为什么没有调用此模板 class 的复制构造函数？

Question

The copy constructor never gets called (I'm using g++ with -std=c++11):
pointers.cpp:

#include <iostream>

template <typename T>
class SharedPtr
{
private:
    int *m_Count;
    T *m_Ptr;

public:
    SharedPtr(T *ptr = nullptr)
        : m_Count(new int), m_Ptr(ptr)
    {
        *m_Count = 1;
    }

    SharedPtr(const SharedPtr &other)
        : m_Count(other.m_Count), m_Ptr(other.m_Ptr)
    {
        std::cout << "Copied: " << *m_Count << std::endl;
        (*m_Count)++;
    }

    ~SharedPtr()
    {
        (*m_Count)--;
        if (*m_Count == 0)
        {
            delete m_Ptr;
            delete m_Count;
        }
    }

    T &operator*()
    {
        return *m_Ptr;
    }
};

main.cpp:

#include "pointers.cpp"
#include <iostream>

class Entity
{
public:
    Entity()
    {
        std::cout << "Entity created!" << std::endl;
    }
    ~Entity()
    {
        std::cout << "Entity destroyed!" << std::endl;
    }
};

int main(int argc, char *argv[])
{
    {
        SharedPtr<Entity> e0;
        std::cin.get();
        {
            SharedPtr<Entity> sharedEntity = new Entity();
            std::cin.get();
            e0 = sharedEntity;
            std::cin.get();
        }
        std::cin.get();
    }

    std::cin.get();
}

I looked at other similar questions asked here, but none of them worked. The copy constructor should be ClassName(const ClassName &other) , right?

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Answer 1

None of the code in main() calls the copy contructor of SharedPtr :

• SharedPtr<Entity> e0; calls SharedPtr(T *ptr = nullptr)
• SharedPtr<Entity> sharedEntity = new Entity(); calls SharedPtr(T *ptr = nullptr)
• e0 = sharedEntity; calls operator=() which is not defined in your SharedPtr