TypeError:找不到C++类型的to_python(按值)转换器

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

本文关键字:python 按值 转换器 to 找不到 C++ 类型 TypeError      更新时间:2023-10-16

我正试图使用Boost.Python将我的C++类公开给Python

struct Base {
virtual ~Base() {};
virtual char const *Hello() {
printf("Base.Hellon");
return "Hello. I'm Base.";
};
};
struct Derived : Base {
char const *Hello() {
printf("Derived.Hellon");
return "Hello. I'm Derived.";
};
Base &test() {
printf("Derived.testn");
// ...
// After some calculation, we get result reference `instance'
// `instance' can be an instance of Base or Derived.
// ...
return instance;
}
};

我想在python中使用以上类,如下所示:

instance = Derived()
// If method test returns an instance of Base
instance.test().Hello() // Result: "Hello. I'm Base."
// If method test returns an instance of Derived
instance.test().Hello() // Result: "Hello. I'm Derived."

我不知道这个问题有什么好的解决办法。我刚试过这个:

struct BaseWrapper : Base, wrapper<Base> {
char const *Hello() {
printf("BaseWrapper.Hellon");
if (override Hello = this->get_override("Hello")) {
return Hello();
}
return Base::Hello();
}
char const *default_Hello() {
printf("BaseWrapper.default_Hellon");
return this->Base::Hello();
}
};
struct DerivedWrapper : Derived, wrapper<Derived> {
char const *Hello() {
printf("DerivedWrapper.Hellon");
if (override Hello = this->get_override("Hello")) {
return Hello();
}
return Derived::Hello();
}
char const *default_Hello() {
printf("DerivedWrapper.default_Hellon");
return this->Derived::Hello();
}
Base &test() {
printf("DerivedWrapper.testn");
if (override Hello = this->get_override("test")) {
return Hello();
}
return Derived::test();
}
Base &default_test() {
printf("DerivedWrapper.default_testn");
return this->Derived::test();
}
};

和他们,我使用以下代码:

BOOST_PYTHON_MODULE(Wrapper) {
class_<BaseWrapper, boost::noncopyable>("Base")
.def("Hello", &Base::Hello, &BaseWrapper::default_Hello);
class_<DerivedWrapper, boost::noncopyable, bases<Base> >("Derived")
.def("Hello", &Derived::Hello, &DerivedWrapper::default_Hello)
.def("test", &Derived::test,  return_value_policy<copy_non_const_reference>());
}

但当我把上面的代码编译成.so文件,并在python 中使用时

derived = Wrapper.Derived() 
derived.test()

它抛出了一个例外:

TypeError: No to_python (by-value) converter found for C++ type: Base
  1. 这篇文章和我的错误一样,但从另一方面来说,它对我没有太大帮助。Boost.Python调用(通过引用):TypeError:找不到C++类型的to_Python(通过值)转换器:

  2. 这篇文章解决了类似的问题,但对我也没有帮助。https://github.com/BVLC/caffe/issues/3494

我有两个问题:

  1. 如果我尝试的方法是正确的,如何解决TypeError问题
  2. 如果我尝试了错误的方法,那么使用boost.python解决问题的最佳方法是什么

此代码适用于我:

struct Base {
virtual ~Base() {};
virtual char const *hello() {
return "Hello. I'm Base.";
};
};
struct Derived : Base {
char const *hello() {
return "Hello. I'm Derived.";
};
Base &test(bool derived) {
static Base b;
static Derived d;
if (derived) {
return d;
} else {
return b;
}
}
};
BOOST_PYTHON_MODULE(wrapper)
{
using namespace boost::python;
class_<Base>("Base")
.def("hello", &Base::hello)
;
class_<Derived, bases<Base>>("Derived")
.def("test", &Derived::test, return_internal_reference<>())
;
}

测试模块:

>>> import wrapper
>>> d = wrapper.Derived()
>>> d.test(True).hello()
"Hello. I'm Derived."
>>> d.test(False).hello()
"Hello. I'm Base."
>>>