为什么在 C++11 或 C++14 中没有定位迭代器?

No, there is no technical reason. As proof, here is a complete implementation of back_emplacer with a demo of your Use Case 1...

#include <iterator>
#include <vector>
#include <iostream>

template<class Container>
class back_emplace_iterator : public std::iterator< std::output_iterator_tag,
                                                   void, void, void, void >
{
protected:
    Container* container;
public:
    typedef Container container_type;

    explicit back_emplace_iterator(Container& x) : container(&x) {}

    template<class T>
    back_emplace_iterator<Container>&
    operator=(T&& t)
    {
        container->emplace_back(std::forward<T>(t));
        return *this;
    }

    back_emplace_iterator& operator*() { return *this; }
    back_emplace_iterator& operator++() { return *this; }
    back_emplace_iterator& operator++(int) { return *this; }
};

template< class Container >
inline back_emplace_iterator<Container>
back_emplacer( Container& c )
{
    return back_emplace_iterator<Container>(c);
}

struct Demo
{
    int i;
    Demo(int i) : i(i) {}
};

int main()
{
    std::vector<int> x = {1,2,3,4,5};

    std::vector<Demo> y;

    std::copy(x.begin(), x.end(), back_emplacer(y));

    for (auto d : y)
        std::cout << d.i << std::endl;
}

可能的已知问题:operator= 的通用引用是否隐藏了隐式生成的复制/移动 operator=?如果是这样，则需要以在重载解析中击败通用引用的方式明确定义这些.