Question

如何从索引1到2开始迭代元组？以下不起作用。

using boost::fusion::cons;
typedef cons<A, cons<B, cons<C, cons<D> > > > MyTuple;
MyTuple tuple_;

template <class T>
struct DoSomething{

  DoSomething(T& t) : t_(&t){ }

  template <class U>
  void operator()(U u){
    boost::fusion::at<mpl::int_<u> >(*t_);
  }
  T* t_;
};

boost::mpl::for_each< boost::mpl::range_c<int, 1, 3> >( DoSomething<MyTuple>(tuple_) );

Answer 1

我不确定您的意图，但以下代码是否符合您的目的？我使用fusion而不是mpl。

struct DoSomething {
  template< class U >
  void operator()( U u ) const {
    std::cout << u << '\n'; // an example
  }
};

using namespace boost::fusion; // Sorry, for brevity

iterator_range<
  result_of::advance_c< result_of::begin< MyTuple >::type, 1 >::type
, result_of::advance_c< result_of::begin< MyTuple >::type, 3 >::type
> ir( advance_c< 1 >( begin( tuple_ ) )
    , advance_c< 3 >( begin( tuple_ ) ) );
for_each( ir, DoSomething() );

希望这有帮助

Answer 2

从你的评论来看，你提到的可能是通过制作一个谓词来实现的确定指定类是否具有成员函数的类，并使用fusion::filter_view。以下代码中的DEF_HAS_MEM_FUNC宏在以下位置进行了解释：

Is it possible to write a template to check for a function's existence?

#include <boost/fusion/include/cons.hpp>
#include <boost/fusion/include/filter_view.hpp>
#include <boost/fusion/include/for_each.hpp>
#include <boost/mpl/placeholders.hpp>
using namespace boost;

#define DEF_HAS_MEM_FUNC( name, func_name, signature )  \
  template< class T >                                   \
  struct name {                                         \
    template< class U, U > struct mfp;                  \
                                                        \
    template< class U >                                 \
    static char f( mfp< signature, &U::func_name >* );  \
                                                        \
    template< class > static char (&f(...))[2];         \
                                                        \
    enum { value = (sizeof( f<T>(0) ) == 1) };          \
  }

DEF_HAS_MEM_FUNC( has_f, f, void(U::*)()const );

struct DoSomething {
  template< class U >
  void operator()( U& u ) const {
    u.f();
  }
};

struct A {};

struct B {
  void f() const {}
};

typedef fusion::cons< A, fusion::cons< B > >  MyTuple;

int main()
{
  MyTuple tuple_;
  fusion::filter_view< MyTuple const, has_f< mpl::_ > > v( tuple_ );
  fusion::for_each( v, DoSomething() );
}

我如何迭代元组

2 个答案: