未定义的基函数引用

Question

给出以下示例：

// Fighter.h
class Fighter {
    public:
        void attack(Fighter * other);
    protected:
        Fighter(const std::string nick) : nickname(nick) { };
        virtual void atk(Fighter * other) = 0;
        const std::string nickname;
};

// Fighter.cpp
#include "Fighter.h"
void Fighter::attack(Fighter * other) {
    std::cout << nickname << " attacks " other->getNickname() << std::endl;
    atk(other);
}

// Warrior.cpp : Fighter
void Warrior::atk(Fighter * other) {
    int dmg = rand() % off;
    int chance = rand() % 6;
    if (chance == 3) {
        dmg += dmg;
        std::cout << nickname << ": Double Damage!";
    }
    other->hit(dmg);
    if (!other->isDead()) {
        other->counter(this);
    }
}

// main.cpp
#include "Fighter.h"
#include "Warrior.h"
Fighter * a = new Warrior(string("A"));
Fighter * b = new Warrior(string("B"));
a->attack(b);  // << LINKER ERROR

我只能做到这一点：

  

未定义参考“战斗机::攻击（战斗机*）＆＃39;

换句话说，我想要一个处理所有派生类的输出的基本方法，所以我不必复制每个子类中的代码？