创建构造函数时未解析的外部符号

Question

在删除构造函数行时，错误消失了。

Dog::Dog(std::string name, double height, double weight, std::string sound) : Animal(name, height, weight) {
    this -> sound = sound;
}


void Dog::ToString() {
    //std::cout << this->name << " is " << this->height << " cm's tall and " << this->weight << " kg's in weight" << std::endl;
    //cannot do the line above as it is private, if it were to be protected we could call it. "sharing with childs of class"

    std::cout << GetName() << " is " << GetHeight() << " cm's tall and " << GetWeight() << " kg's in weight" << std::endl;
}

class Animal {
private:    
    std::string name;
    double height;
    double weight;


    static int numOfAnimals;
    static bool canTalk;

public:
    std::string GetName() {
        return name;
    }

    double GetHeight() {
        return height;
    }

    double GetWeight() {
        return weight;
    }

    void SetName(std::string name) {
        this->name = name;
    }

    void SetHeight(double height) {
        this->height = height; //height that is passed in through parameter becomes the height
    }
    void SetWeight(double weight) {
        this->weight = weight;
    }

    void SetAll(std::string, double, double); 

    Animal(std::string, double, double); //constructor
    Animal(); //for when no parameters are passed
    ~Animal(); //destructor

    static int GetNumOfAnimals() {
        return numOfAnimals;
    }
    void ToString();
};

@inisheer要求的动物类的代码和构造函数

Answer 1

您已经声明了构造函数：

Animal(std::string, double, double); //constructor

但是，您没有定义它，这在这里很重要。编译Dog构造函数后，将对Animal::Animal(std::string, double, double)进行引用，链接器将尝试对此进行引用，但无法对其进行解析。在后续的comment中，您仍然尚未定义此特定的构造函数。

您需要实际定义Animal构造函数和析构函数，例如

Animal(std::string name, double height, double weight) : name(name), height(height), weight(weight) {}

不相关：：这里有潜在的rule-of-three违规隐患，一旦开始与需要特定管理的资源进行交互，您应该更加意识到这一点。您已定义了析构函数，但未定义拷贝构造函数或拷贝赋值运算符。在您的情况下，这很好，因为dtor除了会产生打印副作用之外，没有做任何有趣的事情，但是您应该努力遵循此准则，以避免将来出现麻烦。