您的位置:首页 > 编程语言 > C语言/C++

C++ 构造函数和析构函数是否可以继承?

2015-11-12 10:51 471 查看
先看一个例子:

[cpp] view
plaincopy





#include <iostream>



class A

{

public:

A() { ::std::cout << "constructor: A" << ::std::endl; } // 构造函数不能为 virtual

A(int aa): a(aa) { ::std::cout << "constructor: A, a = " << a << ::std::endl; }

virtual ~A() { ::std::cout << "destructor: A" << ::std::endl; }

public:

int a;

};



class B : public A

{

public:

B() { ::std::cout << "constructor: B" << ::std::endl; }

B(int bb) : b(bb) { ::std::cout << "constructor: B, b = " << b << ::std::endl; }

B(int aa, int bb): A(aa), b(bb) { ::std::cout << "constructor: B, a = " << a << ", b = " << b << ::std::endl; }

~B() { ::std::cout << "destructor: B" << ::std::endl; }

public:

int b;

};



class A2

{

public:

A2() { ::std::cout << "constructor: A2" << ::std::endl; }

explicit A2(int aa) { ::std::cout << "constructor: A2, aa = " << aa << ::std::endl; }

virtual ~A2() { ::std::cout << "destructor: A2" << ::std::endl; }

};



class B2: public A2

{

public:

using A2::A2; // Inheriting constructors(VS2013 not support, GCC before version 4.8 not support)

~B2() { ::std::cout << "destructor: B2" << ::std::endl; }

};



int main()

{

B b_1; // A() --> B() --> ~B() --> ~A() 隐式调用直接父类的无参构造函数

B b_2(22); // A() --> B(int) --> ~B() --> ~A() 隐式调用直接父类的无参构造函数

B b_3(11, 22); // A(int) --> B(int, int) --> ~B() --> ~A() 委托构造函数



// 其实这里调了 B2() 及 B2(int), 只不过没有输出而已

B2 b2_1; // A2() --> B2() --> ~B2() --> ~A2()

B2 b2_2(11); // A2(int) --> B(int) --> ~B2() --> ~A2()

return 0;

}

/*

constructor: A

constructor: B

constructor: A

constructor: B, b = 22

constructor: A, a = 11

constructor: B, a = 11, b = 22

constructor: A2

constructor: A2, aa = 11

destructor: B2

destructor: A2

destructor: B2

destructor: A2

destructor: B

destructor: A

destructor: B

destructor: A

destructor: B

destructor: A

*/

结论:

1. 构造函数不能为 virtual, 构造函数不能继承;

2. 如果子类不显式调用父类的构造函数,编译器会自动调用父类的【无参构造函数】;

3. 继承构造函数(Inheriting constructors)

(1) C++11 才支持;

(2) 实质是编译器自动生成代码,通过调用父类构造函数来实现,不是真正意义上的【继承】,仅仅是为了减少代码书写量(参考 《C++ Primer》)。

FROM: http://blog.csdn.net/duyiwuer2009/article/details/41047609
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: