difference between C++ template partial specialization for function and functor -

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i have been using c++ template class partial specialization function argument while. surprised find same syntax used partial specialize functors.

in case 1 below, easy see f(t) function type. since type, used substitute template parameter, in case 2, semantics of f(t) changed, not type still can pass compiler , work.

i googled few hours, not found valuable info in regard. explain why case 2 worked?

template<class> struct func;   template<class f, class t> struct func<f(t)>              //1. partial specialization function {                              //with signature f(t)    using type = f(t); };  template<class> struct ftor; template<class f, class t> struct ftor<f(t)>              //2. f(t) type? {    using type = f; };  struct foo {   void operator()(int) {} };  int main() {         //1 void(int) function signature     cout<<typeid(typename func<void(int)>::type).name()<<endl;     //2 foo(int)?     cout<<typeid(typename ftor<foo(int)>::type).name()<<endl;     return 0; }

in 2 cases, f(t) signature of function taking t , returning f.

you may rename class func identity , ftor result_of , f returntype better match naming/implementation.


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