Question

我的C ++代码遇到了困难。我是初学者。就像，只有非常基本的C ++知识。所以，我真的无法想办法做到这一点。我想过使用C ++命令制作一个RPG游戏，接近完成它。但不知何故，我无法为英雄保持健康。看一下代码，

#include <iostream>
#include <cstdlib>
#include <ctime>



using namespace std;


class player
{   public:
    int health = 100;
};

int battle();
void death();

int main()
{
   int abc;
   player hero;
   hero.health = abc;
   int a;
   int replay = 1;
   cout << "You have 100 Hp. \n";
   while (replay == 1)
{

srand(time(0));
cout << "\n Press 1 to move forward; 2 To stay. \n";
cin >> a;
    if (a == 2)
    {
        if (rand() % 4 + 1 != 1)
        {
            cout << "You stay at your place. \n";
        }
        else
        {
            cout << "Enemy Attacks! (20 Hp) \n";
            //battle(hero.health);
            //cout << "\n Press 1 to continue. \n";
            cout << "\n Do you want to play again? Press 1 to replay and 0 to quit.\n";
            cin >> replay;
        }
    }
    else if (a == 1)
    {
        if (rand() % 2 + 1 != 1)
        {
            cout << "You moved forward. No one around. \n";
        }
        else
        {
            cout << "You move forward. Enemy attacks! (20 Hp) \n";
            battle(abc);
            cout << "\n Do you want to play again? Press 1 to replay and 0 to quit.\n";
            cin >> replay;
        }
    }
    else
    {
        cout << "Sorry. Please enter a valid move. \n";
    }

}

return 0;
}

int battle(int x)
{
   player enemy;
   enemy.health = 20;
   player hero;
   int y;


while (enemy.health >= 0)
{
int eattack = rand() % 15 + 7;
int attack = rand() % 10 + 1;
int escape = rand() % 4 + 1;
cout << "\n Press 1 to attack. 2 to flee \n";
cin >> y;
if (y == 2)
    {
      if (escape != 1)
      {
        cout << "Can't escape! \n";
        cout << "Enemy attacked! Dealing a damage of: " << eattack << " Hp. \n";
        hero.health = hero.health - eattack;
        cout << "Your Hp is: " << hero.health;
      }
      else
      {
        goto Aftermath;
      }

    }
else if (y != 1)
    {
        cout << "Sorry. Please enter a valid response. \n";
    }
else
    {
        cout << "You attack the enemy. \n";
        cout << "You deal a damage of: " << attack;
        enemy.health = enemy.health - attack;
        if (enemy.health >= 0)
        {
            cout << "\n Enemy attacks you, dealing: " << eattack << " Hp damage.";
            hero.health = hero.health - eattack;
            cout << "\n You have: " << hero.health << " Hp left.";
        }
    }

if ((hero.health <= 0) || (hero.health == 0))
    {
        death();
        enemy.health = -1;
    }
}

if (hero.health > 0)
    {
        cout << "\n Enemy fainted!";
        //cout << "You found Hp Potion! Your Hp was refilled.";
    }

   Aftermath:
      if ((hero.health > 0) && (enemy.health > 0))
       {
          cout << "Escaped Successfully! \n";
       }

     return x;

  }
 void death()
    {
      cout << "You died!";
    }

如你所见，我已经呼吁battle(abc)和battle(hero.health) [我现在评论过]但问题是，它说＆＃34;函数{{1}太多参数}。以前，我只是避免参数和创建对象＆＃34;英雄＆＃34;在战斗方法本身。但是每当你完成一个战斗序列时，它会再次返回并再次声明它，从而使其恢复健康状态。 [查看int battle()]

我真的不了解全局变量等等。我只是想知道是否有解决方法或解决此参数问题的方法。任何其他建议，以健康作为一个常数＆＃39;并且每次都没有宣布也被热烈接受。非常感谢你！

P.S。缩短代码的建议也接受但请，我是初学者。所以先进的策略现在超出了我的技能。我需要时间来掌握概念。

Answer 1

您在main方法之前声明了该函数，然后在main方法之后实现该函数。

问题是，您将函数实现为：

 int battle(int x)

但这与您的声明不符：

 int battle();

只需更改您的函数声明块，以便battle函数与预期的签名匹配：

int battle(int x);
void death();

这将使您的代码编译，但您仍然需要执行一些步骤。

我会给你一个启动器：不是将生命值传递到battle，而是传递整个玩家。

 void battle(Player &player)
 {
     // ...
 }

然后您可以直接在battle功能中修改玩家的生命值。

然后你会用：

来调用它

 battle(hero);

C ++中的参数太多

1 个答案: