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    <blockquote> <p>Why is there no error in the first line but in second?</p> </blockquote> <p>Because that's the way the language is defined. There's no <code>byte + byte</code> operator, which is why the third line fails - <em>both</em> operands are promoted to <code>int</code> automatically. From <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.18.2" rel="nofollow">section 15.18.2 of the JLS</a>:</p> <blockquote> <p>The binary + operator performs addition when applied to two operands of numeric type, producing the sum of the operands.</p> <p>...</p> <p>Binary numeric promotion is performed on the operands (§5.6.2).</p> </blockquote> <p>Now <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.6.2" rel="nofollow">binary numeric promotion</a> always ends up with a value of <code>int</code>, <code>long</code>, <code>float</code> or <code>double</code>... <code>int</code> in this case. So the operation is to add two <code>int</code> values together, and the result is an <code>int</code>.</p> <p>For <code>b++</code>, however, the type is still <code>byte</code>. It's a <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.14.2" rel="nofollow">postfix increment expression (section 15.14.2)</a>:</p> <blockquote> <p>The type of the postfix increment expression is the type of the variable.</p> </blockquote> <p>On a related note, this would be okay:</p> <pre><code>b += 3; </code></pre> <p>That's a <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.26.2" rel="nofollow">compound assignment (section 15.26.2)</a>:</p> <blockquote> <p>A compound assignment expression of the form <code>E1 op= E2</code> is equivalent to <code>E1 = (T) ((E1) op (E2))</code>, where <code>T</code> is the type of <code>E1</code>, except that <code>E1</code> is evaluated only once.</p> </blockquote> <p>Note the casting part, which is why it works.</p>
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    1. PORules for typecasting with byte
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    1. COOhkay ..thanks btw if i want to add two byte type and store their value in byte is their any way?? i did something like this int z = byte(b + b) but this also gave an error...
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    2. CO@Smatik: You just need to cast: `byte z = (byte) (b + b);`
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    3. COoh i wass missing those brackets around byte... i m such a noob
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