Boost.Python和C++ std ::指针向量

Question

我正在使用Boost.Python为我的C++库创建一个包装器,我遇到了一些麻烦,谷歌搜索一整天都没有产生任何结果.例如,我有以下代码:

class Base
{
public:
    virtual void func() = 0;
};

class Derived : public Base
{
public:
    virtual void func()
    {
        cout << "Derived::func()"<< endl;
    }
};


// wrapper for Base
struct BaseWrapper : Base, python::wrapper<Base>
{
    virtual void func()
    {
        this->get_override("func");
    }
};


Base* makeDerived()
{
    return new Derived;
}

vector<Base*>* makeDerivedVec()
{
    vector<Base*> *v = new vector<Base*>;
    v->push_back(new Derived);
    v->push_back(new Derived);
    v->push_back(new Derived);
    return v;
}

BOOST_PYTHON_MODULE(mylib)
{
    // export Base
    class_<BaseWrapper, noncopyable>("Base")
            .def("func", pure_virtual(&Base::func));

    class_<vector<Base*> >("BasePtrVec")
            .def(vector_indexing_suite<vector<Base*> >());

    // export Derived
    class_<Derived, bases<Base> >("Derived")
            .def("func", &Derived::func);

    // export makeDerived()
    def("makeDerived", &makeDerived, return_value_policy<manage_new_object>());

    // export makeDerivedVec()
    def("makeDerivedVec", &makeDerivedVec, return_value_policy<manage_new_object>());
}

所以,我编译它,在python中导入并试试这个:

b = mylib.Base()b.func()

d = mylib.makeDerived()d.func()

正如预期的那样,第一行抛出一个异常,说b.func()是纯虚拟的,第二行打印出来

派生:: FUNC()

那没关系.

但是代码

dlist = mylib.makeDerivedVec()
for d in dlist:
    d.func()

不起作用,Python引发异常:

TypeError: No to_python (by-value) converter found for C++ type: Base*

为什么它正确处理了makeDerived()返回的Base*并且拒绝使用std :: vector中包含的Base*？我怎样才能使它工作？

Answer 1

您可以通过注册Base*为可用于指向 的类型来解决此问题BaseWrapper*：

class_<BaseWrapper, noncopyable, Base*>("Base")
        .def("func", pure_virtual(&Base::func));

但这似乎意味着Base不能有纯虚函数......