Question

我正在学习boost :: mpl，我有以下课程 -

#include <string>

#include <boost/mpl/vector.hpp>
#include <boost/mpl/size.hpp>
#include <boost/mpl/at.hpp>

#include <boost/fusion/include/mpl.hpp>
#include <boost/fusion/container.hpp>


using namespace boost;
using namespace std;

template< typename T > 
class Demo
{
public:
    typedef boost::mpl::size<T> NumDimensions;

    template< size_t D >
    struct Dim
    {
        typedef typename boost::mpl::at_c< T, D >::type Type;
    };

    template< size_t D >
    typename Dim<D>::Type& GetElement()
    {
        return fusion::at_c<D>(elements_);
    }

private:
    typename fusion::result_of::as_vector< T >::type elements_;
};

只要我使用带有默认构造函数（或默认类型）的类型

，这样就可以了

int main(int argc, char *argv[])
{
    typedef Demo< boost::mpl::vector< int, std::string > > D1;
    D1 d;
    D1::Dim<0>::Type &i = d.GetElement<0>();

    i = "Hello World!";

    cout << " " << i << endl;
}

但是，如果我使用没有默认构造函数的类型，则会抛出编译器错误，因为向量初始化失败。是否有一种标准方法可以正确地初始化成员（在构造函数中）而不需要使用指针/引用？

Answer 1

您可以使用fusion::vector的构造函数：

#include <string>

#include <boost/mpl/vector.hpp>
#include <boost/mpl/size.hpp>
#include <boost/mpl/at.hpp>

#include <boost/fusion/include/mpl.hpp>
#include <boost/fusion/container/vector.hpp>
#include <utility>

struct foo {
    explicit foo(int){}
};

template< typename T > 
class Demo
{
public:
    //You don't need to use variadic templates and perfect forwarding
    //but you may need to emulate them to get this effect (depending on
    //where exactly you need to construct Demo).
    //Another alternative would be to pass in a fusion::vector, and
    //copy construct `elements` with that vector.
    template<typename ...Args>
    Demo(Args&& ...args) : elements_(std::forward<Args>(args)...) {}

    typedef boost::mpl::size<T> NumDimensions;

    template< size_t D >
    struct Dim
    {
        typedef typename boost::mpl::at_c< T, D >::type Type;
    };

    template< size_t D >
    typename Dim<D>::Type& GetElement()
    {
        return boost::fusion::at_c<D>(elements_);
    }

private:
    typename boost::fusion::result_of::as_vector< T >::type elements_;
};

int main() {
    Demo<boost::mpl::vector< foo, std::string > > a(foo(10),"hi");
}

如何实例化一个没有默认构造函数的类型的boost :: fusion :: vector成员变量？

1 个答案: