我的代码不会链接：“未定义的对大黄蜂的引用：：大黄蜂()”

Question

我试图为这个编码练习运行我的代码，并且我得到了以下错误:对BumbleBee::BumbleBee()的未定义引用，对GrassHopper::GrassHopper()的未定义引用

我正在练习继承，我使用昆虫类作为基类来继承函数和方法，但是我不明白我做错了什么。任何帮助都是非常感谢的，下面是代码：

#include <iostream>
using namespace std;

// Insect class declaration
class Insect
{
protected:
    int antennae;
    int legs;
    int eyes;

public:

    // default constructor
    Insect();       

    // getters
    int getAntennae() const;
    int getLegs() const;
    int getEyes() const;
};

// BumbleBee class declaration
class BumbleBee : public Insect
{
    public:
    BumbleBee();

    void sting()
    { cout << "STING!" << endl; }
};

// GrassHopper class declaration
class GrassHopper : public Insect
{
    public:
        GrassHopper();
        
        void hop()
        { cout << "HOP!" << endl; }
};


// main
int main()
{
    BumbleBee *bumblePtr = new BumbleBee;
    GrassHopper *hopperPtr = new GrassHopper;

    cout << "A bumble bee has " << bumblePtr->getLegs() << " legs and can ";
    bumblePtr->sting();

    cout << endl;

    cout << "A grass hopper has " << hopperPtr->getLegs() << " legs and can ";
    hopperPtr->hop();

    delete bumblePtr;
    bumblePtr = nullptr;
    delete hopperPtr;
    hopperPtr = nullptr;

    return 0;   
}

// member function definitions
Insect::Insect()
{
    antennae = 2;
    eyes = 2;
    legs = 6;
}
int Insect::getAntennae() const 
{ return antennae; }
int Insect::getLegs() const 
{ return legs; }
int Insect::getEyes() const 
{ return eyes; }


//Desired output
/*
A bumble bee has 6 legs and can sting!

A grass hopper has 6 legs and can hop!

*/

Answer 1

您已经为基类提供了构造函数定义，但没有为子类提供构造函数定义。您还需要从子类构造函数中调用基类构造函数以获得所需的输出。替换下面这样的子类构造函数：

BumbleBee() : Insect() {}

大括号使其成为定义而不是声明，基类构造函数在冒号之后在初始化列表中被调用。