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    <p>Any compiler worth its salt will generate exactly the same machine-language sequence for both constructs for any built-in type (<code>int</code>, <code>float</code>, etc) as long as the statement really is as simple as <code>x = x + a;</code> <em>and optimization is enabled</em>. (Notably, GCC's <code>-O0</code>, which is the default mode, performs <em>anti-optimizations</em>, such as inserting completely unnecessary stores to memory, in order to ensure that debuggers can always find variable values.)</p> <p>If the statement is more complicated, though, they might be different. Suppose <code>f</code> is a function that returns a pointer, then</p> <pre><code>*f() += a; </code></pre> <p>calls <code>f</code> only once, whereas</p> <pre><code>*f() = *f() + a; </code></pre> <p>calls it twice. If <code>f</code> has side effects, one of the two will be wrong (probably the latter). Even if <code>f</code> doesn't have side effects, the compiler may not be able to eliminate the second call, so the latter may indeed be slower.</p> <p>And since we're talking about C++ here, the situation is entirely different for class types that overload <code>operator+</code> and <code>operator+=</code>. If <code>x</code> is such a type, then -- before optimization -- <code>x += a</code> translates to</p> <pre><code>x.operator+=(a); </code></pre> <p>whereas <code>x = x + a</code> translates to</p> <pre><code>auto TEMP(x.operator+(a)); x.operator=(TEMP); </code></pre> <p>Now, if the class is properly written <em>and</em> the compiler's optimizer is good enough, both will wind up generating the same machine language, but it's not a sure thing like it is for built-in types. This is probably what Stroustrup is thinking of when he encourages use of <code>+=</code>.</p>
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    1. POIs x += a quicker than x = x + a?
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    1. COThere is also another aspect - _readability_. C++ idiom for adding `expr` to `var` is `var+=expr` and writing it the other way will confuse readers.
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      1. USTadeusz Kopec
    2. COThis. If you implemented the `+=` operator in a no-AST compiler as in [Crenshaw's tutorial](http://compilers.iecc.com/crenshaw/) it would have a pretty good chance of being faster then the written out version. But all that proves is that you've written an incredibly primitive compiler (even then if you implemented the peephole optimize that Crenshaw suggests you'd probably catch and fix it).
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    3. COIf you find yourself writing `*f() = *f() + a;` you might want to take a good hard look at what you're really trying to achieve...
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      1. USAdam Davis
 

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