问题描述

C++一道关于继承的题目,求大神解答,感激不尽

1. Dynamic_cast

Total: 65 Accepted: 22

Time Limit: 1sec Memory Limit:256MB
Description

Three classes A, B and C are shown below:

class A {
public:
virtual ~A() {};
};
class B: public A {};
class C: public B {};

You are to implement a function string verify(A *), such that it returns "grandpa" if the passed-in argument points to a class A object, and "father" for a class B object , "son" for a class C object.

Your submitted source code should include the whole implementation of the function verify, but without any class defined above.
No main() function should be included.

解决方案

 #include <iostream>
#include <string>
using namespace std;

class A {
public:
virtual ~A() {};
public: virtual string gettype()
{
    return "grandpa";
}
public: static string verify(A * a)
{
    return a->gettype();
}
};

class B: public A
{
public: virtual string gettype()
{
    return "father";
}
};

class C: public B
{
public: virtual string gettype()
{
    return "son";
}
};

int main(int argc, char* argv[])
{
    A a;
    B b;
    C c;
    cout << A::verify(&a) << endl;
    cout << A::verify(&b) << endl;
    cout << A::verify(&c) << endl;
    return 0;
}

解决方案二：

grandpa
father
son
Press any key to continue

解决方案三：

caozhy的答案是正确的

解决方案四：

子类的构造函数需要调用父类的构造函数

解决方案五：

题本身没难度。。请认真审题

解决方案六：

注意看题目：只写一个函数，不能包括以上的类，也不能包括main函数

string verify(A* pa){
    B* pb=dynamic_cast(pa);
    C* pc=dynamic_cast(pa);
    if(pc != NULL)
      return "son";
    if(pb != NULL)
      return "father";
    if(pa != NULL)
      return "grandpa";
    return NULL;
}