继承的构造函数

Posted 2020-10-24 SHAZI

在C++11新标准中，派生类能够重用其直接基类的构造函数

#include <iostream>
using namespace std;

struct Base
{
    Base() { cout << "construct" << endl; }
};
struct Derived : public Base
{
    using Base::Base; //继承Base的构造函数
};

int main()
{
    Derived d;
    return 0;
}

一个构造函数的声明不会改变该函数的访问级别

#include <iostream>
using namespace std;

struct Base
{
    Base() { cout << "construct" << endl; }
};
struct Derived : public Base
{
   private://不会改变继承的构造函数的访问级别，仍然是publiic
     using Base::Base; 
};

int main()
{
    Derived d;
    return 0;
}

当一个基类的构造函数含有默认实参，这些实参并不会被继承。相反，派生类将获得多个继承的构造函数

#include <iostream>
using namespace std;

struct Base
{
    Base(int x=1) { cout << "construct" <<x<< endl; }
};
struct Derived : public Base
{
    //实际上继承了2个构造函数
     using Base::Base; 
};

int main()
{
    Derived d;
    Derived d1(2);
    return 0;
}

如果派生类定义的构造函数与基类的构造函数有相同的参数列表，该构造函数不会被继承

#include <iostream>
using namespace std;

struct Base
{
    Base() { cout << "construct" << endl; }
    Base(int) { cout << "construct int" << endl; }
};
struct Derived : public Base
{

    using Base::Base;
    //不会继承Base的默认构造函数
    Derived() { cout << "Derived class construct" << endl; };
};

int main()
{
    Derived d;
    Derived d(1);
    return 0;
}