byte1 = 1 ( 1 in binary );

You calculated wrong byte1. See above.
I get byte1=1 and byte2=84.

2 != 1.

1 in binary is 1.

public class Main
{ public static void main( final java.lang.String[] args )
 { final java.nio.ByteBuffer buffer = java.nio.ByteBuffer.allocate( 4 );
   buffer.putInt( 340 ).position( 0 );
   for( int i = 0; i < 4; ++i )java.lang.System.out.printf
   ( "%8s%n", java.lang.Integer.toBinaryString( buffer.get() )); }}

      0
      0
      1
1010100

Another thing to watch out for in this type of code...

The >> operator will do a sign extend so if the high bit of the value is
set (aka it's a negative number), it will fill the high bits with 1's
instead of 0's.  So depending on the variable sizes you are working with,
you sometimes have to AND out the high bits after a >> like this:

 short value = (short) 0xFFFF; // 16 ones (aka same as -1 for short);
 short byte1;

 byte1 = (short)((value >> 8) & 0xff);

If your byte1 variable is a byte type, then this is not needed.  But if the
byte1 variable is a short or int, it could be needed depending on the
values you are trying to shift.

clinisbut <clinisbut@gmail.com> wrote in news:291d0703-b1a8-487a-a9c2-
11bd035aa012@i12g2000prf.googlegroups.com:

This question has been answered by others, so I won't repeat their answers.
However, I will point out that:
1)  You are not splitting an Int, you are splitting an int.
2)  An int consists of four bytes.  Splitting an int into two bytes will get
you into trouble for negative numbers and any int larger than 32,767 or
65,535, depending upon your point of view on the sign issues.