如何避免使用临时变量的地址

我的代码目前设置的方式，我觉得我必须在分割错误和使指针指向临时对象之间做出选择。代码如下:

#include <memory>
#include <cstddef>
#include <iostream>
template <typename T> class Node;
template <typename T> class List {
    public:
        typedef std::size_t size_type;
        typedef T value_type;
        typedef T& reference;
        typedef const T& const_reference;
        class iterator {
            public:
                iterator() {
                    isNull = true;
                }
                iterator(const T val) {
                    isNull = false;
                    data = val;
                    prev = this;
                    next = &iterator();  #PROBLEMATIC LINE
                }
                iterator operator++() {
                    if(this->next->isNull)
                        return iterator();
                    iterator old_it = *this;
                    this->prev = &old_it;
                    this->next = this->next->next;
                    this->data = this->next->data;
                    return *this;
                }
                T& operator*() {
                    return data;
                }
                iterator* next;
                iterator* prev;
                T data;
                bool isNull;
        };
        List() {
            _begin = iterator();
            _end = _begin;
        }
        List(size_type n, T val) {
            _begin = iterator(val);
            _end = *(_begin.next);
            _end.prev = &_begin;
        }
        void push_back(T val) {
            iterator temp = iterator();
            _end.data = val;
            _end.next = &temp;
            temp.prev = &_end;
            _end = temp;
        }
        iterator begin() {return _begin;}
        iterator end() {return _end;}
    private:
        iterator _begin;
        iterator _end;
};
int main() {
    List<int> derp= List<int>(3,3);
    List<int>::iterator i = derp.begin();
    std::cout << *i;
    derp.push_back(4);
    std::cout << i.data;
    ++i;
    std::cout << *i;
}

上面的代码接受一个临时变量的地址。当我从

next = &iterator();

*next = iterator();

代码得到分段错误，但我不知道为什么。任何帮助都会很感激。以及您可能碰巧注意到的代码中的任何严重缺陷。