Re: determine if a type is a free function pointer

From:
Fei Liu <feiliu@aepnetworks.com>
Newsgroups:
comp.lang.c++
Date:
Fri, 14 Sep 2007 10:16:13 -0400
Message-ID:
<fce54s$29n$1@aioe.org>
Fei Liu wrote:

Kai-Uwe Bux wrote:

Barry wrote:

Kai-Uwe Bux wrote:

[many lines]

Good job,

boost.is_function seems to do hard coding to meet possible function
argument list length.


Makes me wonder whether there is a bug in my code. I wouldn't be
surprised.
 

though is_function checks function not pointer to function.
with a wrapper this should also work with function check.

template <class T>
struct is_function
{
     static const bool value = is_function_pointer<T*>::value;
};

   SHOW_BOOL( is_function<int(int)>::value );
   SHOW_BOOL( is_function<int(int, int)>::value );
   SHOW_BOOL( is_function<int(*)(int)>::value );
   SHOW_BOOL( is_function<int(*)(int, int)>::value );


I actually think that the code needs some cleanup and that testing for
being
a function is simpler than testing for being a function pointer. So, I
changed it the other way around. Here are the versions I added to my
library:

// is_class_type
// =============

  template < typename T >
  class is_class_type {

    typedef char (&yes) [1];
    typedef char (&no) [2];

    template < typename S >
    static yes check ( int S::* );

    template < typename S >
    static no check ( ... );

  public:

    static bool const value = ( sizeof( check<T>( 0 ) ) == sizeof(yes) );

  }; // is_class_type

// is_pointer_type:
// ================

  struct unused_type {};

  template < typename T >
  struct is_pointer_type {
        static const bool value = false;
        typedef unused_type pointee_type;
      };
    template < typename T >
  struct is_pointer_type<T*> {
        static const bool value = true;
        typedef T pointee_type;
      };
 
// is_function_type
// ================

  template < typename T >
  class is_function_type {
        typedef char (&no) [1];
    typedef char (&yes) [2];
        template < typename S >
    static
    yes check ( S * );
        template < typename S >
    static
    no check ( ... );
      public:
        static bool const value =
      ( ! is_class_type<T>::value )
      &&
      ( sizeof( check<T>( * (T*)(0) ) )== sizeof( yes ) );
      };
 
// is_function_pointer
// ===================
    template < typename T >
  struct is_function_pointer {
        static bool const value =
      ( is_pointer_type<T>::value
        &&
        is_function_type< typename is_pointer_type<T>::pointee_type

::value );

      };

Note that the template is_function_type<> does not need the
is_pointer_type<> template. That is why I think that
is_function_type<> is
conceptually simpler than is_function_pointer<> and my reason to prefer
this implementation to the previous one.

Best

Kai-Uwe Bux


Hmm...are you positive that your implemention meets my requirement?
There are 2 problems after I test run your typetraits implementation:

! is_pointer_type<int>::value = true
is_pointer_type<non_fct_ptr>::value = true
is_pointer_type<fct_ptr>::value = true
is_function_pointer<fct_ptr>::value = true
! is_function_pointer<non_fct_ptr>::value = true <----------- wrong
! is_function_pointer<fct_ptr4>::value = false <----------- wrong


I apologize, the result is indeed correct...I start to think the reason
my initial implementation didn't work was missing the implicit function
type to function pointer type conversion.

Thanks, this discussion is very helpful.
Fei

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