Question

boost :: variant有成员类型，这是某种boost :: mpl结构。有没有办法在编译时获取该结构中的类型索引，所以在运行时的后期我可以做

if(myVariantInstance.which() == typeIndex)
{
   /*...*/
}

而不是

if(myVariantInstance.type() == typeid(ConcreteType))
{
  /*...*/
}

Answer 1

如果你感兴趣，我找到了在没有boost :: mpl的情况下获取boost :: variant类型索引的解决方案。

#include <iostream>
#include <type_traits>

#include <boost/variant/variant.hpp>

using myvariant = boost::variant<int, bool, double, int>;

template <typename T, typename ... Ts>
struct type_index;

template <typename T, typename ... Ts>
struct type_index<T, T, Ts ...>
    : std::integral_constant<std::size_t, 0>
{};

template <typename T, typename U, typename ... Ts>
struct type_index<T, U, Ts ...>
    : std::integral_constant<std::size_t, 1 + type_index<T, Ts...>::value>
{};


template <typename T, typename ... Ts>
struct variant_first_same_type_idx;

template <typename T, typename Head, typename ... Tail>
struct variant_first_same_type_idx<T, boost::variant<Head, Tail ... >>
    : type_index<T, Head, Tail ...>
{};

int main()
{
    std::cout << variant_first_same_type_idx<int, myvariant>::value << std::endl;
    std::cout << variant_first_same_type_idx<bool, myvariant>::value << std::endl;
    std::cout << variant_first_same_type_idx<double, myvariant>::value << std::endl;
}

该程序的输出是：

0
1
2

Answer 2

这有点令人费解，可能有更好的方法，但你可以使用boost :: mpl :: copy。根据你的评论中的例子，这里应该有用的东西：

#include <boost/variant.hpp>
#include <boost/mpl/copy.hpp>
#include <boost/mpl/find.hpp>
#include <boost/mpl/vector.hpp>

typedef boost::mpl::vector<int, long, char> MyMplVector;
typedef boost::mpl::find<MyMplVector, long>::type MyMplVectorIter;
static_assert(MyMplVectorIter::pos::value == 1, "Error");

typedef boost::variant<int, long, char> MyVariant;
typedef boost::mpl::vector<> EmptyVector;
typedef boost::mpl::copy<
  MyVariant::types,
  boost::mpl::back_inserter<EmptyVector>>::type ConcatType;

typedef boost::mpl::find<ConcatType, long>::type MyVariantTypesIter;
static_assert(MyVariantTypesIter::pos::value == 1, "Error");

Answer 3

#include <boost/mpl/index_of.hpp>    
#include <iostream>

typedef boost::variant<int, std::string> VARIANT;
std::ostream &operator<<(std::ostream &_rS, const VARIANT&_r)
{    switch (_r.which())
     {    default:
             return _rS << "what the *";
          case boost::mpl::index_of<VARIANT::types, int>::type::value:
             return _rS << boost::get<int>(_r);
          case boost::mpl::index_of<VARIANT::types, std::string>::type::value:
             return _rS << boost::get<std::string>(_r);
     }
}

PS。我一直对使用访问者访问模式的人感到好奇...

PPS。我知道不需要将输出运算符实现为boost :: variant已经提供了一个-只是出于解释目的...

使用boost :: mpl获取boost :: variant的类型索引

3 个答案: