Java Bytebuffer覆盖字节

Question

I've got a ByteBuffer that contains 1024 bytes.

I need to overwrite a short within the buffer at a certain offset at key times.

I know the ByteBuffer class has putShort() , but this doesn't overwrite the data, it simply adds it in, which is causing buffer overflows.

I'm guessing that there isn't a direct way of doing this using the ByteBuffer , can someone possibly suggest a way to do this?

Thanks

Thanks to everyone that replied, seemed it could be done I was just using the wrong version of putShort(). I guess that's what happens when you stare at the same piece of code for six hours.

Thanks again

Answer 1

Cannot reproduce the problem, all seems OK

    ByteBuffer bb = ByteBuffer.allocate(20);
    bb.putShort(10, (short)0xffff);
    System.out.println(Arrays.toString(bb.array()));

prints

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, -1, 0, 0, 0, 0, 0, 0, 0, 0]

Answer 2

For your special case, you can modify directly the backing array using the array() method.

Then just insert your two bytes at the proper indexes:

if(myBuffer.hasArray()) {
    byte[] array = myBuffer.array();
    array[index] = (byte) (myShort & 0xff);
    array[index + 1] = (byte) ((myShort >> 8) & 0xff);
}

Answer 3

int p = b.position();
b.position( ZePlace );
p.putShort( ZeValue );
b.position( p );

http://docs.oracle.com/javase/7/docs/api/java/nio/Buffer.html#position%28%29

Answer 4

我认为您可以调用接受位置索引的putShort（）这个版本。