CODEWITHSUNDEEP
javaipsubnet
xZise
xZise

Reputation: 2379

Test with Java if two IPs are in the same network

How can I test if two IPs are in the same network according to the subnet mask?

For example I have the IPs 1.2.3.4 and 1.2.4.3: Both are in the same network if the mask is 255.0.0.0 or 255.255.0.0 or even 255.255.248.0 but not if the mask is 255.255.255.0..

Upvotes: 9

Views: 8021

Answers (3)

SkateScout
SkateScout

Reputation: 870

this solution will work with IPv4/IPv6 also.

static boolean sameNetwork(final byte[] x, final byte[] y, final int mask) {
    if(x == y) return true;
    if(x == null || y == null) return false;
    if(x.length != y.length) return false;
    final int bits  = mask &   7;
    final int bytes = mask >>> 3;
    for(int i=0;i<bytes;i++)  if(x[i] != y[i]) return false;
    final int shift = 8 - bits;
    if(bits != 0 && x[bytes]>>>shift != y[bytes]>>>shift) return false;
    return true;
}
static boolean sameNetwork(final InetAddress a, final InetAddress b, final int mask) {
    return sameNetwork(a.getAddress(), b.getAddress(), mask);
}

Upvotes: 0

&#211;scar L&#243;pez
&#211;scar L&#243;pez

Reputation: 236112

Try this method:

public static boolean sameNetwork(String ip1, String ip2, String mask) 
throws Exception {

    byte[] a1 = InetAddress.getByName(ip1).getAddress();
    byte[] a2 = InetAddress.getByName(ip2).getAddress();
    byte[] m = InetAddress.getByName(mask).getAddress();

    for (int i = 0; i < a1.length; i++)
        if ((a1[i] & m[i]) != (a2[i] & m[i]))
            return false;

    return true;

}

And use it like this:

sameNetwork("1.2.3.4", "1.2.4.3", "255.255.255.0")
> false

EDIT :

If you already have the IPs as InetAddress objects:

public static boolean sameNetwork(InetAddress ip1, InetAddress ip2, String mask) 
throws Exception {

    byte[] a1 = ip1.getAddress();
    byte[] a2 = ip2.getAddress();
    byte[] m = InetAddress.getByName(mask).getAddress();

    for (int i = 0; i < a1.length; i++)
        if ((a1[i] & m[i]) != (a2[i] & m[i]))
            return false;

    return true;

}

Upvotes: 15

Voo
Voo

Reputation: 30235

Simple enough: mask & ip1 == mask & ip2 - you have to interpret the IPs all as a single number for that but that should be obvious.

Upvotes: 5

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