Read byte from a long

Question

Hi,
How can I read bytes from a long. i..e read one byte by one byte from long.
Naman Patel

Answer 1

ok here is my problem how do i ignore the sign byte i.e please check this code
long nr=1345345333L, add=7, nr2=0x12345671L;
          long tmp;
          long[] result = new long[2];
          char[] pwd = password.toCharArray();
          for (int pCount = 0; pCount < pwd.length; pCount++){
               if (pwd[pCount] == ' ' || pwd[pCount] == '\t')
                    continue;
               tmp= (long) pwd[pCount];
               nr^= (((nr & 63)+add)*tmp)+ (nr << 8);
               nr2+=(nr2 << 8) ^ nr;
               add+=tmp;
          result[0]=nr & (((long) 1L << 31) -1L); /* Don't use sign bit (str2int) */;
          result[1]=nr2 & (((long) 1L << 31) -1L);
          int i = 0;
          long x = result[0];
          byte b[] = longToBytes(x);
          String str = "";
          i = 0;
          while(i < 4){
               if(b[i] == 0x00) {
                    System.out.println("byte is 0x00");
                    str += "00";
               else {
                    if(b[i] <= 0x0f && b[i] >= 0) {
                         System.out.println("byte is 0x0f" + b);
                         str += "0";
                    break;
               i++;
          i = 0;
          str += Long.toHexString(result[0]);
          x = result[1];
          b = longToBytes(x);
          while(i < 4){
               if(b[i] == 0x00) {
                    System.out.println("byte is 0x00");
                    str += "00";
               else {
                    if(b[i] <= 0x0f && b[i] >= 0) {
                         System.out.println("byte is 0x0f");
                         str += "0";
                    break;
               i++;
          str += Long.toHexString(result[1]);