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14531809
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2013-01-25T23:13:56.163
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2013-01-25T23:31:02.113
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2013-01-25T23:31:02.113
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1932150
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14531766
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2
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The syntactic form <code>T&&</code> in the context of type deduction (which includes template argument deduction, but for instance also the deduction of the type of a variable declared as <code>auto</code>) does not indicate an rvalue reference, but rather what Scott Meyers calls a [universal reference]. Please notice, that only the very particular syntactic form <code>T&&</code> denotes a universal reference, while other, similar forms are not regarded as such. For instance: <pre><code>template<typename T> void foo(T&& t); <-- T&& is a universal reference template<typename T> void foo(T const&& t); <-- T const&& is NOT a universal reference template<typename T> void foo(S<T>&& t); <-- S<T>&& is NOT a universal reference template<typename T> struct S { void foo(T&& t); }; <-- T&& is NOT a universal reference </code></pre> Universal references can bind both to lvalues and to rvalues. If an lvalue of type <code>A</code> is bound, then <code>T</code> is deduced to be <code>A&</code> and the type of the argument resolves into <code>A&</code> (lvalue reference) due to the rule of reference collapsing (<code>A& &&</code> becomes <code>A&</code>). If an rvalue of type <code>A</code> is bound, then <code>T</code> is deduced to be <code>A</code> and the type of the argument resolves into <code>A&&</code> (rvalue reference). [Note: Reference collapsing rule might seem complicated, but they are actually quite easy: to quote Stephan T. Lavavej, "lvalue references are contagious", meaning that when the forms <code>T&& &</code>, <code>T& &</code>, or <code>T& &&</code> get instantiated, they always resolve into <code>T&</code> - only the form <code>T&& &&</code> is resolved into <code>T&&</code>] This is why the <code>std::move</code> function template will be instantiated as follows when the argument is an lvalue (<code>T</code> is deduced to be <code>T&</code>): <pre><code>typename remove_reference<A&>::type&& std::move(A& && a); </code></pre> while it will be instantiated as follows when the argument is an rvalue (<code>T</code> is deduced to be <code>A</code>) <pre><code>typename remove_reference<A>::type&& std::move(A&& a); </code></pre>
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1. POHow are std::move template parameters deduced?
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1. USAndy Prowl
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1. USAndy Prowl
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