在Java中对128位进行按位运算

Question

I have binary representation of two IPv6 addresses

For example: First String is binary representation of

'2001:4E8:0:4000:0:0:0:0'
'00100000000000010000010011101000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000'

Second string binary representation

 '0:0:0:0:ffff:ffff:ffff:ffff'
 '00000000000000000000000000000000000000000000000000000000000000001111111111111111111111111111111111111111111111111111111111111111'

Now I want to do a biwise 'AND' operation on the IPv6 Address and its mask. What would be a good way to achieve this in java?

P:S: Integer.parseInt supports only 32 bit operations

Answer 1

You can use BigInteger 's and() :

BigInteger first = new BigInteger("00100000000000010000010011101000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000",2);
BigInteger second = new BigInteger("00000000000000000000000000000000000000000000000000000000000000001111111111111111111111111111111111111111111111111111111111111111",2);
BigInteger and = first.and(second);

Answer 2

You can use BitSet for this. But for this you will need to convert String to array of bytes

public static byte[] toByteArray(String bytes){
    byte[] bb = new byte[bytes.length()/8];
    byte m1 = (byte) (1 << 7);
    for(int i = 0, j = 0; i < bb.length; i++, j=i*8){
        byte b = bytes.charAt(j) == '1' ? m1 : 0;
        b |= Byte.valueOf(bytes.substring(j + 1, j + 8), 2);
        System.out.println(Integer.toBinaryString(b));
        bb[i] = b;
    }
    return bb;
}

Now you can use it in BitSet

BitSet bs1 = BitSet.valueOf(toByteArray("00000000000000000000000000000000000000000000000000000000000000001111111111111111111111111111111111111111111111111111111111111111"));
BitSet bs2 = BitSet.valueOf(toByteArray("00100000000000010000010011101000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000"));
BitSet bs3 = bs1.and(bs2);

Answer 3

I'm assuming that you started out with the IPv6 address and netmask in standard IPv6 notation ... not in a binary string representation consisting of '0' and '1' characters.

If you use InetAddress.getByName(String) on an IPv6 address literal, you will get an Inet6Address object. If you call getAddress() on this, you will get the raw address as a byte[] . You can do the same to get a byte[] from a netmask.

Then you can loop over the respective byte arrays and use bitwise operations on them.