c++ - how check for member operator(type)? -

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• is possible write template check function's existence? 21 answers

suppose have types bar , foo. how can construct template class has_call_with_arg<> such has_call_with_arg<bar,foo>::value true if , if

bar b; foo f; b(f); 

would compile? looked various related questions (including mentioned above) , tried

template<typename func, typename arg> class has_call_with_arg {   struct bad {};   struct test : func   {     template<typename c>     bad operator()(c const&r);   }; public:   static const bool value =      !std::is_same<bad, typename std::result_of<test(arg const&)>::type >::value; }; 

but didn't work (didn't detect correct match). what's wrong?

template<typename sig, typename functor> struct is_callable; template<typename ret, typename... arg, typename functor>  struct is_callable<ret(arg...), functor> { // partial spec private:     struct no {}; public:     template<typename u> static auto f(std::nullptr_t) -> decltype(std::declval<u>()(std::declval<arg>()...));     template<typename u> static no f(...);     static const int value = std::is_convertible<decltype(f<functor>(nullptr)), ret>::value; }; 

i created content for tutorials, explain construction of trait (non-variadic form first).