Boost.Python.ArgumentError:python str未转换为std :: string

时间:2012-09-05 08:50:49

标签: c++ boost boost-python

我收到一个错误,我无法通过Boost.Python包装一个相当简单的c ++类。 首先,课程:

#include <boost/python.hpp>
#include <boost/shared_ptr.hpp>
#include <vector>
#include <string>

class token {
    public:
    typedef boost::shared_ptr<token> ptr_type;

    static std::vector<std::string> empty_context;
    static std::string empty_center;

    token(std::string& t = empty_center,
            std::vector<std::string>& left = empty_context,
            std::vector<std::string>& right = empty_context)
        : center(t), left_context(left), right_context(right) {}
    virtual ~token(void)  {}  

    std::vector<std::string> left_context;
    std::vector<std::string> right_context;
    std::string center;
};

std::string token::empty_center;
std::vector<std::string> token::empty_context;

通过

暴露给python
BOOST_PYTHON_MODULE(test) {
    namespace bp = boost::python;
    bp::class_<token, token::ptr_type>("token")
    .def(bp::init<std::string&, bp::optional<std::vector<std::string>&,
            std::vector<std::string>&> >())
    ;   
}

然后,当尝试在python中使用它时:

Python 2.7.2 (default, Aug 19 2011, 20:41:43) [GCC] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> from test import token
>>> word = 'aa'
>>> t = token(word)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
Boost.Python.ArgumentError: Python argument types in
    token.__init__(token, str)
did not match C++ signature:
    __init__(_object*, std::string {lvalue})
    __init__(_object*, std::string {lvalue}, std::vector<std::string,     std::allocator<std::string> > {lvalue})
    __init__(_object*, std::string {lvalue}, std::vector<std::string,     std::allocator<std::string> > {lvalue}, std::vector<std::string, std::allocator<std::string>     > {lvalue})
    __init__(_object*)
>>> 

有人能指出我的来源吗?不应该Boost.Python负责将python的str转换为std::string吗?

涉及的软件/库的版本:

Boost version: 1.46.1
Python version: 2.7.2
Compiler: g++ (SUSE Linux) 4.6.2
Makefile generation: cmake version 2.8.6
Make: GNU Make 3.82

1 个答案:

答案 0 :(得分:9)

您的参数类型为std::string &,是std::string的可修改引用。 Python字符串是不可变的,因此无法转换为可修改的引用。您的参数类型应为const std::string &