C ++：如何在不定义参数化构造函数的情况下正确初始化基类成员变量？

Question

场景1：没有编译问题。当使用初始化列表在派生类中初始化基类时

class Base
{
    public:
        int x;
};
class D:public Base
{
    public:
        int y;
        D(int y1):Base{y1+1},y{y1}{}
};

int main()
{
    D d(5);
    return 0;
}

方案2：不编译并要求使用参数化的构造函数。注意基类中的虚拟析构函数 $ g ++ -o main * .cpp main.cpp：在构造函数“ D :: D（int）”中： main.cpp：16：34：错误：没有匹配的函数可以调用“ Base :: Base（）”          D（int y1）：基数{y1 + 1}，y {y1} {}                                   ^ main.cpp：5：7：注意：候选对象：Base :: Base（）  班级基础        ^ ~~~ main.cpp：5：7：注意：候选人期望0个参数，提供1个 main.cpp：5：7：注意：候选者：constexpr Base :: Base（const Base＆） main.cpp：5：7：注意：参数1没有从'int'到'const Base＆'的已知转换

class Base
{
    public:
    int x;
    virtual ~Base(){}    
};
class D:public Base
{
    public:
        int y;
        D(int y1):Base{y1+1},y{y1}{}
};

int main()
{
    D d(5);
    return 0;
}

Answer 1

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