定义为基类的 unique_ptr 然后作为派生类强制转换[重复]

Question

我希望看到来自两个不同派生类的两个对象打印，关键是在定义为基类后将对象强制转换为派生类。这是代码：

#include <iostream>
#include "memory"

using namespace std;

template<class T>
class Base
{
public:
    Base() {};
    ~Base() {};
    void Print()
    {
        std::cout << "Print from Base" << std::endl;
    }
};

class Derived1: public Base<int>
{
public:
    Derived1() {}
    void Print()
    {
        std::cout << "Print from Derived1" << std::endl;
    }
};

class Derived2: public Base<int>
{
public:
    Derived2() {}
    void Print()
    {
        std::cout << "Print from Derived2"<< std::endl;
    }
};

class user
{
public:
    user();
    user(int type)
    :m_type(type)
    {
        switch(type) // change should be done here?
        {
        case 1:
            m_ptr = make_unique<Derived1>(); // m_ptr still base? why not derived
            break;
        case 2:
            m_ptr = make_unique<Derived2>();;
            break;
        default:
            break;
        }
    }

    ~user() {};
    void UserPrint()
    {
        m_ptr->Print();
    }
private:
    std::unique_ptr<Base<int>> m_ptr;
    int m_type;
};

int main()
{
    user a(1);
    user b(2);

    a.UserPrint();
    b.UserPrint();

    return 0;
}

我期待看到

Print from Derived1
Print from Derived2

但相反，我看到了

Print from Base
Print from Base

如何更改代码以使其正常工作？我更愿意更改

user(int type)

的构造函数。

Answer 1

我认为您对 C++ 中的重写方法有一些误解，因为您缺少（假定的）虚拟方法上的 virtual 关键字：

#include <iostream>
#include "memory"

using namespace std;

template<class T>
class Base
{
public:
    Base() {};
    ~Base() {};
    virtual void Print()
    {
        std::cout << "Print from Base" << std::endl;
    }
};

class Derived1: public Base<int>
{
public:
    Derived1() {}
    void Print() override
    {
        std::cout << "Print from Derived1" << std::endl;
    }
};

class Derived2: public Base<int>
{
public:
    Derived2() {}
    void Print() override
    {
        std::cout << "Print from Derived2"<< std::endl;
    }
};

class user
{
public:
    user();
    user(int type)
    :m_type(type)
    {
        switch(type) // change should be done here?
        {
        case 1:
            m_ptr = make_unique<Derived1>(); // m_ptr still base? why not derived
            break;
        case 2:
            m_ptr = make_unique<Derived2>();;
            break;
        default:
            break;
        }
    }

    ~user() {};
    void UserPrint()
    {
        m_ptr->Print();
    }
private:
    std::unique_ptr<Base<int>> m_ptr;
    int m_type;
};

int main()
{
    user a(1);
    user b(2);

    a.UserPrint();
    b.UserPrint();

    return 0;
}