关于nontrivial default constructor(二)

情况一：

如果一个没有任何构造函数的class派生自一个带有默认构造函数的base class，那么派生类的构造函数被视为nontrivial，因此会被编译器合成出来。它将调用基类的默认构造函数。

class Base
{
public:
Base()
{
cout << "Base constructor ..." << endl;
}
};

class Derived : public Base
{
public:

};

int main(void)
{
Derived derived;
return 0;
}

结果如下：
                                

   
    

情况二：

如果设计者提供多个构造函数，但就是没有default constructor， 编译器会扩张每一个构造函数。但它不会合成一个新的默认构造函数。

class Base
{
public:
Base()
{
cout << "Base constructor ..." << endl;
}
};

class Derived : public Base
{
public:
int a;
Derived(int x)
{
a = x;
}
};

int main(void)
{
//Derived derived;    //出现编译错误
Derived derived(10);
cout<<derived.a<<endl;
return 0;
}

当然，如果此时定义了默认构造函数，如下，

class Base
{
public:
Base()
{
cout << "Base constructor ..." << endl;
}
};

class Derived : public Base
{
public:
int a;
Derived(int x)
{
a = x;
}
Derived()
{
cout<<"user-defined default constructor ..."<<endl;
a = 0;
}
};

int main(void)
{
Derived derived0;
cout<<derived0.a<<endl;
Derived derived1(10);
cout<<derived1.a<<endl;

return 0;
}

如果类内还存在着带有默认构造函数的member class object，则这些默认构造函数也会被调用——在基类构造函数之后

class Base
{
public:
Base()
{
cout << "Base constructor ..." << endl;
}
};

class B
{
public:
B()
{
cout << "B constructor ..." << endl;
}
};
class Derived : public Base
{
public:
int a;
B b;
Derived(int x)
{
a = x;
}
Derived()
{
cout<<"user-defined default constructor ..."<<endl;
a = 0;
}
};

int main(void)
{
Derived derived0;
cout<<derived0.a<<endl;
Derived derived1(10);
cout<<derived1.a<<endl;

return 0;
}

结果如下：