c++ - 显式传递基类型

Question

我有一个显式函数，它引用一个类的基类型。传递它的正确方法是什么？

我目前正在进行静态转换:

#include <iostream>

using namespace std;

struct Base
{
    Base() { cout << "Base Constructor" << endl; }
    Base(Base const& c) { cout << "Base-Base Constructor" << endl; }
};

struct Derived : public Base
{
    Derived() { cout << "Derived Constructor" << endl; }
    explicit Derived(Base const& c) { cout << "Derived-Base Constructor" << endl; }
    Derived(Derived const& c) { cout << "Derived-Derived Constructor" << endl; }
};

int main()
{
    Base B;
    cout << "\n";
    Derived D;
    cout << "\n";
    Base* test1 = new Derived(D);
    cout << "\n";
    Base* test3 = new Derived(static_cast<Base>(D));
    cout << "\n";
    Base* test2 = new Derived(B);
    cout << "\n";


    return 0;
}

但这会调用基类的复制构造函数。

我可以通过 *static_cast<Base*>(&D) ，但这似乎有点骇人听闻。我觉得我只是忽略了一种简单的方法来做到这一点。谢谢。

Answer 1

使用这个:

static_cast<Base&>(D)

或者这个:

static_cast<const Base&>(D)