c++ - 将基类传递给期望派生的构造函数时，为什么它们都销毁了自己？

Question

我知道您不能期望有一个派生引用指向它的基类，因为它会失去功能，但我很好奇实际会发生什么以了解更多有关事件序列的信息。

#include <iostream>

struct Base
{
    Base() { std::cout << "Base constructor" << std::endl; }
    Base(const Base&) { std::cout << "Base copy constructor" << std::endl; } 
    ~Base() { std::cout << "Base destructor" << std::endl; }
};

struct Derived : public Base
{
    Derived(const Base& b) : Base(b) {  std::cout << "Derived constructor" << std::endl; }
    ~Derived() { std::cout << "Derived destructor" << std::endl; }
};


struct OwnsDerived
{
    OwnsDerived(const Derived& derivedObject) : derivedRef(derivedObject) { std::cout << "OwnsDerived created..." << std::endl; }

    const Derived& derivedRef;
};


int main () {
    std::cout << "Main starts: " << std::endl;
    const Base b;

    std::cout << "\nMain creating object that owns a reference to derived: " << std::endl;
    OwnsDerived s(b);

    std::cout << "\nMain ends: " << std::endl;
    return 0; 
}

我得到的结果是:

Main starts: 
Base constructor

Main creating object that owns a reference to derived: 
Base copy constructor  
Derived constructor   <-- why is a derived trying to be created when passing the wrong type in the constructor
OwnsDerived created...
Derived destructor   //<-- these are treated as temporaries 
Base destructor      //<--

Main ends: 
Base destructor

即使我将引用更改为对象的拷贝，它的行为也类似，这也让我感到惊讶。

我非常感谢任何指示或任何资源来了解这些构造函数/析构函数的更细微的机制!

Answer 1

OwnsDerived 构造函数通过引用 const 获取 Derived，但您传递的是 Base 实例。因此，编译器将尝试将 Base 实例转换为 Derived。这可以通过调用 Derived 中的(隐式)转换构造函数来实现。因此，创建了 Derived 的临时实例。

要修复隐式转换，请将 Derived 构造函数声明为 explicit .

struct Derived : public Base
{
    explicit Derived(const Base& b);
    //...
};