关于C++虚析构函数

在上一篇博文《关于C++对象的自杀 》末尾提到了基类的虚析构函数 ，于是再添一篇。

什么时候析构函数需要加个virtual 来修饰呢？

当要通过指向基类的指针删除子类对象时。

Why?

这是为了保证基类和子类的析构函数都得到调用。

个人理解 ：因为基类类型的指针指向的是子类对象中的基类部分，如果析构函数不为虚函数，则无法调用到子类的析构函数。

如下代码：

#include <iostream> using namespace std; class ClsBase{ public: ClsBase(){ cout << "ClsBase constructor." << endl; } ~ClsBase(){ cout << "ClsBase destructor." << endl; } }; class ClsDerived : public ClsBase { public: ClsDerived(){ cout << "ClsDerived constructor." << endl; } ~ClsDerived(){ cout << "ClsDerived destructor." << endl; } }; int main(){ ClsBase *p = new ClsDerived(); delete p; return 0; }

输出为：

ClsBase constructor. ClsDerived constructor. ClsBase destructor.

说明 了基类析构函数不为虚函数时，delete指向基类的指针时无法调用到子类的析构函数。

将基类析构函数修改为虚函数，得到如下代码：

#include <iostream> using namespace std; class ClsBase{ public: ClsBase(){ cout << "ClsBase constructor." << endl; } virtual ~ClsBase(){ cout << "ClsBase destructor." << endl; } }; class ClsDerived : public ClsBase { public: ClsDerived(){ cout << "ClsDerived constructor." << endl; } ~ClsDerived(){ cout << "ClsDerived destructor." << endl; } }; int main(){ ClsBase *p = new ClsDerived(); delete p; return 0; }

输出为：

ClsBase constructor. ClsDerived constructor. ClsDerived destructor. ClsBase destructor.

此时，子类的析构函数得到调用。

个人理解 ：

这是因为基类中将析构函数声明为虚函数，则析构函数（的索引/指针）会被放在虚表vtbl中。

在继承过程中，vtbl属于基类部分，即基类类型的指针p可以指到的范围。

并且，析构函数的虚函数性质会在继承过程中得到保持（虽然函数名不同），于是子类的析构函数也位于虚表中，确保delete p时可以调用到子类的析构函数。

于是，析构函数自底向上得到调用了。

最后引用C++ FAQ的一个段落：

BTW, if you're interested, here are the mechanical details of why you need a virtual destructor when someone says delete using a Base pointer that's pointing at a Derived object. When you say delete p , and the class of p has a virtual destructor, the destructor that gets invoked is the one associated with the type of the object *p , not necessarily the one associated with the type of the pointer. This is A Good Thing. In fact, violating that rule makes your program undefined. The technical term for that is, "Yuck."