从较小的字节数组（Java）进行长时间转换

Question

I am trying to convert a byte array to long, but receive BufferUnderflowException

    ByteBuffer buffer = ByteBuffer.allocate(Long.BYTES);
    byte[] arg1 = new byte[] {0x04, (byte)0xB0};
    buffer.put(arg1, 0, arg1.length);
    buffer.flip();
    long val = buffer.getLong();

In debug mode, I looked inside buffer. It has an internal byte array, and it fills the unspecified bytes with "0". So why does the exception occur? How can I fix it?

Answer 1

The spec of getLong() specifically says that it throws BufferUnderflowException if there are fewer than eight bytes remaining in this buffer . Your buffer has only two bytes.

Answer 2

It seems like you should use Long.BYTES to fill your buffer . If your byte[] is exhausted you could switch to 0. Like,

ByteBuffer buffer = ByteBuffer.allocate(Long.BYTES);
byte[] arg1 = new byte[] { 0x04, (byte) 0xB0 };
for (int i = 0; i < Long.BYTES; i++) {
    int pos = Long.BYTES - i - 1;
    byte val = (pos < arg1.length) ? arg1[pos] : 0;
    buffer.put(val);
}
buffer.flip();
long val = buffer.getLong();
System.out.println(val);

Output is

45060

Going the other way

System.out.println(Integer.toHexString(45060));
b004

Answer 3

An ArrayList may have an Object[16] , but that still doesn't mean that its logical size is bigger than the number of elements you've added. You need eight bytes in the buffer to read a long out of it.

If you are trying to use that byte array as an unsigned short, just write leading zero bytes to the buffer. Not sure why you want a long over an int if it's just two bytes, though.

Answer 4

The problem here is with -: buffer.flip();

flip is used to flip the buffer from read to write mode or write to read mode by resetting the limit as the current position.

If you comment out the line, it will work fine.