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    <p>The problem is that </p> <pre><code>p.emplace_back(v[i]); </code></pre> <p>passes an lvalue to <code>emplace_back</code>, which means that your move constructor (which expects an rvalue reference) won't work. </p> <p>If you actually want to move values from one container to another, you should explicitly call <a href="http://en.cppreference.com/w/cpp/utility/move" rel="nofollow"><code>std::move</code></a>:</p> <pre><code>p.emplace_back(std::move(v[i])); </code></pre> <p>(The idea behind a move constructor like <code>obj(obj&amp;&amp; tmp)</code> is that <code>tmp</code> should be an object that isn't going to be around for much longer. In your <strong>first</strong> loop, you pass a temporary object to <code>emplace_back</code>, which is fine -- a rvalue reference can bind to a temporary object and steal data from it because the temporary object is about to disappear. In your <strong>second</strong> loop, the object that you pass to <code>emplace_back</code> has a name: <code>v[i]</code>. That means it's not temporary, and could be referred to later in the program. That's why you have to use <a href="http://en.cppreference.com/w/cpp/utility/move" rel="nofollow"><code>std::move</code></a> to tell the compiler "<em>yes, I really meant to steal data from this object, even though someone else might try to use it later.</em>")</p> <hr> <p><strong>Edit:</strong> I'm assuming that your rather unusual usage of <code>emplace_back</code> is a relic of having to craft a little example for us. If that isn't the case, see @jogojapan's answer for a good discussion about why using a <code>std::vector</code> move constructor or repeated calls to <code>push_back</code> would make more sense for your example.</p>
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    1. POC++11 - emplace_back between 2 vectors doesn't work
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    1. COI was convinced that v[i] it's an Rvalue, since it's a reference to a position of a well defined type, strange to know that it is an lvalue. but your solution works ...
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    2. CO@user1849534: I've updated the answer to try to explain it a bit more. The gist of it is that `v[i]` is an lvalue because it's a named object that you could refer to later.
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