how to perform xor operation correctly?

Question

i have a binary string and i would like to perform a xor operation consequentially on several bits of that string. my string is :

011001100011100000000011

i am trying to perform the calculation using the next line of code:

private String ParityCalc(String str){
    char[] cA = str.toCharArray();
    int[] D = new int[6];
    D[0] = D29^cA[0]^cA[1]^cA[2]^cA[4]^cA[5]^cA[9]^cA[10]^cA[11]^cA[12]^cA[13]^cA[16]^cA[17]^cA[19]^cA[22];
    D[1] = D30^cA[1]^cA[2]^cA[3]^cA[5]^cA[6]^cA[10]^cA[11]^cA[12]^cA[13]^cA[14]^cA[17]^cA[18]^cA[20]^cA[23];
    D[2] = D29^cA[0]^cA[2]^cA[3]^cA[4]^cA[6]^cA[7]^cA[11]^cA[12]^cA[13]^cA[14]^cA[15]^cA[18]^cA[19]^cA[21];
    D[3] = D30^cA[1]^cA[3]^cA[4]^cA[5]^cA[7]^cA[8]^cA[12]^cA[13]^cA[14]^cA[15]^cA[16]^cA[19]^cA[20]^cA[22];
    D[4] = D30^cA[0]^cA[2]^cA[4]^cA[5]^cA[6]^cA[8]^cA[9]^cA[13]^cA[14]^cA[15]^cA[16]^cA[17]^cA[20]^cA[21]^cA[23];
    D[5] = D29^cA[2]^cA[4]^cA[5]^cA[7]^cA[8]^cA[9]^cA[10]^cA[12]^cA[14]^cA[18]^cA[21]^cA[22]^cA[23];
    for (int i = 0; i < 6; i++){
        if (D[i] == 48){
            D[i] = 0;
        } else if (D[i] == 49){
            D[i] = 1;
        }
    }
    StringBuilder parity = new StringBuilder();
    parity.append(D[0]).append(D[1]).append(D[2]).append(D[3]).append(D[4]).append(D[5]);
    D29 = D[4];
    D30 = D[5];
    return parity.toString();
}

the result that i am getting for the final parity is: 100000. the correct result should be: 001001.

the D29 and D30 are parity bits carried on from previous calculations, both are integers.

what am i doing wrong and how can i fix it? i should probably do it as a bitwise operation but i cant seem to figure it out. any help would be appreciated.

Answer 1

That would be my approach:

private String ParityCalc(String str){
    int input = Integer.parseInt(str,2);
    int[] D = new int[6];
    D[0] = input & (int)0x4b3e37; // Mask for indices 0,1,2,4,5,9,10,11,12,13,16,17,19,22
    D[0] = (Integer.bitCount(D[0])&0x1)^D29; // Parity of masked input XOR D29

// D[1-5] accordingly

    StringBuilder parity = new StringBuilder();
    parity.append(D[0]).append(D[1]).append(D[2]).append(D[3]).append(D[4]).append(D[5]);
    D29 = D[4];
    D30 = D[5];
    return parity.toString();
}

Mask: 0,1,2,4,5,9,10,11,12,13,16,17,19,22

3 3         2         1
210987654321098765432109876543210 "Position"
000000000010010110011111000110111 BIN
    0   0   4   B   3   E   3   7 Hex (4 digits bin = 1 Hex)