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    <blockquote> <blockquote> <p>I am used to optimizations being constrained such that they cannot change observable behaviour.</p> </blockquote> </blockquote> <p>This is correct. As a general rule -- known as the <em>as-if</em> rule -- compilers can change code if the change is not observable.</p> <blockquote> <blockquote> <p>This restriction does not seem to apply to RVO. </p> </blockquote> </blockquote> <p>Yes. The clause quoted in the OP gives an exception to the <em>as-if</em> rule and allows copy construction to be omitted, even when it has side effects. Notice that the RVO is just one case of copy-elision (the first bullet point in C++11 12.8/31).</p> <blockquote> <blockquote> <p>Do I ever need to worry about the side effects mentioned in the standard?</p> </blockquote> </blockquote> <p>If the copy constructor has side effects such that copy elision when performed causes a problem, then you should reconsider the design. If this is not your code, you should probably consider a better alternative.</p> <blockquote> <blockquote> <p>What do I as a programmer need to do (or not do) to allow this optimization to be performed?</p> </blockquote> </blockquote> <p>Basically, if possible, return a local variable (or temporary) with the same cv unqualified type as the function return type. This allows RVO but doens't enforce it (the compiler might not perform RVO).</p> <blockquote> <blockquote> <p>For example, does the following prohibit the use of copy elision (due to the move):</p> </blockquote> </blockquote> <pre><code>// notice that I fixed the OP's example by adding &lt;double&gt; std::vector&lt;double&gt; foo(int bar){ std::vector&lt;double&gt; quux(bar, 0); return std::move(quux); } </code></pre> <p>Yes, it does because you're not returning the name of a local variable. This</p> <pre><code>std::vector&lt;double&gt; foo(int bar){ std::vector&lt;double&gt; quux(bar,0); return quux; } </code></pre> <p>allows RVO. One might be worried that if RVO is not performed then moving is better than coping (which would explain the use of <code>std::move</code> above). Don't worry about that. All major compilers will do the RVO here (at least in release build). Even if a compiler doesn't do RVO but the conditions for RVO are met then it will try to do a move rather than a copy. In summary, using <code>std::move</code> above will certainly make a move. Not using it will likely neither copy nor move anything and, in the worst (unlikely) case, will move.</p> <p>(<strong>Update:</strong> As haohaolee's pointed out (see comments), the following paragraphs are not correct. However, I leave them here because they suggest an idea that might work for classes that don't have a constructor taking a <code>std::initializer_list</code> (see the reference at the bottom). For <code>std::vector</code>, haohaolee found a workaround.)</p> <p>In this example you can force the RVO (strict speaking this is no longer RVO but let's keep calling this way for simplicity) by returning a <em>braced-init-list</em> from which the return type can be created:</p> <pre><code>std::vector&lt;double&gt; foo(int bar){ return {bar, 0}; // &lt;-- This doesn't work. Next line shows a workaround: // return {bar, 0.0, std::vector&lt;double&gt;::allocator_type{}}; } </code></pre> <p>See this <a href="https://stackoverflow.com/questions/19427196/construction-helper-function-make-xyz-where-xyx-is-a-class-with-noncopy-constrai">post</a> and <a href="https://stackoverflow.com/users/46642/r-martinho-fernandes">R. Martinho Fernandes</a>'s brilliant <a href="https://stackoverflow.com/a/19428052/1137388">answer</a>.</p> <p>Be carefull! Have the return type been <code>std::vector&lt;int&gt;</code> the last code above would have a different behavior from the original. (This is another story.)</p>
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    1. COI'm curious about the other story. I know that if it was vector<int>, { bar, 0 } would create a vector with 2 elements, but how can I force a multiple parameters call not a initializer list call?
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    2. COAnd for `std::vector<double>`, I didn't get it working on GCC 4.8.1. It creates a vector with two elements bar and 0 unless adding a third parameter `std::vector<double>::allocator_type()`
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    3. CO@haohaolee: You're right. It doesn't work as I expected because the constructor taking `std::initializer_list` takes precedence. Your workaround to change this priority works. However, it's a bit weird to invove `allocator_type` to control overload resolution. I'm not blaming your approch, it's more a C++ (core and library) issue. I'll update the post to reflect these new findings. Thanks.
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