Question

我试图习惯上课。在这里，我创建了一个名为Animal的基类和一个名为Dog的派生类。

我原本能够让基类单独工作，但是当我尝试添加派生类时，事情变得混乱而且我遇到了错误。这是代码，如果你能让我知道我做错了什么，那就太棒了！

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

class Animal{
protected:

    int height, weight;
    string name;

public:

    int getHeight() { return height; };
    int getWeight() { return weight; };
    string getName() { return name; };

    Animal();
    Animal(int height, int weight, string name);
};

Animal::Animal(int height, int weight, string name){
    this->height = height;
    this->weight = weight;
    this->name = name;
}


class Dog : public Animal{
private:

    string sound;

public:

    string getSound() { return sound; };
    Dog(int height, string sound);
};

Dog::Dog(int height, string sound){
    this->height = height;
    this->sound = sound;
}

int main()
{
    Animal jeff(12, 50, "Jeff");
    cout << "Height:\t" << jeff.getHeight << endl;
    cout << "Weight:\t" << jeff.getWeight << endl;
    cout << "Name:\t" << jeff.getName << endl << endl;

    Dog chip(10, "Woof");
    cout << "Height:\t" << chip.getHeight() << endl;
    cout << "Sound:\t" << chip.getSound() << endl;
}

Answer 1

未定义Animal类的默认构造函数。你需要：

Animal::Animal() : height(0), weight(0) // Or any other desired default values
{
}

您还应该在基类上有一个虚拟析构函数。

class Animal
{
public:
    ~Animal() {} // Required for `Animal* a = new Dog(...); delete a;`
                 // deletion via base pointer to work correctly
};

编辑：

删除Animal（）后，我收到一条错误消息“动物”：没有合适的默认构造函数

您需要实现默认构造函数（参见上文）。没有它，int成员将不会被初始化并具有未定义的值。

C ++派生类错误

1 个答案: