C ++面向对象的问题

时间:2014-09-25 19:03:02

标签: c++ oop

我正在尝试学习C ++ OOP,我制作了以下代码:

的main.cpp

#include <iostream>
#include <string>
#include "monster.h"
using namespace std;

int main(int argc, char** argv) {
    Monster monster("Wizard",150,50);
    Monster monster2("Gorgoyle",450,15);

    cout << monster2.getHealth() << endl;
    monster.attack(monster2);
    cout << monster2.getHealth() << endl;
}

monster.h

#ifndef MONSTER_H
#define MONSTER_H
#include <iostream>
#include <string>
using namespace std;

class Monster
{
public:
    Monster(string name_, int health_, int damage_);
    ~Monster();
    int attack(Monster opponet);
    int getHealth();
    string name;
    int damage;
    int health = 0;
    int getDamage();
    void setHealth(int health_);
    void setDamage(int damage_);
    void setName(string name);
    void doDamageToOpponent(Monster opponent);
    string getName();
};

#endif

monster.cpp

#include "monster.h"

Monster::Monster(string name_, int health_, int damage_) {
    health = health_;
    setDamage(damage_);
    setName(name_);
}

Monster::~Monster() { }

int Monster::attack(Monster opponent) {
    doDamageToOpponent(opponent);
}

void Monster::doDamageToOpponent(Monster opponent) {
    int newHealth = opponent.getHealth() - this->getDamage();
    opponent.setHealth(newHealth);
}

int Monster::getHealth() {
    return health;
}

int Monster::getDamage() {
    return damage;
}

void Monster::setHealth(int health_) {
    health = health_;   
}

void Monster::setDamage(int damage_) {
    this->damage = damage_;
}

void Monster::setName(string name_) {
    this->name = name_;
}

string Monster::getName() {
    return name;
}

现在我的问题是,当我运行这段代码时,我希望有一个monster2对象可以保留400个生命值,但它仍然是450:S

为了达到这个目的,必须做些什么?我注意到它在doDamageToOppoenet中可以是400,但是当它离开那个块时,它仍然是450.请帮助我!感谢。

3 个答案:

答案 0 :(得分:2)

您按价值传递对象

void Monster::doDamageToOpponent(Monster opponent) <- This should be by reference
int Monster::attack(Monster opponent) <- idem

这意味着:您正在创建Monster对象的新副本,以便在您正在调用的函数中造成损害,然后实际处理该副本损坏但保留原来具有未触及值的对象。

以下签名将起作用:

void Monster::doDamageToOpponent(Monster& opponent)
int Monster::attack(Monster& opponent)

如果您想了解更多相关信息,请阅读以下内容:Passing stuff by referencePassing stuff by value

答案 1 :(得分:2)

原因是函数attackdoDamageToOpponent正在获取参数的副本,因为您通过值传递 。然后发生的事情是你改变函数内传递的Monsters的副本。函数返回后,这些副本会死亡(因为它们是函数的本地副本)并且原始的感兴趣的各方都没有发生任何事情。

尝试通过引用传递参数。引用就好像它是原始变量一样。考虑:

int a = 0;
int &refa = a; /* refa acts as real "a", it refers to the same object "a" */
int b = a;     /* this is your case */
b = 6;         /* b will be changed, but "a" not */
refa = 6;      /* a is changed, really "a", refa is just different name for "a" */

尝试:

int Monster::attack( Monster &opponent){
    doDamageToOpponent( opponent);
}

void Monster::doDamageToOpponent( Monster &opponent){
    int newHealth = opponent.getHealth() - this->getDamage();
    opponent.setHealth( newHealth);
}

答案 2 :(得分:1)

你正在通过值传递对手,即功能:

int Monster::attack(Monster opponent);

实际上会收到对手的副本并修改该副本。每次有一个修改某个对象的函数时,你需要通过引用传递要修改的对象,或者将指针传递给它,例如,

int Monster::attack(Monster& opponent);

int Monster::attack(Monster* opponent);

我建议使用const T&作为输入参数,使用T*作为输出参数,所以在这种情况下,后一种形式。我推荐后者用于输出参数的原因是因为它使调用者更明确:

monster.attack(&monster2); // passing a pointer: monster2 will be modified.