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 11:56:44 -0400
Message-ID:
<46EAAF3C.3070703@aepnetworks.com>
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 ) );


This statement is slightly confusing to me. So you convert the function
to function pointer, convert 0 to function pointer, dereference the
function pointer back to a function. If it was indeed a function
pointer, the first inner check struct convert it to function pointer
again. Is my understanding correct? Now I know why managers don't like C++.

Fei

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