我正在尝试制作一个非常简单的基于文本的游戏,当我尝试从外部函数访问动态结构时遇到错误。我初始化了我的头文件中的所有变量和结构,并在main函数中声明了一个动态alloc。但我仍然有错误。我错过了什么吗?这是我的代码。
==================主要功能“new.cpp”====================
#include <stdlib.h>
#include <iostream>
#include <string>
#include "game.h"
using namespace std;
difficulty _df_1;
player* player_1 = new player[3];
int main()
{
//initialize player stats
player_1->hp = 100;
player_1->life = 3;
player_1->mana = 50;
player_1->player_id = 1;
player_1->level = 1;
player_1->player_name= "emmet";
//..end
int turn=1;
cout << "What is your name? <<<";
getline(cin,player_1->player_name,'\n');
cout << "Choose a difficulty level: [0]Easy [1]Normal [2]Godlike" << endl;
int ch;
cin >> ch;
switch(ch)
{
case 0:
cout << "Scardy Cat chose EASY." << endl;
break;
case 1:
cout << "A really nice way to start. NORMAL" << endl;
break;
case 2:
cout << "Overly Manly Man is playing GODLIKE." << endl;
break;
default: cout << "I wonder how you can play this game if you can even read simple instructions."<< endl;return 0; break;
}
while(turn == 1)
{
char ch;
cout << "What do you want to do now? \n <<<<";
cin >> ch;
cin.ignore(5,'\n');
switch(ch)
{
case 'a': case 'A':
player_stat();
break;
case 'v': case 'V':
cheat_menu();
break;
case 'x': case 'X':
return 0;
break;
case '`':
break;
default: cout << "We were unable to process your request. Please try again" << endl; break;
}
}
delete player_1;
return 0;
}
void cheat_menu()
{
cout << "CHEATERS WILL ROT IN THE DEEPEST DEPTHS OF TARTARUS." << endl;
cout << "Enter Code:" << endl;
string cheat;
getline(cin,cheat,'\n');
if(cheat == "poo")
{
system("sleep 3");
cout << "Cheat Activated.." << endl;
player_1->hp += 1000;
player_1->level += 10;
player_1->mana += 1000;
player_stat();
}
else
{
cout << "Wrong cheat code.." << endl;
}
//system("sleep 3");
//system("clear");
}
==================结束主要功能==============
=======外部功能“player_stat.cpp”===========
#include <iostream>
#include "game.h"
using namespace std;
void player_stat()
{
cout << "Name: " << player_1->player_name << endl
<< "Hp: " << player_1->hp << endl
<< "Life: " << player_1->life << "\t Mana: " << player_1->mana << endl
<< "Level: " << player_1->level << "\t XP: " << player_1->xp
<< endl;
//system("sleep 3");
//system("clear");
}
==================结束外部功能==============
==========头文件“game.h”===================
#ifndef _GAME_
#define _GAME_
#include "player_stat.cpp"
using namespace std;
//function prototypes...
void player_stat();
void cheat_menu();
//structs for player and NPC
struct player
{
string player_name;
int life;
double atk;
double hp;
double mana;
int player_id;
int level;
long int xp;
string weapon_name;
double weapon_damage;
};
enum difficulty {EASY,NORMAL,GODLIKE};
#endif
===========结束头文件=======================
这些是我得到的错误。请不要介意int main()中的其他省略函数。 :P
In file included from game.h:4
from new.cpp
in function `void player_stat()':
`player_1' undeclared (first use in this function)
(Each undeclared identifier is reported only once for each function it appears in.)
At global scope:
`player*player_1' used prior to declaration
答案 0 :(得分:0)
添加
extern player * player_1;
到game.h
这将使所有模块中的player_1变量可用。
也删除
#include“player_stat.cpp”
你不应该在头文件中包含任何cpp文件
答案 1 :(得分:0)
使用extern
存储说明符声明变量:
extern player* player_1;
这告诉编译器player_1
变量是在其他地方定义的,链接器稍后会解析它。
另外正如我所说,你实际上分配了三个播放器对象。如果你只需要一个,那么不要分配三个:
player* player_1 = new player;
但是,这也不需要,声明一个正常的非指针变量也可以正常工作:
player player_1;
然后在其他文件中使用extern
告诉编译器该变量是在其他地方定义的:
extern player player_1;