Java语言按位左移的字节序问题

Question

I am trying to translate this simple function from Go to Javascript:

func ShiftLeft(b []byte) []byte {
    l := len(b)
    if l == 0 {
        panic("shiftLeft requires a non-empty buffer.")
    }

    output := make([]byte, l)

    overflow := byte(0)
    for i := int(l - 1); i >= 0; i-- {
        output[i] = b[i] << 1
        output[i] |= overflow
        overflow = (b[i] & 0x80) >> 7
    }

    return output
}

My first attempt was this:

function makeEmpty(size) {


  var result = [];

  for (var i = 0; i < size; i++) {
    result.push(0x00);
  }

  return result;

}

function shiftLeft (b) {

  var len = b.length;

  if (len == 0) {
    throw 'shiftLeft requires a non-empty buffer';
  }

  var output = makeEmpty(len);

  var overflow = 0;

  for (var i = len - 1; i >= 0; i--) {
    output[i] = b[i] << 1;
    output[i] |= overflow;
    overflow = (b[i] & 0x80) >> 7;
  }

  return output;

}

However, this does not work. Given the following test case:

function fromOctal(str) {

  var bytes = [parseInt(str, 2)];

  return bytes;

}

console.log(shiftLeft(fromOctal("10000000"))

The Javascript version returns [256], but the expected result is "00000000" or [0].

What am I getting wrong here? I think it might have to do with endianness, but I have no idea about how to deal with this kind of issue consistently.

Answer 1

Your mistake appears to be in assuming that the elements of your array are 8-bit integers, but the result of bitwise operators in JavaScript are 32-bit integers and so the most significant bit remains when you do the left shift.

I believe that adding a bit mask by changing:

output[i] = b[i] << 1;

to:

output[i] = (b[i] << 1) & 0xFF;

should fix your issue.

http://jsfiddle.net/MTj63/

As a side note, I think your fromOctal() function should actually be named fromBinary().