boost::python：如何覆盖静态属性？

Question

我有一个基类和一个派生类，它们都有一个同名的静态方法。是否可以在保持名称相同的同时公开它们？这会编译，但在导入模块时会引发异常。

struct Base
{
    static std::string say_hi() { return "Hi"; }
};

struct Derived : public Base
{
    static std::string say_hi() { return "Hello"; }
};

BOOST_PYTHON_MODULE(HelloBoostPython)
{
    namespace py = boost::python;

    py::class_<Base>("Base").add_static_property("say_hi", &Base::say_hi);

    py::class_<Derived, py::bases<Base>>("Derived").add_static_property("say_hi", &Derived::say_hi);
}

导入时：

>>> import HelloBoostPython
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
SystemError: initialization of HelloBoostPython raised unreported exception

将最后一行更改为不同的名称可以，但我宁愿覆盖基类的属性：

py::class_<Derived, py::bases<Base>>("Derived").add_static_property("say_hello", &Derived::say_hi);

这可行，但我得到的是类方法而不是属性：

BOOST_PYTHON_MODULE(HelloBoostPython)
{
    namespace py = boost::python;

    py::object base = py::class_<Base>("Base");
    base.attr("say_hi") = Base::say_hi;

    py::object derived = py::class_<Derived, py::bases<Base>>("Derived");
    derived.attr("say_hi") = Derived::say_hi;
}

这将为我提供属性，但如果静态方法不是常量，则在一般情况下不起作用：

BOOST_PYTHON_MODULE(HelloBoostPython)
{
    namespace py = boost::python;

    py::object base = py::class_<Base>("Base");
    base.attr("say_hi") = Base::say_hi();

    py::object derived = py::class_<Derived, py::bases<Base>>("Derived");
    derived.attr("say_hi") = Derived::say_hi();
}

嗯

Answer 1

查看Boost Python 如何暴露静态属性以及它在Python 中的行为方式，看起来这是Boost Python 中的一个错误。

当我尝试导入有问题的示例时出现以下错误：

>>> import HelloBoostPython
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: can't set attribute
>>>

无法设置属性，因为派生类中存在冲突。 Derived 类首先从 Base 继承 say_hi 属性，然后尝试定义它自己的。这可以通过将Derived 类中的say_hi 更改为say_hello 来轻松验证，正如作者所做的那样：

py::class_<Derived, py::bases<Base>>("Derived").add_static_property("say_hello", &Derived::say_hi);

然后我们可以看到如下数据描述符help(HelloBoostPython.Derived)：

    class Derived(Base)
...
...
     |  ----------------------------------------------------------------------
     |  Data descriptors defined here:
     |
     |  say_hello
     |
     |  ----------------------------------------------------------------------
     |  Data and other attributes defined here:
     |
     |  __instance_size__ = 24
     |
     |  ----------------------------------------------------------------------
     |  Data descriptors inherited from Base:
     |
     |  say_hi
...
...

但是当我们在python中定义类似的时候：

class Base(object):

    def get_say_hi(self):
        return "hi"

    say_hi = property(get_say_hi)

class Derived(Base):

    def get_say_hi(self):
        return "hello"

    say_hi = property(get_say_hi)

我们可以看到这里我们只有say_hi定义在Derived中：

class Derived(Base)
...
...
 |  ----------------------------------------------------------------------
 |  Data descriptors defined here:
 |
 |  say_hi
 |
 |  ----------------------------------------------------------------------
 |  Data descriptors inherited from Base:
 |
 |  __dict__
 |      dictionary for instance variables (if defined)
 |
 |  __weakref__
 |      list of weak references to the object (if defined)
(END)

从Derived 中删除say_hi 后：

class Base(object):

    def get_say_hi(self):
        return "hi"

    say_hi = property(get_say_hi)

class Derived(Base):
    None

我们从Base继承它然后：

class Derived(Base)
...
...
 |  ----------------------------------------------------------------------
 |  Data descriptors inherited from Base:
 |
 |  __dict__
 |      dictionary for instance variables (if defined)
 |
 |  __weakref__
 |      list of weak references to the object (if defined)
 |
 |  say_hi
(END)

Answer 2

问题在于，当 Boost.Python 的元类尝试声明和设置Derived.say_hi 时，它解析为Base.say_hi.__set__，因为它是从Base 继承的现有类描述符。一个解决方案可能是自己将属性绑定到您的 Derived 类。例如

namespace py = boost::python;

py::class_<Base>("Base").add_static_property("say_hi", &Base::say_hi);

py::class_<Derived, py::bases<Base>> Derived_("Derived");

PyDict_SetItemString(
    py::downcast<PyTypeObject>(Derived_.ptr())->tp_dict,
    "say_hi",
    Derived_.attr("__base__").attr("__dict__")["say_hi"].attr("__class__")(
        py::make_function(&Derived::say_hi)
    ).ptr()
);

这样say_hi 直接绑定到Derived 类作为Boost.Python.StaticProperty 实例。现在是以下代码：

print(Base().say_hi, Base.say_hi, Derived().say_hi, Derived.say_hi)

会正确打印：

Hi Hi Hello Hello