使用ByteBuffer在Java中发布复制字节字符串

Question

I'm trying to duplicate a byte String that I produce in Objective-C (on iOS) in Java but am having trouble. Here's the string I want to produce:

"\x01\x00\x00\x00\x01\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00"

I get that string from an array of 4 integers - [1,1,0,0] - where each integer has 4 bytes. After looking at a bunch of questions on here, I've attempted constructing that string in Java using ByteBuffer as follows:

ByteBuffer bytes = ByteBuffer.allocate(16);
bytes.putInt(1);
bytes.putInt(1);
bytes.putInt(0);
bytes.putInt(0);

String byteString = new String(bytes.array());

However that gives me:

"\x00\x00\x00\x01\x00\x00\x00\x01\x00\x00\x00\x00\x00\x00\x00\x00"

When I unpack that, the array of ints I get is [16777216, 16777216, 0, 0] . Obviously I'm doing something wrong, and I'm hoping someone can point me in the right direction.

Answer 1

iOS is little-endian, so the least significant byte of a 4-byte integer comes first.

Java ByteBuffer defaults to big-endian , so the opposite applies.

The initial order of a byte buffer is always BIG_ENDIAN

You can change this with

bytes.order(ByteOrder.LITTLE_ENDIAN);

Answer 2

What you want is:

ByteBuffer bytes = ByteBuffer.allocate(16);
bytes.putInt(16777216);
bytes.putInt(16777216);
bytes.putInt(0);
bytes.putInt(0);

String byteString = new String(bytes.array());

The endianness of the platforms are different, so when you put 4 bytes in, the bytes are reversed.