我有一个抽象类的Player及其子AI和Human。总的来说,当我创建两个对象Human和AI时,它可以很好地工作。但是一旦在等待播放器指针类型的函数中将它们用作参数时,它们的类型就不再是AI和Human,而是两者都是Player对象。
Game.hpp:
#include "Player.hpp"
#include "Human.hpp"
class Game {
private:
Player *j1, *j2;
public:
Game();
Game(Player*, Player*);
void setP1(Player*);
void setP2(Player*);
Player* getP1();
Player* getP2();
};
Game.cpp:
#include "Game.hpp"
Game::Game(){
}
Game::Game(Player *pp1, Player *pp2){
p1 = pp1;
p2 = pp2;
}
void Game::setP1(Player *pp1){
p1 = pp1;
}
void Game::setP2(Player *pp2){
p2 = pp2;
}
Player* Game::getP1(){
return p1;
}
Player* Game::getP2(){
return p2;
}
Player.hpp:
#ifndef PLAYER_H
#define PLAYER_H
#include <string>
using std::string;
class Player {
protected:
string nom;
int age;
public:
Player();
Player(string, int);
void setNom(string);
void setAge(int);
string getNom();
int getAge();
virtual void upAge() = 0;
};
#endif
这是main.cpp:
#include "Player.hpp"
#include "Human.hpp"
#include "Game.hpp"
#include <iostream>
#include <string>
using std::cout;
using std::endl;
using std::string;
int main(){
Player *j;
Human h;
Game Game;
cout << typeid(h).name() << endl;
Game.setJ1(&h);
cout << typeid(Game.getJ1()).name() << endl;
return 0;
}
我希望两个提示显示相同的结果。但是第一个显示“人类”,第二个显示“玩家”。我该如何处理?
编辑1:添加了Player.hpp文件。
答案 0 :(得分:0)
基类Player
必须包含一个虚函数才能将类型名称作为派生类。
从cpp参考中检查以下示例中的typeid。
#include <iostream>
#include <string>
#include <typeinfo>
struct Base {}; // non-polymorphic
struct Derived : Base {};
struct Base2 { virtual void foo() {} }; // polymorphic
struct Derived2 : Base2 {};
int main() {
// Non-polymorphic lvalue is a static type
Derived d1;
Base& b1 = d1;
std::cout << "reference to non-polymorphic base: " << typeid(b1).name() << '\n';
Derived2 d2;
Base2& b2 = d2;
std::cout << "reference to polymorphic base: " << typeid(b2).name() << '\n';
}
可能的输出:
reference to non-polymorphic base: 4Base
reference to polymorphic base: 8Derived2