在Boost Python中公开C ++接口 [英] Exposing C++ interface in boost python

问题描述

示例代码说明:

struct Base
{
  virtual int foo() = 0;
};

struct Derived : public Base
{
  virtual int foo()
  {
    return 42;
  }
};

Base* get_base()
{
  return new Derived;
}

BOOST_PYTHON_MODULE(libTestMod)
{
  py::class_<Base>("Base", py::no_init)
    .def("foo", py::pure_virtual(&Base::foo));

  py::def("get_base", get_base, py::return_internal_reference<>()); //ignore mem leak
}

• Base :: foo在python中不会被覆盖
• Base:foo将用c ++实现，但不应公开给python

尝试了以上代码，但无法编译.

Tried the above code but fails to compile.

更新: 编译错误:

/path/to/boostlib/boost/1.53.0-0/common/include/boost/python/object/value_holder.hpp:66:11: error: cannot declare field 'boost_1_53_0::python::objects::value_holder<Base>::m_held' to be of abstract type 'Base'
Main.C:59:8: note:   because the following virtual functions are pure within 'Base':
Main.C:61:15: note:         virtual int Base::foo()

推荐答案

抽象C ++类无法以这种方式暴露给Boost.Python. Boost.Python

Abstract C++ classes cannot be exposed in this manner to Boost.Python. The Boost.Python tutorial gives examples as to how to expose pure virtual functions. In short, when decorating methods with boost::python::pure_virtual, a wrapper type needs to be created to allow C++ to polymorphic resolve the virtual function, and the virtual function implementation will delegate resolving the function polymorphically in the Python object's hierarchy.

struct BaseWrap : Base, boost::python::wrapper<Base>
{
  int foo()
  {
    return this->get_override("foo")();
  }
};

...

boost::python::class_<BaseWrap>("Base", ...)
  .def("foo", boost::python::pure_virtual(&Base::foo))
  ;

有关详细信息，当通过boost::python::class_公开类型时，HeldType默认为公开的类型，并且HeldType是在Python对象中构造的. class_ 文档状态:

For details, when a type is exposed via boost::python::class_, HeldType defaults to the type being exposed, and the HeldType is constructed within a Python object. The class_ documentation states:

模板参数:

• T:正在包装的类
• HeldType:指定包装在T实例[...]中的Python对象中实际嵌入的类型.默认为T.
• T: The class being wrapped
• HeldType: Specifies the type that is actually embedded in a Python object wrapping a T instance [...]. Defaults to T.

因此，boost::python::class_<Base>将失败，因为T = Base和HeldType = Base，以及Boost.Python将尝试将HeldType的对象实例化为表示Base实例的Python对象.此实例化将失败，因为Base是抽象类.

Hence, the boost::python::class_<Base> will fail, because T = Base and HeldType = Base, and Boost.Python will try to instantiate an object of HeldType into a Python object that represents an instance of Base. This instantiation will fail as Base is an abstract class.

这是一个完整的示例，显示了BaseWrap类的使用.

Here is a complete example showing the use of a BaseWrap class.

#include <boost/python.hpp>

struct Base
{
  virtual int foo() = 0;
  virtual ~Base() {}
};

struct Derived : public Base
{
  virtual int foo()
  {
    return 42;
  }
};

Base* get_base()
{
  return new Derived;
}

namespace python = boost::python;

/// @brief Wrapper that will provide a non-abstract type for Base.
struct BaseWrap : Base, python::wrapper<Base>
{
  BaseWrap() {}

  BaseWrap(const Base& rhs)
    : Base(rhs)
  {}

  int foo()
  {
    return this->get_override("foo")();
  }
};

BOOST_PYTHON_MODULE(example)
{
  python::class_<BaseWrap>("Base")
    .def("foo", python::pure_virtual(&Base::foo));
    ;

  python::def("get_base", &get_base,
              python::return_value_policy<python::manage_new_object>());
}

及其用法:

>>> import example
>>> class Spam(example.Base):
...     pass
... 
>>> Spam().foo()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
RuntimeError: Pure virtual function called
>>> class Egg(example.Base):
...     def foo(self):
...         return 100
... 
>>> e = Egg()
>>> e.foo()
100
>>> d = example.get_base()
>>> d.foo()
42

---

可以通过在没有默认初始化器(boost::python::no_init)和不可复制(boost::noncopyable)的情况下公开Boost.Python中的抽象类.缺少初始化器会阻止从其派生Python类型，从而有效地防止覆盖.此外，由Derived在C ++中实现Base::foo()的实现细节是无关紧要的.如果Python根本不了解foo()方法，请省略通过def()公开它.

---

It is possible to expose an abstract class in Boost.Python by exposing it with no default initializer (boost::python::no_init) and non-copyable (boost::noncopyable). The lack of an initializer prevents Python types from deriving from it effectively preventing overriding. Additionally, the implementation detail that Base::foo() is implemented within C++ by Derived is inconsequential. If Python should not know about a foo() method at all, then omit exposing it via def().

#include <boost/python.hpp>

struct Base
{
  virtual int foo() = 0;
  virtual ~Base() {}
};

struct Derived
  : public Base
{
  virtual int foo()
  {
    return 42;
  }
};

struct OtherDerived
  : public Base
{
  virtual int foo()
  {
    return 24;
  }
};

Base* get_base()
{
  return new Derived;
}

Base* get_other_base()
{
  return new OtherDerived;
}

BOOST_PYTHON_MODULE(example)
{
  namespace python = boost::python;
  python::class_<Base, boost::noncopyable>("Base", python::no_init)
    ;

  python::class_<Derived, python::bases<Base> >("Derived", python::no_init)
    .def("foo", &Base::foo)
    ;

  python::class_<OtherDerived, python::bases<Base> >(
      "OtherDerived", python::no_init)
    ;

  python::def("get_base", &get_base,
              python::return_value_policy<python::manage_new_object>());

  python::def("get_other_base", &get_other_base,
              python::return_value_policy<python::manage_new_object>());
}

互动用法:

>>> import example
>>> b = example.get_base()
>>> b.foo()
42
>>> b = example.get_other_base()
>>> b.foo()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'OtherDerived' object has no attribute 'foo'

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