Base class of non exposed derived class in Python side?
- Ngôn ngữ chính
- C++
- Star
- 537
- Fork
- 223
- Merge trung bình
- 11 giờ 22 phút
- Pull request đã merge (30 ngày)
- 2
Mô tả
Hello, I have a use case where a polymorphism factory can instantiate object of a non exposed class. I have a multi-level inheritance scheme and the last level is not exposed to Python. What I try to understand is why the Python object of the last level takes the type of the top-most class in the inheritance hierarchy and not the one just above?
Here is a minimal example to reproduce the behavior:
C++ side
```
class Base
{
public:
Base() {}
virtual ~Base() = default;
};
class Derived: public Base
{
public:
Derived(int value) { m_value = value; }
~Derived() = default;
int getValue() { return m_value; }
private:
int m_value;
};
class DerivedTwo: public Derived
{
public:
DerivedTwo(int value): Derived(value) {}
~DerivedTwo() = default;
};
std::shared_ptr factory()
{
return std::make_shared(9);
}
std::shared_ptr factoryTwo()
{
return std::make_shared(99);
}
BOOST_PYTHON_MODULE(mymodule)
{
using namespace boost::python;
class_, boost::noncopyable>("Base", init<>());
class_, std::shared_ptr, boost::noncopyable>("Derived", init())
.def("get_value", &Derived::getValue)
;
def("factory", &factory);
def("factory_two", &factoryTwo);
}
```
Python side
```
import mymodule
obj = mymodule.factory()
print(type(obj))
obj2 = mymodule.factory_two()
print(type(obj2))
```
Output
```
<-- WHY?
```
Obviously, if I expose DerivedTwo class, all is OK.
Thanks for your help.
Hướng dẫn đóng góp
Chưa lập chỉ mục được hướng dẫn đóng góp cho kho mã nguồn này
Đánh giá
Issue này chưa được đánh giá.