Bit Shifting Problem

0xffff is just the hexadecimal representation of  the decimal number 65535.

The hex digit 'f' corresponds to '15' in decimal or '1111' in binary.

So 0xffff (the 0x is just a prefix) is equivalent to '11111111 11111111' in binary.

When you AND an int (using the & operator) with 0xffff it masks out the 2 most signifcant bytes and leaves you with the contents of the 2 least signifcant bytes.

When  performing bit arithmetic, it's common to use hexadecimal nunbers because you can see immediately what their binary representation is.

Bit arithmetic can be done with both the 'int' and 'long' types (as well as their unsigned counterparts), so there's no need for these particular manipulations when using the 'long' type like there is with 'short'. 

The problem is that 'short' operands are automatically promoted to 'int' before the bit shifting operators are applied and an 'int' result is then produced. However, this code should give you the result you want:

         short s1 = 256;
         short s2 = (short)(s1 << 8 & 0xffff); // 0