为什么boost :: python迭代器会跳过第一个元素？

尝试在C ++中使用Python实现可迭代对象时遇到一个奇怪的问题（使用boost :: python）。Python似乎总是取消引用前面的一个元素，因此，结果是它跳过了第一个元素，同时也取消了对“ end”元素的引用。我也不确定是否正确选择了返回值策略，但是如果我将int替换为std :: string作为元素类型，这似乎是唯一可以正常工作的策略。故意选择了迭代器标记-我打算实现可迭代对象以访问只能被遍历一次的资源。

C ++代码：

#include <Python.h>
#include <boost/python.hpp>

#include <iostream>
#include <iterator>

int nextInstance{0};

class Foo
{
public:
    class iterator : public std::iterator<std::input_iterator_tag, int>
    {
    public:
        iterator() = delete;
        iterator& operator=(const iterator&) = delete;

        iterator(const iterator& other)
        :
            instance_(nextInstance++),
            pos_(other.pos_)
        {
            std::cout << instance_ << " copy ctor " << other.instance_ << " (" << pos_ << ")\n";
        }

        explicit iterator(int pos)
        :
            instance_(nextInstance++),
            pos_(pos)
        {
            std::cout << instance_ << " ctor (" << pos_ << ")\n";
        }

        bool operator==(iterator& other)
        {
            std::cout << instance_ << " operator== " << other.instance_ << " (" << pos_ << ", " << other.pos_ << ")\n";

            return pos_ == other.pos_;
        }

        int& operator*()
        {
            std::cout << instance_ << " operator* (" << pos_ << ")\n";

            return pos_;
        }

        iterator operator++(int)
        {
            ++pos_;

            std::cout << instance_ << " operator++ (" << pos_ << ")\n";

            return *this;
        }

        ~iterator()
        {
            std::cout << instance_ << " dtor\n";
        }

    private:
        const int instance_;
        int       pos_{0};

    };

    iterator begin()
    {
        std::cout << "begin()\n";

        return iterator(0);
    }

    iterator end()
    {
        std::cout << "end()\n";

        return iterator(3);
    }
};

BOOST_PYTHON_MODULE(pythonIterator)
{
    boost::python::class_<Foo, boost::noncopyable>("Foo", boost::python::init<>())
        .def("__iter__", boost::python::iterator<Foo, boost::python::return_value_policy<boost::python::copy_non_const_reference>>{});
}

Python代码：

#!/usr/bin/python

import pythonIterator

foo = pythonIterator.Foo()

for i in foo:
    print i

输出：

end()
0 ctor (3)
begin()
1 ctor (0)
2 copy ctor 1 (0)
3 copy ctor 0 (3)
1 dtor
0 dtor
4 copy ctor 2 (0)
5 copy ctor 3 (3)
3 dtor
2 dtor
4 operator== 5 (0, 3)
4 operator++ (1)
6 copy ctor 4 (1)
6 operator* (1)
6 dtor
1
4 operator== 5 (1, 3)
4 operator++ (2)
7 copy ctor 4 (2)
7 operator* (2)
7 dtor
2
4 operator== 5 (2, 3)
4 operator++ (3)
8 copy ctor 4 (3)
8 operator* (3)
8 dtor
3
4 operator== 5 (3, 3)
5 dtor
4 dtor