Question

这涉及一些相当棘手的遗产，但在这里忍受我。我的问题不是一个特定的错误，而只是“我将如何具体地做这个？”

这个想法是有一个抽象的基类食物（注意这个问题都是过于简化的）

//parent of Animal
//parent of Plant
//~Food()
//Food()
#pragma once

class Food
{
public:
    Food(){}
    ~Food(){}



};

从那里出来的是动物和植物。我现在不太担心工厂动物需要拥有虚拟功能Hunt and Eat

#pragma once
#include "Food.h"


class Animal : public Food
{
//eat() which accepts a Food* type as an argument. it is an abstract virtual in this class, it has the form
//bool eat(Food* food)

//hunt() which accepts an stl list of Food* pointers to Food type objects. the food list is declared globally in main and passed here. it has the form
//hunt(list<Food*> &foodlist)
};

从那以后会有更多的课程;草食动物，食肉动物，杂食动物（继承食肉动物和食草动物）。这是草食动物

//Child of Animal
//Parent of Lemur, Koala, Squirrel, Omnivore

//~Herbivore()
//hunt(list<Food*&foodList):bool (only eats plant types)
#pragma once
#include "Animal.h"
#include <iostream>
#include <string>
#include <list>
using namespace std;

class Herbivore : public virtual Animal
    {
    public:

        Herbivore() {}
        ~Herbivore(){}
        //eat() and hunt() are probably virtual here as well, as they aren't used directly, only the lower classes directly access them


};

从那些是最底层的子类，它们都大致有这种形式。这是一只松鼠

//child of Herbivore
//leaf node

#pragma once
#include "Animal.h"
#include "Herbivore.h"

class Squirrel : public Herbivore
{ 

        //bool eat() is fully defined here instead of being virtual.
        //bool hunt() is fully defined here instead of being a virtual.

       //both have the same argument lists as the virtuals in Animal



};

这是主要的

list<Food*> Food_list; //global list of Food items that will be passed to hunt()
int main()
{

    list<Food*>::iterator it = Food_list.begin();

    (*it)->eat(*it); //passing the iterator to itself as a test. this seems to work ok
    (*it)->hunt(Food_list); //however this, in my code, refuses to work for a few reasons

};

所以基本上所有东西都来自食物...但这是一件坏事。

我已经尝试了以下几个问题

我在Animal中尝试了虚拟功能的初始版本，并没有任何食物，它抱怨食物没有功能搜索

error C2039: 'hunt' : is not a member of 'Food'

....我认为这是公平的，虽然不应该看松鼠而不是食品类？

我尝试在食物中制作一个纯粹的虚拟食物用于进食和狩猎，从那时起，每次尝试实例化任何类型的叶子类（如松鼠或老虎或其他任何东西）都会返回“无法实例化抽象类”错误。

error C2259: 'Squirrel' : cannot instantiate abstract class

我尝试在食物中进食和狩猎不那么抽象，比如狩猎（list＆amp; foodlist），但后来它说“语法错误，标识符'列表'”，就像它不知道列表是什么.. ..即使我加入Food.h后

error C2061: syntax error : identifier 'list'

并且所有这些错误都与错误“'Food :: hunt'配对：函数不带1个参数”

error C2660: 'Food::hunt' : function does not take 1 arguments

我的整体问题是，你如何将这个抽象的虚函数从Animal转换为它的叶子类？你怎么称呼它？基本上我所尝试的一切都是悲惨地失败 * 不要担心里面吃什么（）或打猎（），我只是在寻找适当的声明 *

这个项目的github也可以在这里找到 https://github.com/joekitch/OOP_JK_Assignment_4 如果需要的话

Answer 1

一些想法，

我假设草食动物在某处被定义......
使用虚拟析构函数
当某些东西告诉你它无法实例化时，正在调用一个Food（），Animal（）构造函数。

示例代码：

class Food
{
public:
    Food(){ }
    virtual ~Food(){ }
};

class Animal : public Food
{
 Animal() : Food() { }
 virtual Animal() { } //Cause C++

 virtual bool eat(Food* food) = 0;
 virtual hunt(list<Food*> &foodlist) = 0;
};

class Squirrel : public Herbivore
{ 
 Squirrel() : Herbivore() { }
 ~Squirrel() { } //not virtual

  bool eat(Food *food) { //stuff };
  void hunt(list<Food *> &foodlist) { //stuff };
};

list<Animal*> animal_list; //global list of Food items that will be passed to hunt()
int main()
{
    animal_list.push_back(new Squirrel()); // Make sure you fill the array?

    list<Food*>::iterator it = Food_list.begin();

    (*it)->eat(*it); //passing the iterator to itself as a test. this seems to work ok
    (*it)->hunt(animal_list); //however this, in my code, refuses to work for a few reasons

};

Answer 2

我找到的解决方案涉及动态铸造。基本上，你需要将迭代器指针DOWN从Food *类型转换为更低的类似Herbivore或Animal类型，无论哪种类型必须具有你想要在其中完全定义的函数

Herbivore* temp = dynamic_cast<Herbivore*>(*it)
if ( temp ){
        cout << "iterator thing is a Herbivore " << endl;
        temp->hunt(Food_list);
    cout << "iterator thing is of the type " << typeid(temp).name() << endl;}
    else cout << "iterator is not a Herbivore " << endl;}

以上代码将尝试将转换为Herbivore 类型。如果它成功（也就是说，它的父类是Herbivore），那么Temp将被转换为左边指定的Herbivore 类型。如果失败，temp将是NULL类型。这个临时指针指向与 it指针相同的东西....但它只是被视为草食动物而不是食物*。

错误C2259：无法实例化抽象类;或者“我到底应该如何使用虚拟功能呢？”

2 个答案: