Question

这是一个抽象基类和一个具体的子类，我想通过Cython向Python公开：

class NodeDistance {

protected:
    const Graph& G;

public:
    NodeDistance(const Graph& G);
    virtual ~NodeDistance();
    virtual void preprocess() = 0;
    virtual double distance(node u, node v) = 0;
};


class NeighborhoodDistance: public NetworKit::NodeDistance {

public:
    NeighborhoodDistance(const Graph& G);
    virtual ~NeighborhoodDistance();
    virtual void preprocess();
    virtual double distance(node u, node v);
};

这是我第一次尝试声明Cython类的接口。为了避免在cppclass es和Python包装器类之间发生命名冲突，我将每个Class声明为_Class，后跟其正确的名称"Namespace::Class"。

cdef extern from "../cpp/distmeasures/NodeDistance.h":
    cdef cppclass _NodeDistance "NetworKit::NodeDistance":
        _NodeDistance(_Graph G) except +
        void preprocess() except +
        double distance(node, node) except +


cdef extern from "../cpp/distmeasures/NeighborhoodDistance.h":
    cdef cppclass _NeighborhoodDistance(_NodeDistance) "NetworKit::NeighborhoodDistance":
        _NeighborhoodDistance(_Graph G) except +
        void preprocess() except +
        double distance(node, node) except +

但是现在我在尝试表达_NeighborhoodDistance是_NodeDistance的子类时遇到语法错误。我做错了什么？

Error compiling Cython file:
------------------------------------------------------------
...
        void preprocess() except +
        double distance(node, node) except +


cdef extern from "../cpp/distmeasures/NeighborhoodDistance.h":
    cdef cppclass _NeighborhoodDistance(_NodeDistance) "NetworKit::NeighborhoodDistance":
                                                   ^
------------------------------------------------------------

_NetworKit.pyx:1698:52: Syntax error in C++ class definition

Answer 1

我认为你甚至不能在Cython 0.20.1中表达基类和重命名的组合。您可以不重命名类并在cdef extern from：

中指定命名空间

# C++ classes shown at the end
cdef extern from "example.hpp" namespace "example":
    cdef cppclass Base:
        void some_method() except +

    cdef cppclass Derived(Base):
        void some_method() except +

...或者不指定继承：

cdef extern from "example.hpp" namespace "example":
    cdef cppclass Base "example::Base":
        void some_method() except +

    cdef cppclass Derived "example::Derived":
        void some_method() except +

无论哪种方式，Cython似乎都不完全理解C ++继承，你需要一个显式的强制转换：

def test():
    cdef Derived d
    cdef Base *p = <Base *>&d
    p.some_method()

这很难看，因为演员有效地关闭了C ++中的类型检查，但是要小心它可以安全使用。（在其他情况下，Cython的类型检查需要在C / C ++中不需要的强制转换，这是非常不幸的。）

供参考，以下是我使用的课程：

// example.hpp
#include <cstdio>

namespace example {
    struct Base {
        virtual void some_method() = 0;
        virtual ~Base() = 0;
    };

    struct Derived {
        virtual void some_method()
        {
            std::puts("Hello!");
        }
        ~Derived()
        {
        }
    };
}

用于声明具有别名的类层次结构的Cython语法

1 个答案: