Finding Block Check Character (BCC) from given HEX code

Question

I am communicating with a device via serial in Android. The result I am getting is:

01H 50H 30H 02H 28H 34H 45H 38H 39H 42H 42H 41H 43H 29H 03H 48H

where last hex code ie: 48H is Block Check Character(BCC).

And as per document it says that BCC is checksum byte from the second byte to the byte of the last but one.

I tired following method to get the BCC but it gives me 38H which does not matches with the device output.

byte[] inputByteArray = {0x50,0x30,0x02,0x28,0x34,0x45,0x38,0x39,0x42,0x42,0x41,0x43,0x29,0x03};
int sum = 0;
        for (byte b : inputByteArray) {
            sum = (sum + b) & 0xFF;
        }
        int compliment = (sum ^ 0xFF);
        int adding1 = compliment + 1;
        sum = (((sum ^ 0xFF) + 1) & 0xFF);

I am not able to figure it out that how it is calculated from the given hex output from the device. I need to find the logic since, the same way I need to calculate the other BCC values for my next request.

Any kind of help will really be appreciated.

Answer 1

First get the sum of all byte in the array

byte[] inputByteArray = 
  {0x50,0x30,0x02,0x28,0x34,0x45,0x38,0x39,0x42,0x42,0x41,0x43,0x29,0x03};

int sumByte = 0;
for (int x = 1; x < inputByteArray .length; x++) {
    sumByte += inputByteArray [x];
}
//sumByte = 712

A method to calculate BCC

public int getBCCbyte(int sumByte){
    double BCCInDouble = 0;       
    int ToFindBCC = sumByte;

    //Get the Binary String of the SUM 
    String binary = Integer.toBinaryString(ToFindBCC);//(1011001000)

    String finalBinaryBCC = binary.substring(binary.length() - 7);//(1001000)

    //Convert the binary String to char array and calculate the decimal value 
    char[] binaryList = finalBinaryBCC.toCharArray();//[1,0,0,1,0,0,0]
    for (int k = 0; k <= 6; k++) {
        if (binaryList[k] == '1') {
            BCCInDouble += Math.pow(2, 6 - k);
        }
    }
    return (int) BCCInDouble;//72
}

Here 72 is the decimal value for 48H.

Hope this will be helpful for others.