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    <p>That's because <code>b += 1.0;</code> is equivalent to <code>b = (int) ((b) + (1.0));</code>. The <a href="https://docs.oracle.com/javase/specs/jls/se8/html/jls-5.html#jls-5.1.3" rel="nofollow noreferrer">narrowing primitive conversion (JLS 5.1.3)</a> is hidden in the compound assignment operation.</p> <h3><a href="https://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.26.2" rel="nofollow noreferrer">JLS 15.26.2 Compound Assignment Operators</a> (JLS Third Edition):</h3> <blockquote> <p>A compound assignment expression of the form <em>E1 op= E2</em> is equivalent to <em>E1 = (T)((E1) op (E2))</em>, where <em>T</em> is the type of <em>E1</em>, except that <em>E1</em> is evaluated only once.</p> <p>For example, the following code is correct:</p> <pre><code>short x = 3; x += 4.6; </code></pre> <p>and results in <code>x</code> having the value <code>7</code> because it is equivalent to:</p> <pre><code>short x = 3; x = (short)(x + 4.6); </code></pre> </blockquote> <p>This also explains why the following code compiles:</p> <pre><code>byte b = 1; int x = 5; b += x; // compiles fine! </code></pre> <p>But this doesn't:</p> <pre><code>byte b = 1; int x = 5; b = b + x; // DOESN'T COMPILE! </code></pre> <p>You need to explicitly cast in this case:</p> <pre><code>byte b = 1; int x = 5; b = (byte) (b + x); // now it compiles fine! </code></pre> <hr> <p>It's worth noting that the implicit cast in compound assignments is the subject of <strong>Puzzle 9: Tweedledum</strong> from the wonderful book <a href="http://www.javapuzzlers.com/" rel="nofollow noreferrer"><em>Java Puzzlers</em></a>. Here are some excerpt from the book (slightly edited for brevity):</p> <blockquote> <p>Many programmers think that <code>x += i;</code> is simply a shorthand for <code>x = x + i;</code>. This isn't quite true: if the type of the result is wider than that of the variable, the compound assignment operator performs a silent narrowing primitive conversion.</p> <p>To avoid unpleasant surprises, <strong>do not use compound assignment operators on variables of type</strong> <code>byte</code>, <code>short</code>, or <code>char</code>. When using compound assignment operators on variables of type <code>int</code>, ensure that the expression on the right-hand side is not of type <code>long</code>, <code>float</code>, or <code>double</code>. When using compound assignment operators on variables of type <code>float</code>, ensure that the expression on the right-hand side is not of type <code>double</code>. These rules are sufficient to prevent the compiler from generating dangerous narrowing casts.</p> <p>For language designers, it is probably a mistake for compound assignment operators to generate invisible casts; compound assignments where the variable has a narrower type than the result of the computation should probably be illegal.</p> </blockquote> <p>The last paragraph is worth noting: C# is a lot more strict in this regard (see <a href="http://msdn.microsoft.com/en-us/library/aa691316%28VS.71%29.aspx" rel="nofollow noreferrer">C# Language Specification 7.13.2 Compound assignment</a>).</p>
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    1. COThanks. It's hard to search for documentation when the keyword you want to use is something like "+=". :P
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    2. COAll `op=` operators are called "compound assignment" in Java. But yes, I do think that querying for symbols instead of keywords is hard in most search engines.
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    3. COAny googlers here? I'd like a search box which also takes into account symbols like + = < and > :)
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