Java中的位操作C源代码

Question

I try to calculate the checksum of a Sega Genesis rom file in Java. For this i want to port a code snipped from C into Java:

static uint16 getchecksum(uint8 *rom, int length)
{
  int i;
  uint16 checksum = 0;

  for (i = 0; i < length; i += 2)
  {
    checksum += ((rom[i] << 8) + rom[i + 1]);
  }

  return checksum;
}

I understand what the code does. It sums all 16bit numbers (combined from two 8 bit ones). But what i didn't understand is what's happening with the overflow of the uint16 and how this transfers to Java code?

Edit: This code seems to work, thanks:

int calculatedChecksum = 0;
int bufferi1=0;
int bufferi2=0;
bs = new BufferedInputStream(new FileInputStream(this.file));

bufferi1 = bs.read();
bufferi2 = bs.read();
while(bufferi1 != -1 && bufferi2 != -1){
    calculatedChecksum += (bufferi1*256 + bufferi2);
    calculatedChecksum = calculatedChecksum % 0x10000;
    bufferi1 = bs.read();
    bufferi2 = bs.read();
}

Answer 1

Simply put, the overflow is lost. A more correct approach (imho) is to use uint32 for summation, and then you have the sum in the lower 16 bits, and the overflow in the upper 16 bits.

Answer 2

static int checksum(final InputStream in) throws IOException {
  short v = 0;
  int c;
  while ((c = in.read()) >= 0) {
    v += (c << 8) | in.read();
  }
  return v & 0xffff;
}

This should work equivalently; by using & 0xffff , we get to treat the value in v as if it were unsigned the entire time, since arithmetic overflow is identical wrt bits.

Answer 3

You want addition modulo 2 ¹⁶ , which you can simply spell out manually:

checksum = (checksum + ((rom[i] << 8) + rom[i + 1])) % 0x10000;
//                                                   ^^^^^^^^^