C++中继承情况下的构造函数

如果需要向基类构造函数传递实参，需要将基类包含在派生类的构造函数初始化列表中。如下：

 

 

#include <iostream>

using namespace std;

 

class Base

{

private:

    int value;

public:

    Base(int i = 0) : value(i)

    {

        cout << "con in base. value is " << value << endl;

    }

};

 

class Derived : public Base

{

private:

    int value;

public:

    Derived(int i = 1) : value(i), Base(i)

    {

        cout << "con in derived. value is " << value << endl;

    }

};

 

int main()

{

    Derived a;

}

上面的程序的输出结果为：

con in base. value is 1

con in derived. value is 1

若将程序改写成：

#include <iostream>

using namespace std;

class Base

{

private:

    int value;

public:

    Base(int i = 0) : value(i)

    {

        cout << "con in base. value is " << value << endl;

    }

};

class Derived : public Base

{

private:

    int value;

public:

    Derived(int i = 1) : value(i) //, Base(i)

    {

        cout << "con in derived. value is " << value << endl;

    }

};

int main()

{

    Derived a;

}

则输出结果为：

con in base. value is 0

con in derived. value is 1

若进一步讲程序改写成：

#include <iostream>

using namespace std;

class Base

{

private:

    int value;

public:

    Base(int i) : value(i)

    {

        cout << "con in base. value is " << value << endl;

    }

};

class Derived : public Base

{

private:

    int value;

public:

    Derived(int i = 1) : value(i)//, Base(i)

    {

        cout << "con in derived. value is " << value << endl;

    }

};

int main()

{

    Derived a;

}

则程序编译时会报错：

error: no matching function for call to 'Base::Base()'

因为派生类的构造函数会调用基类的默认构造函数，而在上面的代码中Base类没有定义默认构造函数，故报错。