std :: shared_ptr和继承

ssb 发表于 Dev

我在shared_ptr继承的类之间进行自动类型转换时遇到一些问题。

我的类结构如下，一个基类Base和两个派生类Derived1和Derived2。

// Base class
class Base {
protected:
  ...
  ...
public:
  Base() = default;
  virtual ~Base() = default;
  virtual void run() = 0;
  ...
  ...
};

// Derived class
class Derived1: Base {
protected:
  ...
  ...
public:
  Derived1() = default;
  virtual ~Derived1() = default;
  void run() {...}
  ...
  ...
};

// Derived class
class Derived2: Base {
protected:
  ...
  ...
public:
  Derived2() = default;
  virtual ~Derived2() = default;
  void run() {...}
  ...
  ...
};

我有一个功能 doSomething()

void doSomething(std::shared_ptr<Base> ptr) {
  ptr->run();
  ...
}

我用派生类调用函数，如下所示：

doSomething(make_shared<Derived1>())
doSomething(make_shared<Derived2>())

但我得到一个错误，说-

no viable conversion from 'shared_ptr<class Derived1>' to 'shared_ptr<class Base>'
no viable conversion from 'shared_ptr<class Derived1>' to 'shared_ptr<class Base>'

我究竟做错了什么？仅用于static_pointer_cast基本类型是否安全？喜欢 -

doSomething(static_pointer_cast<Base>(make_sahred<Derived2>()))

解决方案我的错...问题是我私下继承了基类。

比尔·林奇

据我所知，您提供的代码可以很好地编译：http : //ideone.com/06RB2W

#include <memory>

class Base {
    public:
        Base() = default;
        virtual ~Base() = default;
        virtual void run() = 0;
};

class Derived1: public Base {
    public:
        Derived1() = default;
        virtual ~Derived1() = default;
        void run() {}
};

class Derived2: public Base {
    public:
        Derived2() = default;
        virtual ~Derived2() = default;
        void run() {}
};

void doSomething(std::shared_ptr<Base> ptr) {
    ptr->run();
}

int main() {
    doSomething(std::make_shared<Derived1>());
    doSomething(std::make_shared<Derived2>());
}