相关疑难解决方法(0)

考虑一个类层次结构,其中A是基类并B派生自A.

如果未定义复制构造函数B,编译器将合成一个.调用时,此复制构造函数将调用基类复制构造函数(即使是合成的构造函数,如果用户没有提供).

#include <iostream>

class A {
    int a;
public:
    A() {
        std::cout << "A::Default constructor" << std::endl;
    }

    A(const A& rhs) {
        std::cout << "A::Copy constructor" << std::endl;
    }
};

class B : public A {
    int b;
public:
    B() {
        std::cout << "B::Default constructor" << std::endl;
    }
};

int main(int argc, const char *argv[])
{
    std::cout << "Creating B" << std::endl;
    B b1;
    std::cout << "Creating …

就像在标题中一样,如何从派生类复制构造函数中调用基类复制构造函数？

根据我的理解，当创建派生类的对象时，基类构造函数会自动调用（如果存在没有参数的构造函数）。复制构造函数似乎不是这种情况：

#include <iostream>

class Base
{
public:

    Base() 
    { 
        std::cout << "Base Constructor called" << std::endl; 
    }   

    Base(const Base& ref)
    {
        std::cout << "Base Copy Constructor called" << std::endl; 
    }
};

class Derived : Base
{
public:

    Derived() 
    { 
        std::cout << "Derived Constructor called" << std::endl; 
    }

    Derived(const Derived& ref) //: Base(ref) // <- without this Base copy constructor doesnt get called
    {
        std::cout << "Derived Copy Constructor called" << std::endl; 
    }
};

int main()
{
    Derived d1;

    Derived d2 …