dynamic_cast只能正确执行

Question

问题的结构是这样的 食物是一个抽象基类;植物和动物直接从它那里继承。 食草动物，食肉动物和杂食动物继承动物， 而水果和坚果和叶从植物 继承狐猴，考拉和松鼠继承草食动物

总体而言，它是一个炎热的烂摊子，但它是必要的锻炼。 整个项目可以在GitHub https://github.com/joekitch/OOP_JK_Assignment_4/blob/master/OOP_JK_Assignment_4/Lemur.h 完整的类图中也对GitHub的

，但这里有相关的位和鲍勃（至少，那些我认为是相关的） 首先是食品类，它包含了几乎没有什么

#pragma once 
#include <string> 
#include <list> 
using namespace std; 

class Food 
{ 
public: 

    Food(){ } 
    virtual ~Food(){ } 


}; 

下一个是动物，它包含了追捕（的虚函数），吃（）函数

#pragma once 
#include "Food.h" 
#include "Animal.h" 
#include "Plant.h" 
#include <iostream> 
#include <string> 
#include <list> 
using namespace std; 

class Animal : public Food 
{ 
public: 
    Animal(void) : name(), alive(true), age(0), calories(0), weight(0) { } 
     Animal(string& animal_name, int animal_age, int animal_calories, double animal_weight) : 
     name(animal_name), alive(true), age(animal_age), calories(animal_calories), weight(animal_weight), maxcalories(animal_calories) {} 

    virtual ~Animal(){} 

    virtual bool eat(Food* food){return false;}; 
    virtual bool hunt(list<Food*> &foodlist){return false;}; 


    virtual void PrintSelf(){}; 

    virtual string& getName(){ 
     return name; 
    }; 


     std::string name; 
     bool alive; 
     int age, calories, maxcalories; 
     double weight; 
}; 

这是Herbivore，它完全定义了hunt（）函数，这是问题开始的地方（请参阅我在hunt（）中的注释）。狩猎需要在全球范围内主要宣布的类型Food *的列表（）

#pragma once 
#include "Animal.h" 
//#include "Lemur.h" 
#include "Plant.h" 
#include "Fruit.h" 
#include "Leaf.h" 
#include "Nut.h" 
#include <iostream> 
#include <string> 
#include <list> 
#include <typeinfo> 
using namespace std; 

class Herbivore : public virtual Animal 
    { 
    public: 
     Herbivore() {} 
     virtual ~Herbivore(){} 
     virtual bool eat(Food* food) {cout << "herbivore.h eat() called" << endl; return true;}; 



     bool hunt(list<Food*> &foodlist) //herbivore version of hunt() 
     { 
      int fruitcounter=0; 
      int plantcounter=0; 
      string name; 
      for (list<Food*>::iterator it = foodlist.begin(); it != foodlist.end(); it++) 
      { 
       if (Plant* temp = dynamic_cast<Plant*>(*it)) 
       { 
//this is there the problems start. the above dynamic cast SHOULD make temp 
//non-null if the thing i'm looking at is a child of plant (that is, if the thing 
//in the food list is a fruit or a nut or a leaf). And indeed it does...but the 
//next dynamic cast (in the eat() function of Lemur) doesn't detect any fruits.... 

        plantcounter++; 

        if (eat(*it)) 
        fruitcounter++; 
        //return true; 
       } 

      } 
      cout << "there are " << fruitcounter << " fruits and " << plantcounter << " plants in the list." << endl; 
      return false; 
     }; 


}; 

这里是Fruit类。没有什么特别明显的，我只是把它放在这里只是在情况下，他们可以帮助解决这个问题

#pragma once 
#include <iostream> 
#include <string> 
#include "Plant.h" 
using namespace std; 

class Fruit : public Plant { 
    public: 
     Fruit (std::string& plant_name, int energy_value): 
      Plant (plant_name, energy_value){} //constructor pased to base class 

     ~Fruit(){ } //destructor 

     //inherits basically everything from the Plant base class, makes leae nodes in the class tree easy to write and access 



    }; 

现在这里是真正的麻烦制造者。这个狐猴完全定义了eat（）函数，并且从狩猎（）传递给它的食物*，从草食动物中调用，并且对它做了更多的测试，看看它是否是水果（这是唯一的植物狐猴可以吃）

#pragma once 
#include "Animal.h" 
#include "Herbivore.h" 
#include "Plant.h" 
#include "Fruit.h" 
#include "Leaf.h" 
#include "Nut.h" 
#include <iostream> 
#include <string> 
#include <list> 
#include <typeinfo> 
using namespace std; 

class Lemur : public Herbivore 
{ 
    public: 
     Lemur(void) : name(), alive(true), age(0), calories(0), weight(0) {} 
     Lemur(string& animal_name, int animal_age, int animal_calories, double animal_weight) : 
     name(animal_name), alive(true), age(animal_age), calories(animal_calories), weight(animal_weight), maxcalories(animal_calories) {} 


     ~Lemur(){} 



     bool eat(Food* food) 
     { 




      if (Fruit* temp = dynamic_cast<Fruit*>(food)) 
      { 
       //PROBLEM, it sees every plant as a fruit in this 
//case...at least according to a typeinfo().name() that i have run in here. However the temp 
//always returns null, so this proper if statement never actually happens, so it never sees 
//any fruit, even though there's a whole bunch in the list (500 of them). what's wrong? 
       cout << "it's a fruit" << endl; 
       return true; 
      } 
      else 
      { 
       //cout << "not a fruit" << endl; 
       return false; 
      } 


     } 


    void PrintSelf() 
    { 
     cout << "i am a " << age << " year old, " << weight << " kilogram " << name << " with " << calories << " calories." << endl; 
    }; 

    string& getName(){ 
     return name; 
    }; 

     std::string name; 
     bool alive; 
     int age, calories, maxcalories; 
     double weight; 


}; 

，你可以看到，将dynamic_cast不会返回一个非空气温，即使我已经证实，它遍历该列表。我也使用计数器变量来追踪它的进度，奇怪的是它说有1500个植物在列表中......但是0个水果......

我在构造我的演员是错的吗？我的遗传是关闭的吗？做什么？

编辑;我添加的虚拟析构函数为每个类，所以这是不看的github回购问题

Answer 1

，看起来问题出在哪里main.cpp你对象添加到Food_list：

else if (FileIn_type == Plant_type) 
{ 
    Plants++; 
    FileIn >> FileIn_name >> FileIn_calories; 
    Food_list.push_back(new Plant (FileIn_name, FileIn_calories)); 
} 

您只能创建一个要添加到列表中的Plant对象。所以这就是所有这些对象的动态类型。

仅仅在main.cpp的代码位之上，您可以创建级联的if/else语句，这些语句在动物名称字符串中创建不同的动物类型。你需要为植物做类似的事情。