I have a an array of byte, size n, that really represents an array of short of size n/2. Before I write the array to a disk file I need to adjust the values by adding bias values stored in another array of short. In C++ I would just assign the address of the byte array to a pointer for a short array with a cast to short and use pointer arithmetic or use a union.
How may this be done in Java - I'm very new to Java BTW.
You could do the bit-twiddling yourself but I'd recommend taking a look at the ByteBuffer and ShortBuffer classes.
byte[] arr = ...
ByteBuffer bb = ByteBuffer.wrap(arr); // Wrapper around underlying byte[].
ShortBuffer sb = bb.asShortBuffer(); // Wrapper around ByteBuffer.
// Now traverse ShortBuffer to obtain each short.
short s1 = sb.get();
short s2 = sb.get(); // etc.
while (sb.hasRemaining()) if you want to loop over all of them docs.oracle.com/javase/6/docs/api/java/nio/…You can wrap your byte array with java.nio.ByteBuffer.
byte[] bytes = ...
ByteBuffer buffer = ByteBuffer.wrap( bytes );
// you may or may not need to do this
//buffer.order( ByteOrder.BIG/LITTLE_ENDIAN );
ShortBuffer shorts = buffer.asShortBuffer( );
for ( int i = 0, n=shorts.remaining( ); i < n; ++i ) {
final int index = shorts.position( ) + i;
// Perform your transformation
final short adjusted_val = shortAdjuster( shorts.get( index ) );
// Put value at the same index
shorts.put( index, adjusted_val );
}
// bytes now contains adjusted short values
The correct way to do this is using shifts. So
for (int i = 0; i < shorts.length; i++) {
shorts[i] = (short)((bytes[2*i] << 8) | bytes[2*i + 1]);
}
Also, it depends on the endian-ness of the stream in many respects. This may work better
