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    <p>The issue here is that bitwise operators work on ints or longs, not bytes. <code>b &amp; 0xF</code> is essentially treated as <code>((int)b) &amp; ((int)0xF)</code>. You can trace it all from the JLS definitions of each operation.</p> <ul> <li>First <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.22.1" rel="nofollow">JLS 15.22.1</a> (which defines <code>&amp;</code> and <code>|</code>) explains that when both operands are convertible to integer primitive types, "binary numeric promotion is first performed on the operands (§5.6.2)."</li> <li><a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.6.2" rel="nofollow">JLS 5.6.2</a>, in turn, says that unless either operand is a <code>float</code>, <code>double</code> or <code>long</code>, both values are widened to <code>int</code>.</li> <li>Finally, widening is defined in <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.2" rel="nofollow">JLS 5.1.2</a> and states that "widening conversion of a signed integer value to an integral type T simply sign-extends the two's-complement representation of the integer value to fill the wider format." Bytes are signed (<a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-4.html#jls-4.2" rel="nofollow">JLS 4.2</a>).</li> </ul> <p>So, your <code>b</code> byte is widened to an <code>int</code> using sign extension before being AND'd or OR'ed with the right operand.</p> <p>Note that this would imply that the result of <code>b &amp; 0F</code> should be an <code>int</code>, not a <code>byte</code>. This is in fact the case (meaning that your explicitly casting it to <code>int</code> is superfluous). You can test this by auto-boxing it to an <code>Object</code> and then checking that object's type:</p> <pre><code>byte b = -1; Object o = (b &amp; 0xF); System.out.println(o.getClass()); // prints "class java.lang.Integer", not "class java.lang.Byte" </code></pre>
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