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    <p>Altough both, <code>return std::move(local)</code> and <code>return local</code>, do work in sense of that they do compile, their behavior is different. And probably only the latter one was intended.</p> <p>If you write a function which returns a <code>std::vector&lt;string&gt;</code>, you have to return a <code>std::vector&lt;string&gt;</code> and exactly it. <code>std::move(local)</code> has the type<code>std::vector&lt;string&gt;&amp;&amp;</code> which is <strong>not</strong> a <code>std::vector&lt;string&gt;</code> so it has to be converted to it using the move constructor.</p> <p>The standard says in 6.6.3.2:</p> <blockquote> <p>The value of the expression is implicitly converted to the return type of the function in which it appears.</p> </blockquote> <p>That means, <code>return std::move(local)</code> is equalvalent to</p> <pre><code>std::vector&lt;std::string&gt; converted(std::move(local); // move constructor return converted; // not yet a copy constructor call (which will be elided anyway) </code></pre> <p>whereas <code>return local</code> only is</p> <pre><code>return local; // not yet a copy constructor call (which will be elided anyway) </code></pre> <p>This spares you one operation.</p> <hr> <p>To give you a short example of what that means:</p> <pre><code>struct test { test() { std::cout &lt;&lt; " construct\n"; } test(const test&amp;) { std::cout &lt;&lt; " copy\n"; } test(test&amp;&amp;) { std::cout &lt;&lt; " move\n"; } }; test f1() { test t; return t; } test f2() { test t; return std::move(t); } int main() { std::cout &lt;&lt; "f1():\n"; test t1 = f1(); std::cout &lt;&lt; "f2():\n"; test t2 = f2(); } </code></pre> <p>This will output</p> <pre><code>f1(): construct f2(): construct move </code></pre>
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    1. POAre returned locals automatically xvalues
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    1. COThanks for your input.
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    2. CONote that this is only true in the presence of NVRO/elision. Without elision, `f1` will also have to move.
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      1. USNicol Bolas
    3. CO@NicolBolas: Yes, that's the "not yet a copy constructor call" part written above (well, actually it has to be the move constructor call). However, this affects both, `f1` and `f2`, so even then the `f1` has one constructor call less.
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