Re: Trying to apply SFINAE

From:

Kai-Uwe Bux <jkherciueh@gmx.net>

Newsgroups:

comp.lang.c++

Date:

Sat, 04 Oct 2008 22:37:53 -0400

Message-ID:

<gc9963$htd$1@aioe.org>

Francesco wrote:

Hi to all,
maybe something like the following might work.
Relying on the fact that an iterator should be either
- a pointer
- a class with some inner typedefs (here iterator_category)
The inspector class can be customized to better suit your needs...
Maybe there is a simpler way... don't know :-)
Bye,
Francesco

// code
#include <iterator>
#include <iostream>
#include <vector>

// ala boost::enable_if
template< typename T, bool K >
class CTypeEnable;

template< typename T >
class CTypeEnable< T, false >
{};

template< typename T >
class CTypeEnable< T, true >
{
public:
typedef T tResult;
};

// ala boost::??
template< typename T >
struct CIsIter
{
typedef char (&tYes)[1];
typedef char (&tNo)[2];

template< typename T2 >
static typename CTypeEnable< tYes,
sizeof( typename T2::iterator_category ) >::tResult
Check( T2 * );

template< typename T2 >
static tYes Check( T2 ** );

static tNo Check( ... );

enum { kResult = sizeof( Check( (T*) NULL ) )
== sizeof( tYes ) };
};

// SFINAE with return type

template< typename T >
typename CTypeEnable< void, !CIsIter< T >::kResult >::tResult F( T )
{ std::cout << "NO ITERATOR\n"; }

template< typename T >
typename CTypeEnable< void, CIsIter< T >::kResult >::tResult F( T )
{ std::cout << "ITERATOR\n"; }

struct A
{};

int main()
{
F( 10 );
F( ( int*)NULL);
std::vector< int > vec;
F( vec.begin() );
F( A() );
}
//end code

That idea can work a little simpler:

struct yes_type { char dummy; };
struct no_type { yes_type a; yes_type b; };

template < typename T >
struct is_iterator_type {

  template < typename S >
  static
  yes_type check ( S*, typename S::iterator_category* ptr = 0 );

  template < typename S >
  static
  yes_type check ( S** );

  static
  no_type check ( ... );

public:

  static bool const value =
    sizeof( check((T*)0) ) == sizeof( yes_type );

};

#include <iostream>
#include <vector>

template < typename T >
void show ( void ) {
  if ( is_iterator_type<T>::value ) {
    std::cout << "is iterator type\n";
  } else
    std::cout << "is not iterator type\n";
}

#define SHOW(T) \
    std::cout << #T << " "; \
  show<T>();

int main() {
  SHOW( std::vector<int>::const_iterator );
  SHOW( std::vector<int> );
  SHOW( int* );
  SHOW( int );
}

Best

Kai-Uwe Bux