当我尝试在C ++中使用Python实现Python的可迭代对象时(使用boost :: python),我遇到了一个奇怪的问题。 Python似乎总是取消引用前面的一个元素,因此,在结果中它跳过第一个元素并且还取消引用“end”元素。我也不确定我的返回值策略是否正确选择,但如果我用std :: string替换int作为元素类型,它是唯一似乎正常工作的策略。故意选择了迭代器标记 - 我打算实现可迭代对象来访问只能遍历一次的资源。
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
答案 0 :(得分:4)
您的增量后运算符中存在错误。具体来说,您实施的是预 - 增量,而非发布 - 增量:
iterator operator++(int)
{
++pos_;
return *this; // return value *after* having incremented it
}
正确的实施方式是:
iterator operator++(int)
{
iterator tmp(*this);
++pos_;
return tmp; // return saved tmp *before* having incremented it
}
在修复之后:
>>> list(pythonIterator.Foo())
... snip lots of output ...
[0, 1, 2]
答案 1 :(得分:3)
所以,让我通过建议使用Boost Iterator为您处理迭代器复杂性来回报您的服务:
<强> Live On Coliru 强>
#include <Python.h>
#include <boost/python.hpp>
#include <boost/iterator/iterator_facade.hpp>
class Foo
{
public:
struct iterator : boost::iterator_facade<iterator, int, boost::single_pass_traversal_tag, int>
{
iterator(int i) : current_(i) {}
bool equal(iterator const& other) const { return current_ == other.current_; }
int dereference() const { return current_; }
void increment() { ++current_; }
private:
int current_;
};
iterator begin() { return 0; }
iterator end() { return 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::return_by_value>>{});
}
打印:
$ ./test.py
0
1
2
当然,使迭代器返回副本的选择受到源范围缺失的启发。 (显然iterator_facade
完全适合于左值参考(如果你需要)