从 C++ 中的派生 class 获取共享指针

Question

我想从派生的 class 中得到 shared_ptr。
std::enable_shared_ptr_from_this 由基础 class 继承并获得 std::shared_ptr<Base>，但不是 std::shared<Derived>。我可以使用 std::reinterpret_pointer_cast 或 std::dynamic_pointer_cast 但是

这么多代码
多重继承安全吗？
或者这是唯一的方法？

示例代码：

class Base : public std::enable_shared_from_this<Base> {

};

class Derived : public Base {
public:
    std::shared_ptr<Derived> GetPointer() const {
        return shared_from_this(); // but it returns std::shared<Base> type
    }
};

提前致谢！

Answer 1

按照您目前的方式进行操作非常安全。

#include <memory>

class Base : public std::enable_shared_from_this<Base>{
public:
    virtual int32_t getid() = 0;
};
class Derived : public Base{
public:
    int32_t id;
    Derived(int32_t id){
        this->id = id;
    }
    ~Derived(){
        
    }
    int32_t getid(){
        return id;
    }
};
int main(){
    std::shared_ptr<Derived> child = std::make_shared<Derived>(1033);
    std::shared_ptr<Base> parent = child;//This needs no extra work
    std::shared_ptr<Derived> secondchildinstance = std::static_pointer_cast<Derived>(parent);//this needs casting
    if(secondchildinstance!=nullptr){
        //shared_from_this can be called directly from the child without an explicit method
        std::shared_ptr<Base> secondparentinstance = secondchildinstance->shared_from_this();
        
        //shared_from_this would require downcasting to get the child
        std::shared_ptr<Derived> thirdchildinstance =  std::static_pointer_cast<Derived>(secondchildinstance->shared_from_this());//requires casting
        
        if(thirdchildinstance!=nullptr){
            printf("use count:%ld id:%d",thirdchildinstance.use_count(),thirdchildinstance->getid());
        }
    }
    return 0;
}

您可能可以通过包装方法使 down-casting 从 parent 到 child 的工作更轻松。

需要注意的是，您不需要显式方法来调用 shared_from_this，因为 parent 已经继承了它，您可以调用 child->shared_from_this() 直接但它会给出需要向下转换的基础 class 的共享实例。

从 C++ 中的派生 class 获取共享指针

Getting shared pointer from this of derived class in C++

c++

stl

shared-ptr