C++ 构造函数和析构函数是否可以继承？

先看一个例子：

#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》）。