[Boost-users] Class template spec with enable_if

From: Agoston Bejo (gusz1_at_[hidden])
Date: 2004-12-21 14:53:45

Hello again!
See the example below. (Platform: VC++7.1)
This error is strange for me because it only occurs in the case of class
template
specialization.
The compiler deduces the type of is_arithmetic<T>::value falsely, not
as bool, althought it is known at compile time, isn't it?

The error messages (both referring to the erroneous line marked with
"ERROR"):

error C2296: '&&' : illegal, left operand has type 'is_arithmetic<T>::value'
error C2297: '&&' : illegal, right operand has type
'is_arithmetic<T2>::value'

Thx,
Agoston

Example:

//-------------------------------------------------------

#include <iostream>
#include <boost/utility/enable_if.hpp>
#include <boost/type_traits.hpp>
#include <boost/mpl/and.hpp>

using namespace std;
using namespace boost;
using namespace mpl;

// -----------------

template<typename T1, typename T2, typename TEnabler = void>
struct Enabler
{
 static const bool value = false;
};

template<typename T1, typename T2>
struct Enabler<T1, T2,
 typename enable_if_c<
  is_arithmetic<T1>::value && is_arithmetic<T2>::value
  >::type

   // ERROR!!!

>
{
 static const bool value = true;
};
*/
/*
//WOULD WORK:
 typename enable_if<
  and_< is_arithmetic<T1>, is_arithmetic<T2> >
  >::type
*/

// -----------------

template<typename T1, typename T2>
 typename enable_if_c<
  is_arithmetic<T1>::value && is_arithmetic<T2>::value
  >::type
f(T1 t1, T2 t2) { cout << t1 << " " << t2 << endl; }

// -----------------

int _tmain(int argc, _TCHAR* argv[])
{
 f(5.0, 6);
 return 0;
}