c++ - 如何在取消引用另一种类型后获得你得到的类型？

Question

我如何在编译时找到取消引用后得到的类型？

#include <vector>
#include <memory>

template <class TIteratorToPointerContainer>
class Something
{
public:
    typedef /*the thing you get by dereferencing TIteratorToPointer*/ TPointer;
    typedef /*the thing you get by dereferencing TPointer*/           TValue;
};

int main()
{
    Something<
              typename std::vector< std::shared_ptr<int> >::iterator 
              >::TPointer pointer;
                      // "pointer" is of type std::shared_ptr<int>
    Something< 
              typename std::vector< std::shared_ptr<int> >::iterator 
              >::TValue value;
                    // "value" is of type int
    return 0;
}

我可以使用 C++11 功能。

根据答案编辑:

typedef typename TIteratorToPointerContainer::value_type TPointer;
typedef typename TPointer::element_type TValue;

适用于 std::vector< std::shared_ptr<int> >但不适用于 std::vector< int* > .

Answer 1

#include <type_traits>
#include <utility>

template <class TIteratorToPointerContainer>
class Something
{
private:
     using TPointer_ = decltype( *std::declval<TIteratorToPointerContainer>() );
     using TValue_ = decltype( *std::declval<TPointer>() );
public:
    using TPointer = typename std::remove_reference<TPointer_> :: type;
    using TValue = typename std::remove_reference<TValue_> :: type;
};