Python ctype，C++对象销毁

考虑以下python ctype-c++绑定：

// C++
class A
{
public:
    void someFunc();
};

A* A_new() { return new A(); }
void A_someFunc(A* obj) { obj->someFunc(); }
void A_destruct(A* obj) { delete obj; }

# python
from ctypes import cdll

libA = cdll.LoadLibrary(some_path)

class A:
    def __init__(self):
        self.obj = libA.A_new()

    def some_func(self):
        libA.A_someFunc(self.obj)

删除C++对象的最佳方式是什么，因为不再需要该对象。

[编辑] 我添加了建议的删除函数，但问题仍然是由谁以及何时调用该函数。应该尽可能方便。

---

解决方案

您可以实现__del__方法，该方法调用您必须定义的析构函数：

C++

class A
{
public:
    void someFunc();
};

A* A_new() { return new A(); }
void delete_A(A* obj) { delete obj; }
void A_someFunc(A* obj) { obj->someFunc(); }

Python

from ctypes import cdll

libA = cdll.LoadLibrary(some_path)

class A:
    def __init__(self):
        fun = libA.A_new
        fun.argtypes = []
        fun.restype = ctypes.c_void_p
        self.obj = fun()

    def __del__(self):
        fun = libA.delete_A
        fun.argtypes = [ctypes.c_void_p]
        fun.restype = None
        fun(self.obj)

    def some_func(self):
        fun = libA.A_someFunc
        fun.argtypes = [ctypes.c_void_p]
        fun.restype = None
        fun(self.obj)

还请注意，您遗漏了__init__方法上的self参数。此外，您必须显式指定返回类型/参数类型，因为ctype默认为32位整数，而在现代系统上指针可能为64位。

有些人认为__del__是邪恶的。或者，您可以使用with语法：

class A:
    def __init__(self):
        fun = libA.A_new
        fun.argtypes = []
        fun.restype = ctypes.c_void_p
        self.obj = fun()

    def __enter__(self):
        return self

    def __exit__(self, exc_type, exc_val, exc_tb):
        fun = libA.delete_A
        fun.argtypes = [ctypes.c_void_p]
        fun.restype = None
        fun(self.obj)

    def some_func(self):
        fun = libA.A_someFunc
        fun.argtypes = [ctypes.c_void_p]
        fun.restype = None
        fun(self.obj)

with A() as a:
    # Do some work
    a.some_func()