delete函数的动态解析

operator delete(void* )是做为静态函数来使用的,即使不将他定义为静态函数,编译器也会把它做为静态函数来处理,因此我们通常应该显式地将函数operator new()和operator delete()声明为静态函数.他们永远都不能是非静态成员函数.

 

class B{
public:
    virtual ~B(){
        cout<<"destruct B"<<endl;
    }
    static void operator delete(void*){
        cout<<"B delete"<<endl;
    }
    static void operator delete[](void*){
        cout<<"B delete[]"<<endl;
    }
};
class D:public B{
public:
    virtual ~D(){
        cout<<"destruct D"<<endl;
    }
    static void operator delete(void*){
        cout<<"D delete"<<endl;
    }
    static void operator delete[](void*){
        cout<<"D delete[]"<<endl;
    }
};
int main(){
    B* test=new D;
    delete test;

}

 

例子中的结果显示:
destruct D
destruct B
D delete

当delete test时,函数会自动调用D的delete函数.从表面上看,delete,delete[]这类静态函数是具有多态的特性的.