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copied!<p>So, I'm going to cheat by changing your syntax.</p> <pre><code>template<typename T> using Ptr = T::template ptr<T>; template<typename T> using WPtr = T::template wptr<T>; struct Pointerable { template<typename T> using ptr = std::shared_ptr<T>; template<typename T> using wptr = std::weak_ptr<T>; }; class Foo : public Pointerable { }; Ptr<Foo> x = std::make_shared<Foo>( ... ); WPtr<Foo> y = x; </code></pre> <p>here, the only purpose of <code>Pointerable</code> is to tag your types that you want to be used. <code>Ptr<Foo></code> and <code>WPtr<Foo></code> could be written to work on any class:</p> <pre><code>template<typename T> using Ptr = std::shared_ptr<T>; template<typename T> using WPtr = std::weak_ptr<T>; </code></pre> <p>but I suppose you want to be able to tag objects as "managed by smart pointers" for whatever reason.</p> <p>We could even make <code>Pointerable</code> completely empty, if you don't like that <code>template</code> boilerplate in it:</p> <pre><code>struct Pointerable {}; </code></pre> <p>and add a traits class:</p> <pre><code>template<typename T> struct is_pointerable : std::is_base_of<Pointerable, T> {}; </code></pre> <p>which we then use:</p> <pre><code>template<typename T, typename=typename std::enable_if< is_pointerable<T>::value >::type > using Ptr = std::shared_ptr<T>; </code></pre> <p>but I think the boilerplate works better -- SFINAE <code>using</code> seems overkill.</p>

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