[Boost-users] [mpl] lazy condition in enable_if

Subject: [Boost-users] [mpl] lazy condition in enable_if
From: Sam Fisher (infest21h_at_[hidden])
Date: 2015-01-28 14:22:37

Hi,

I need to run boost::mpl::for_each<> function with template condition, like:

boost::mpl::for_each< vector_of_types, 
boost::mpl::make_identity<boost::mpl::_1> > (Functor< condition >());

I'm trying to implement lazy binding in enable_if.

Visual studio fails with error "'X<IsFoo<boost::mpl::_>>::operator ()' : 
ambiguous call to overloaded function"

minimal test code which fail on boost 1.57

#include <boost/mpl/for_each.hpp>
#include <boost/mpl/lambda.hpp>
#include <boost/mpl/bind.hpp>
#include <boost/mpl/vector.hpp>
#include <boost/core/enable_if.hpp>

#include <iostream>
#include <type_traits>

template <int> struct dummy { dummy(int) {} }; // from doc

struct Foo {};
struct Bar {};

template <typename T>
struct IsFoo
{
   typedef typename std::is_same<T, Foo>::type type;
};

#define generate_type(_fn) \
typename boost::mpl::apply<boost::mpl::bind< \
   boost::mpl::lambda< _fn <boost::mpl::_> >::type, \
   boost::mpl::bind<typename boost::mpl::lambda<_Pred>::type, U> \
     > >::type

#define enable_type generate_type(boost::enable_if)

#define disable_type generate_type(boost::disable_if)

template <typename _Pred>
struct X
{
   template <typename U>
   enable_type operator()(boost::mpl::identity<U>, dummy<0> = 0) const
   {
     std::cout << "enable: " << typeid(U).name() << std::endl;
   }

   template <typename U>
   disable_type operator()(boost::mpl::identity<U>, dummy<1> = 0) const
   {
     std::cout << "disable: " << typeid(U).name() << std::endl;
   }
};

int main()
{
   X<IsFoo<boost::mpl::_> > x;
   x(boost::mpl::identity<Foo>());
   x(boost::mpl::identity<Bar>());

   return 0;
}