2字节数组的差异

Question

I have a byte array which can have up to 4000 elements. These elementes can be byte/boolean(1 byte) int(2 byte) or long/float(4byte). They are not mixed, so one byte array only contains one data type

If the array changes I want to get the position which is affected.. but if arr[5] changes, it is for example the second real value which has changed! but if it is a boolean it is the 6th value.

so far I have done this

private void diff(byte[] a, byte[] b){
if (a.length == b.length) {
        for (int i = 0; i < getCount(); i++) {
            if (!get(MyType.real,i, b).equals(get(MyType.real,i, a))) {
                //difference
            }
        }
        this.data = data;
    }
}

private Object get(MyType type, int idx, byte[] data) {
    logger.entry(idx, data);
    ByteBuffer buffer = ByteBuffer.wrap(data)
            .order(ByteOrder.LITTLE_ENDIAN);
    switch (type) {
    case BOOL:
        return logger.exit(buffer.get(idx) > 0);
    case DWORD:
        return logger.exit(buffer.asIntBuffer().get(idx));
    case INT:
        return logger.exit(buffer.asShortBuffer().get(idx));
    case REAL:
        return logger.exit(buffer.asFloatBuffer().get(idx));
    case TIME:
        return logger.exit(buffer.asIntBuffer().get(idx));
    default:
        return logger.exit(buffer.get(idx));
    }
}

but this takes quite a while. Any Ideas to improve the performance?

Answer 1

I think your code is too complicated. You could find the absolute position of the difference and then divide it by the size of the array type.

Example code:

private int diffpos(byte[] a, byte[] b, int typeLenght){
    if (a.length == b.length) {
        for (int i = 0; i < a.length; i++) {
            if (a[i] != b[i]) {
                return i / typeLenght;
            }
        }
    }
    return 0;
}

Answer 2

Find the first difference by comparing bytes and then divide this index by the size of a single element in the buffer:

int pos = findFirstDifference(a, b);
pos /= type.elementSize();