C++中构造函数与析构函数能否为虚函数
看书时发现，C++中的基类的构造函数不能为虚函数（VC6.0中为虚函数是不能通过编译的），析构函数应该为虚函数（MFC中CObject的析构函数即为虚函数）。
通过以下面的代码，来看看这样的说法对不对：
测试一：

#include <iostream>using namespace std;class Base{public:Base(){cout << "Base::Base()" <<endl;}~Base(){cout << "Base::~Base()" <<endl;}};class Derived : public Base{public:Derived(){cout << "Derived::Derived()" <<endl;}~Derived(){cout << "Derived::~Derived()" <<endl;}};int main(int argc, char* argv[]){Derived d;// 依次调用Base::Base(),Derived::Derived()                // 程序结束时依次调用Derived::~Derived(),Base::~Base() return 0;// 注意现在基类的析构函数不是虚函数，但是按照调用顺序不会造成内存泄露。}

测试二：

#include <iostream>using namespace std;class Base{public:Base(){cout << "Base::Base()" <<endl;}~Base(){cout << "Base::~Base()" <<endl;}};class Derived : public Base{public:Derived(){cout << "Derived::Derived()" <<endl;}~Derived(){cout << "Derived::~Derived()" <<endl;}};int main(int argc, char* argv[]){Derived *d2 = new Derived(); // 依次调用Base::Base(),Derived::Derived()Base *pBase = d2;delete pBase; // 只调用了Base::~Base()，而没有调用Derived::~Derived() // 因为这里基类的析构函数不是虚函数 return 0;}

测试三：

#include <iostream>using namespace std;class Base{public:Base(){cout << "Base::Base()" <<endl;}virtual ~Base(){cout << "Base::~Base()" <<endl;}};class Derived : public Base{public:Derived(){cout << "Derived::Derived()" <<endl;}~Derived(){cout << "Derived::~Derived()" <<endl;}};int main(int argc, char* argv[]){Derived *d2 = new Derived(); // 依次调用Base::Base(),Derived::Derived()Base *pBase = d2;delete pBase; // 依次调用Derived::~Derived(),Base::~Base(), // 因为这里基类的析构函数是虚函数 return 0;}

总结：
(1) 假如你不打算使用多态（用基类指针指向派生类实例），基类析构函数是否为虚函数是无关紧要的。
(2) 假如你打算使用多态，一定要注意了，让基类的析构函数为虚函数。
PS： 对于非虚函数的调用，完全取决于指针的类型。因为非虚函数的地址在编译阶段就会根据指针的类型确定下来。举一个简单的例子，派生类有一个非虚函数f(),基类中不存在这样的函数。使用一个基类指针指向一个派生类实例后，假如你打算使用这个基类指针调用f()，编译器会让你打消这个念头（编译无法通过）。