Re: Introspecting function pointer - detecting if a parameter is passed as const

From:

cbarron3@ix.netcom.com (Carl Barron)

Newsgroups:

comp.lang.c++.moderated

Date:

Sat, 14 Jul 2007 14:34:31 CST

Message-ID:

<1i180mr.1uu3e3h1gs25jmN%cbarron3@ix.netcom.com>

Jens Breitbart <jbreitbart@gmail.com> wrote:

Hello all,

I just read section 22.6 of the book "C++ Templates - The Complete
Guide" by
Vandevoord & Josuttis about introspecting function pointers and I have
a question. Is it possible to detect if a parameter of a function is
qualified with const?

Here is an example of what I would like to do (based on the code from
the book stated above) .

void f1 ( int a );
void f2 ( const int a );

template <typename FunctorT>
void first_param_is_const ( FunctorT func ) {
    std::cout << "The first parameter is ";
    if ( FunctorT::Param1TisConst ) {
       std::cout << "const" << std::endl;
       return;
    }
    std::cout << "non-const" << std::endl;
}

    well determining if a type is const is simple,using
partial template specification.

   template <class T>
   struct is_const
   {
      static const bool value = false;
   };

   template <class T>
   struct is_const<T const>
   {
      static const bool value = true;
   };

   to get the first parameter type of a generalized functor
is a problem unless it contains certain nested typedefs so
I will assum Functor has a nested publically accessable typedef
for param1_type.

   then it is simple.

   template <class Functor>
   void test(Functor f)
   {
     std::coutt << " the first parameter is "
     if(!is_const<typename Functor::param1_type>::value)
     {
         std::cout << "not ";
     }
     std::cout << "constant.\n";
  }

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