如何通过以下c++代码使用SWIG处理python中抽象方法的映射:
class A : Base {
virtual int f() = 0;
};
class B : public A {
int f() { return 10 };
};
class C : public A {
int f() { return 20 };
};
std::map< std::string, std::shared_ptr<A>> my_map;
在python中,我也想做类似的事情:
my_B = B()
my_map["foo"] = my_B
或更简单:
my_map["foo"] = B()
为了使用跨语言多态性,我必须精确说明A或B可能是导演类。
我的问题:
问题,例如删除my_B。我如何轻松转移
从python到C++的“my_B”所有权?
非常感谢你的帮助
一种。
最佳答案
这是一个跟踪构建/销毁的工作示例,以显示共享指针的引用计数正在工作:
测试
#include <map>
#include <memory>
#include <string>
#include <iostream>
class A {
public:
virtual int f() = 0;
A() { std::cout << "A()" << std::endl; }
virtual ~A() { std::cout << "~A()" << std::endl; }
};
class B : public A {
public:
int f() { return 10; }
B() { std::cout << "B()" << std::endl; }
virtual ~B() { std::cout << "~B()" << std::endl; }
};
class C : public A {
public:
int f() { return 20; }
C() { std::cout << "C()" << std::endl; }
virtual ~C() { std::cout << "~C()" << std::endl; }
};
std::map< std::string, std::shared_ptr<A>> my_map;
测试
%module test
%{
#include "test.h"
%}
%include <std_map.i>
%include <std_shared_ptr.i>
%include <std_string.i>
// declare all visible shared pointers so SWIG generates appropriate wrappers
// before including the header.
%shared_ptr(A)
%shared_ptr(B)
%shared_ptr(C)
%include "test.h"
// Declare the template instance used so SWIG will generate the wrapper.
%template(Map) std::map<std::string, std::shared_ptr<A>>;
输出:
>>> import test
>>>
>>> m=test.cvar.my_map # global variables are in module's cvar.
>>> m['foo'] = test.C()
A()
C()
>>> m['foo'].f()
20
>>> del m['foo'] # only reference, so it is freed
~C()
~A()
>>> b = test.B() # 1st reference
A()
B()
>>> m['bar'] = b # 2nd reference
>>> del m['bar'] # NOT freed.
>>> del b # now it is freed.
~B()
~A()
关于python - SWIG C++/Python:如何处理抽象类的shared_ptr的std::map,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61484768/