我的代码不会链接：“对Bumblebee :: Bumblebee（）的“未定义”引用。

Question

我正在尝试为此编码练习运行我的代码，并且我会遇到此错误：对bumblebee :: bumblebee（），对grasshopper :: grasshopper（）

我正在练习继承，我将昆虫类用作基础类来继承功能和方法，但我不明白我在做什么错。感谢任何帮助，这里是代码：

#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

您为基类提供了一个构造函数定义，但没有提供子类。您还需要从子类构造函数调用基类构造函数以获取所需的输出。
替换这样的子类构造函数：

BumbleBee() : Insect() {}

卷曲括号使其成为定义，而不是声明，并且在结肠后的初始化列表中调用基类构造函数。