什么是c++中的继承构造函数-优快云博客

先来看一段简单的代码：

#include <iostream>
#include <string>

class Base {
public:
    Base() {
        std::cout << "Base() is called" << std::endl;
    }
    Base(int num_) {
        std::cout << "Base(int num_) is called" << std::endl;
        num = num_;
    }

    int number() const {
        return num;
    }

protected:
    int num = 10;
};

class Derived : public Base {
public:
    Derived(const std::string &address_) {
        std::cout << "Derived(const std::string &address_) is called" << std::endl;
        address = std::move(address_);
    }

private:
    std::string address;
};

int main() {
    Derived derived("White House");
    std::cout << "num is " << derived.number() << std::endl;

    return 0;
}

这段代码定义一个父类Base和其子类Derived，并在Base中定义了两个构造函数，Derived中定义了一个构造函数。程序的输出如下：

Derived中的构造函数会首先调用父类的默认构造函数Base()构造父类对象，然后再执行其构造函数剩余的语句，num的输出为默认值10。

现在将Derived构造方式做一些修改：

int main() {
    Derived derived;
    std::cout << "num is " << derived.number() << std::endl;

    return 0;
}

编译出错：

Derived已经自定义了一个构造函数，编译器不会再为Derived合成一个默认构造函数，只能显示用std::string参数构造Derived对象。

在Derived的定义中增加一行代码using Base::Base;并作如下修改：

class Derived : public Base {
    using Base::Base;
public:
    Derived(const std::string &address_) {
        std::cout << "Derived(const std::string &address_) is called" << std::endl;
        address = std::move(address_);
    }

    std::string& get_address() {
        return address;
    }

private:
    std::string address = "Haihu Zhuangyuan";
};

int main() {
    Derived derived;
    std::cout << "num is " << derived.number() << std::endl;
    std::cout << "address is " << derived.get_address() << std::endl;

    return 0;
}

此时程序可以编译通过，执行输出如下：

Derived的构造函数没有被调用，反而是父类的默认构造函数可以被调用，似乎子类可以直接调用父类的构造函数构造对象（此时构造父类对象，子类的数据成员使用默认值）。

如果不适用using Base::Base;，要得到上面同样的输出，程序可以这样修改：

class Derived : public Base {
public:
    Derived(const std::string &address_) {
        std::cout << "Derived(const std::string &address_) is called" << std::endl;
        address = std::move(address_);
    }

    Derived() : Base() {
        
    }

    std::string& get_address() {
        return address;
    }

private:
    std::string address = "Haihu Zhuangyuan";
};

上面的程序自定义了一个默认构造函数，在默认构造函数中显示调用父类的默认构造函数，当然也可以这样修改：

class Derived : public Base {
public:
    Derived(const std::string &address_) {
        std::cout << "Derived(const std::string &address_) is called" << std::endl;
        address = std::move(address_);
    }

    Derived() = default;

    std::string& get_address() {
        return address;
    }

private:
    std::string address = "Haihu Zhuangyuan";
};

让编译器合成一个子类的默认构造函数，合成的构造函数会调用父类的默认构造函数。

接着对Derived对象的构造方式做一下修改：

int main() {
    Derived derived(80);
    std::cout << "num is " << derived.number() << std::endl;
    std::cout << "address is " << derived.get_address() << std::endl;

    return 0;
}

编译错误：

Derived中缺少对应的构造函数，如果仍然增加using Base::Base;语句呢？

class Derived : public Base {
    using Base::Base;
public:
    Derived(const std::string &address_) {
        std::cout << "Derived(const std::string &address_) is called" << std::endl;
        address = std::move(address_);
    }

    Derived() = default;

    std::string& get_address() {
        return address;
    }

private:
    std::string address = "Haihu Zhuangyuan";
};

int main() {
    Derived derived(80);
    std::cout << "num is " << derived.number() << std::endl;
    std::cout << "address is " << derived.get_address() << std::endl;

    return 0;
}

此时编译通过，输出如下：