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POWhat is the behaviour of compiler generated move constructor?
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copied!<p>Does <code>std::is_move_constructible<T>::value == true</code> imply that <code>T</code> has a usable move constructor? If so, what is the default behaviour of it?</p> <p>Consider the following case:</p> <pre><code>struct foo { int* ptr; }; int main() { { std::cout << std::is_move_constructible<foo>::value << '\n'; foo f; f.ptr = (int*)12; foo f2(std::move(f)); std::cout << f.ptr << ' ' << f2.ptr << '\n'; } return 0; } </code></pre> <p>and the output is:</p> <pre><code>1 0000000C 0000000C </code></pre> <p>I thought that <code>f.ptr</code> should be <code>nullptr</code>. So in this case, </p> <ol> <li>Is <code>f2</code> move constructed ?</li> <li>If so, shouldn't the rvalue be invalidated?</li> <li>How can I know if instances of a class can be properly move-constructed (invalidate the old one)?</li> </ol> <p>(I'm using VS11.)</p> <h2>Update</h2> <p>The default behaviour of move constructor is same as a copy constructor, is it correct? If it's true,</p> <ol> <li>We always expect a move ctor to steal the resources of the moved-from object, while the default one does not behave as expected, so what's the point of having a default move ctor? </li> <li>How can I know if a class has a custom move constructor (which can be guaranteed to behave properly)?</li> </ol> <p>It seems that <code>foo f2(std::move(f));</code> calls the copy ctor when I declared one, see:</p> <pre><code>struct foo { int* ptr; foo() {} foo(const foo& other) { std::cout << "copy constructed\n"; } }; int main() { { std::cout << std::is_move_constructible<foo>::value << '\n'; foo f; foo f2(std::move(f)); } system("pause"); return 0; } </code></pre> <p>Now the output is:</p> <pre><code>1 copy constructed </code></pre> <p>If <code>foo</code> has a move constructor, then wouldn't <code>foo f2(std::move(f))</code> call it?</p> <p>So now my questions is: How to know if a class has a move ctor, and if it has one, how can I explicitly call it?</p> <h2>What I'm trying to do is…</h2> <pre><code>template<typename T, bool has_move_ctor> struct MoveAux; template<typename T> struct MoveAux<T, true> { static void doMove(T* dest, T* src) { new(dest) T(std::move(*src)); //move ctor } }; template<typename T> struct MoveAux<T, false> { static void doMove(T* dest, T* src) { new(dest) T(*src); //copy ctor src->~T(); } }; template<typename T> inline doMove(T* dest, T* src) { MoveAux<T,/*a trait*/>::doMove(dest, src); } </code></pre> <p>So I thought <code>std::is_move_constructible<T>::value</code> can be passed to the template, while now I see that this trait only cares if <code>T t(T())</code> is a valid expression, it may call <code>T::T(const T&)</code>. Now assume that <code>T</code> is a custom class, then I want the above templates to behave like:</p> <ol> <li>If I don't declare a move ctor, I want that template method calls the <code>MoveAux<T,false>::doMove</code>.</li> <li>If I declared one, I need it calls to <code>MoveAux<T,true>::doMove</code>.</li> </ol> <p>Is it possible to make this work?</p>

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