有没有办法打破boost :: mpl for_each？

Question

简单的问题,让我给出一些背景知识:

我有一个mpl::vector类型,其中每个类型都有一个id,在运行时我使用mpl::for_each迭代这个向量并找到给定id的匹配类型.但是一旦发现,继续循环是没有意义的,所以 - 问题是,有没有办法摆脱它(没有抛出异常)？

Answer 1

实现像find_if我更改for_each(调用它exec_if)以获取bool模板参数之类的东西.该bool如果执行应与下一个序列指示,或影响提前返回.

#include <iostream>

#include <boost/mpl/vector.hpp>
#include <boost/mpl/is_sequence.hpp>
#include <boost/mpl/begin_end.hpp>
#include <boost/mpl/apply.hpp>
#include <boost/mpl/bool.hpp>
#include <boost/mpl/next_prior.hpp>
#include <boost/mpl/deref.hpp>
#include <boost/type_traits/is_same.hpp>
#include <boost/mpl/assert.hpp>

namespace mpl = boost::mpl;

template< bool done = true >
struct exec_if_impl
{
  template<typename Iterator, typename LastIterator, typename Pred, typename Exec>
  static void execute(Iterator*, LastIterator*, Pred const&, Exec const&)
  {
  }
};

template<>
struct exec_if_impl<false>
{
  template<typename Iterator, typename LastIterator, typename Pred, typename Exec>
  static void execute(Iterator*, LastIterator*, Pred const& f, Exec const& e)
  {
    typedef typename mpl::deref<Iterator>::type item;

    if (!f(static_cast<item*>(0)))
    {
      typedef typename mpl::next<Iterator>::type iter;
      exec_if_impl<boost::is_same<iter, LastIterator>::value>
        ::execute(static_cast<iter*>(0), static_cast<LastIterator*>(0), f, e);
    }    
    else
      e(static_cast<item*>(0));
  }
};

template<typename Sequence, typename Pred, typename Exec>
inline
void exec_if(Pred const& f, Exec const& e, Sequence* = 0)
{
  BOOST_MPL_ASSERT(( mpl::is_sequence<Sequence> ));

  typedef typename mpl::begin<Sequence>::type first;
  typedef typename mpl::end<Sequence>::type last;

  exec_if_impl<boost::is_same<first,last>::value>
    ::execute(static_cast<first*>(0), static_cast<last*>(0), f, e);
}

namespace msg
{
  struct m1 { enum { TYPE = 1 }; static const char* name() { return "m1"; } };
  struct m2 { enum { TYPE = 2 }; static const char* name() { return "m2"; } };
  struct m3 { enum { TYPE = 3 }; static const char* name() { return "m3"; } };
  struct m4 { enum { TYPE = 4 }; static const char* name() { return "m4"; } };
  struct m5 { enum { TYPE = 5 }; static const char* name() { return "m5"; } };
}

struct checker
{
  checker(int chk_type) : type(chk_type) {}

  template <typename Mtype>
  bool operator()(Mtype* = 0) const
  {
    return Mtype::TYPE == type;
  }

  int type;
};

struct exec
{
  template <typename Mtype>
  void operator()(Mtype* = 0) const
  {
    std::cout << Mtype::name() << " executed" << std::endl;
  }
};

int main(void)
{
  typedef mpl::vector<msg::m1, msg::m2, msg::m3, msg::m4, msg::m5> mseq;

  checker chk(3); 

  exec_if<mseq>(chk, exec());

  return 0;
}

我将其更改为exec_if,所以现在当谓词匹配时,将使用类型触发执行的仿函数 - 这正是我所需要的.