ByteBuffer Little Endian insert not working

Question

I have to make a two way communication between a legacy system and an android device. The legacy system uses little endian byte ordering. I have successfully implemented the rec

Answer 1

On a related note:

This code:

 int unicodePointsLen = textContent.length() * 2;
 ByteBuffer unicodePointsBuffer = ByteBuffer.allocateDirect(unicodePointsLen);
 short unicodePointValue;
 for (int i = 0; i < textContent.length(); i++) 
 {  
     unicodePointValue = (short)textContent.charAt(i);
     unicodePointsBuffer.put((byte)(unicodePointValue & 0x00FF)).put((byte)(unicodePointValue >> 8));
 }

Is about 25% faster than this:

 int unicodePointsLen = textContent.length() * 2;
 ByteBuffer unicodePointsBuffer = ByteBuffer.allocateDirect(unicodePointsLen);
 unicodePointsBuffer.order(ByteOrder.LITTLE_ENDIAN);
 for (int i = 0; i < textContent.length(); i++) 
 {  
     unicodePointsBuffer.putShort((short)textContent.charAt(i));  
 }

Using JDK 1.8.

I am trying to pass unicode points from JAVA to C++ through JNI and the first method is the fastest I found. Curious that it is faster than the second snippet.

Answer 2

You are, for some strange reason, reinitializing your byte buffers and throwing away the previous copies where you had changed the endian order. The following code works just fine for me:

ByteBuffer byteBuffer = ByteBuffer.allocate(4);
byteBuffer.order(ByteOrder.BIG_ENDIAN);
byteBuffer.putInt(88);
byte[] result = byteBuffer.array();
System.out.println(Arrays.toString(result));

Prints [0, 0, 0, 88]

ByteBuffer byteBuffer = ByteBuffer.allocate(4);
byteBuffer.order(ByteOrder.LITTLE_ENDIAN);
byteBuffer.putInt(88);
byte[] result = byteBuffer.array();
System.out.println(Arrays.toString(result));

Prints [88, 0, 0, 0]