题目内容

写结果。classvehicle{intMaxSpeed;intWeight;public:vehicle(){MaxSpeed=0;Weight=0;}virtual~vehicle(){Stop();}virtualvoidStop(){cout<<"VS";}};classbicycle:virtualpublicvehicle{intHeight;public:~bicycle(){Stop();}voidStop(){cout<<"BS";}};classmotorcar:virtualpublicvehicle{intSeatNum;public:~motorcar(){Stop();}voidStop(){cout<<"MAS";}};classmotorcycle:publicbicycle,publicmotorcar{public:~motorcycle(){Stop();}voidStop(){cout<<"MYS";}};intmain(){vehicle*ptr;motorcycled;ptr=&d;ptr->Stop();}

查看答案
更多问题

写结果。classA{public:virtualvoidPrint()=0;};classA1:publicA{public:voidPrint(){cout<<"A1P";} };classB1:publicA{public:voidPrint(){cout<<"B1P";} };voidfun(A*p){p->Print();}intmain(){A*ptr;A1a1;B1b1;ptr=&a1;fun(ptr);ptr=&b1;fun(ptr);return 0;}

写结果。class Base{ public: virtual void Set(int b){x=b;}virtual int Get() { return x;} private: int x;};class Derived: public Base{ public: void Set(int d){y = d;} int Get() {return y;} private: int y;};int main( ){ Base B_obj; Derived D_obj; Base *p=&B_obj; p->Set(10);cout<Get(); p=&D_obj;p->Set(20);cout<< p->Get();p->Base::Set(30);cout<< p->Base::Get();p->Set(p->Get()+20);cout<< p->Get();return 0;}

如果希望利用高级ACL来识别源IP地址为172.16.10.1且目的IP地址是172.16.20.0/24网段的IP报文并执行“拒绝”的动作,那么应该采用下面哪一条规则?( )

A. rule deny source 172.16.10.1 0.0.0.0
B. rule deny source 172.16.10.1 0.0.0.0 destination 172.16.20.0 0.0.0.255
C. rule deny tcp source 172.16. 10. 1 0.0.0.0 destination 172.16.20.0 0.0.0.255
D. rule deny ip source 172. 16.10. 1 0.0.0.0 destination 172.16.20.0 0.0.0.255

下面关于NAT说法错误的是()

A. 静态NAT本质是1对1映射
B. NAPT通过一个IP地址和不同端口号绑定,实现多对1映射
C. NAT的功能是将私网IP地址转换为公网IP地址,并不能达到节约IP地址的目的
D. 动态NAT一般结合ACL进行联动配置

答案查题题库