipAddressStringSubnetError2.contains( ipAddressStringSubnetError1 ) ) );

Output 10.162.155.1-255 contains 10.162.155.1-51 false

It should return true but it’s returning false.But if ipAddressStringSubnetError1 range is 10.162.155.2-51 then it returns true.The problem only occurs when both ranges start with same address(10.162.155.1).

NOTE: Same problem is there for IPV6 also.

• Case-2

IPAddressString ipAddressStringSubnetError3 =
                new IPAddressString( "2001:0db8:85a3:0000:0000:8a2e:0370:7334/120" );
        IPAddressString ipAddressStringSubnetError4 =
                new IPAddressString( "2001:0db8:85a3:0000:0000:8a2e:0370:7334/128" );
        System.out.println( String.format( "%s contains %s %s", ipAddressStringSubnetError3,
                ipAddressStringSubnetError4, ipAddressStringSubnetError3.contains( ipAddressStringSubnetError4 ) ) );

        System.out.println( String.format( "%s contains %s %s", ipAddressStringSubnetError4,
                ipAddressStringSubnetError3, ipAddressStringSubnetError4.contains( ipAddressStringSubnetError3 ) ) );

Output

2001:0db8:85a3:0000:0000:8a2e:0370:7334/120 contains 2001:0db8:85a3:0000:0000:8a2e:0370:7334/128 true

2001:0db8:85a3:0000:0000:8a2e:0370:7334/128 contains 2001:0db8:85a3:0000:0000:8a2e:0370:7334/120 true

Subnet Calculator

IP Address:	2001:0db8:85a3:0000:0000:8a2e:0370:7334/120
Full IP Address:	2001:0db8:85a3:0000:0000:8a2e:0370:7334
Total IP Addresses:	256
Network:	2001:0db8:85a3:0000:0000:8a2e:0370:7300
IP Range:	2001:0db8:85a3:0000:0000:8a2e:0370:7300 - 2001:0db8:85a3:0000:0000:8a2e:0370:73ff


IP Address:	2001:0db8:85a3:0000:0000:8a2e:0370:7334/128
Full IP Address:	2001:0db8:85a3:0000:0000:8a2e:0370:7334
Total IP Addresses:	1
Network:	2001:0db8:85a3:0000:0000:8a2e:0370:7334
IP Range:	2001:0db8:85a3:0000:0000:8a2e:0370:7334 - 2001:0db8:85a3:0000:0000:8a2e:0370:7334

In the case2 both are returning true, but 2001:0db8:85a3:0000:0000:8a2e:0370:7334/128 contains 2001:0db8:85a3:0000:0000:8a2e:0370:7334/120 should return false.

Note: Same problem is there with IPV4 address also when address not end with zero.

-Case-3

IPAddressString ipAddressStringSubnetError5 = new IPAddressString( "192.13.1.0/25" );
        IPAddressString ipAddressStringSubnetError6 = new IPAddressString( "192.13.1.1-255" );
        System.out.println( String.format( "%s contains %s %s", ipAddressStringSubnetError5,
                ipAddressStringSubnetError6, ipAddressStringSubnetError5.contains( ipAddressStringSubnetError6 ) ) );
        System.out.println( String.format( "%s contains %s %s", ipAddressStringSubnetError6,
                ipAddressStringSubnetError5, ipAddressStringSubnetError6.contains( ipAddressStringSubnetError5 ) ) );

Output

192.13.1.0/25 contains 192.13.1.1-255 false
192.13.1.1-255 contains 192.13.1.0/25 false

In case3 i think 192.13.1.1-255 contains 192.13.1.0/25 should return true.

Thanks, Vamsi.