基于控制台的RPG游戏实现代码,C代码库,德仔网

代码语言

代码分类

【C】基于控制台的RPG游戏实现

作者:舞维天下 / 发布于2013/12/26/ 837

基于控制台设计的RPG游戏，键盘的WSAD(上下左右)，当遇到NPC(屏幕上N代替)的时候可以对话并接收任务，找到一只猪(屏幕上Z代替)后游戏结束。主要使用到了逻辑逻辑判断、循环、数据结构(结构体、数组)

#include <stdio.h>
/* 为了支持颜色 */
#include <windows.h>

// 地图区域
const int MAP_X_LENGTH = 24;
const int MAP_Y_LENGTH = 18;

/* 声明函数 */
void game();
int deriction_down();
int deriction_up();
int deriction_left();
int deriction_right();
void role_up();
void role_display();
void gameinfo();
int NPC(int npc);






// 屏幕缓冲区
int display[18][24] = {
  {3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3},
  {3,1,2,0,0,0,0,0,2,0,0,3,0,0,0,0,0,0,0,0,0,0,0,0},  
  {3,0,2,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},  
  {3,0,2,2,2,0,2,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},  
  {3,0,0,0,2,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},  
  {3,0,2,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},  
  {3,0,2,2,2,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},  
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},  
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},  
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,4,0,0,0,0,0,0,0,0,0},  
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},  
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3},  
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3},  
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3},  
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3},  
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3},  
  {3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3},  
  {3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3}

};

// 描述位置结构体
struct Position{
 int x;// x
 int y;// y
};

// 角色
struct Role{
 struct Position oldP;// 旧位置
 struct Position newP;// 新位置
 void (* display)();  
};

struct Role role = {
  {1, 1},
  {1, 1}
};

int main(){
 role.display = &role_display;
 
 while(1){
   game();
 }
 return 0;
}

void gameinfo(){
 SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), FOREGROUND_INTENSITY |
      FOREGROUND_RED | FOREGROUND_GREEN );
  
 printf("=======================\n");
  printf("=====欢迎来玩游戏======\n");
 printf("=======================\n");
  SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE),FOREGROUND_INTENSITY |
      FOREGROUND_RED | FOREGROUND_BLUE);
}


void game(){
 role.display();
  gameinfo();
 // 遍历map中的数据
 int i,j;
  for( i=0; i< MAP_Y_LENGTH ; i++){
  for( j=0; j< MAP_X_LENGTH ; j++){
    int temp = display[i][j];
    
    if(temp == 1){
      printf("&curren;");
    }else if(temp == 2){
    printf("&prod;");
      }else if(temp == 3){
        printf("■");
      }else if(temp == 4){
    printf("N ");       
   }else if(temp == 5){// 猪
      printf("Z ");
   }else{
      printf(" ");
    }
  }
  printf("\n");
 }
 
 char a = getch();
 
 // 记录旧位置
 role.oldP.y = role.newP.y;
 role.oldP.x = role.newP.x;
 switch(a){
  case 's': // 向下
   if(NPC(deriction_down()) == 0){
    role.newP.y = role.newP.y + 1;// 向下走  
   }
  break;
  case 'd': // 向右
   if(NPC(deriction_right()) == 0){
    role.newP.x = role.newP.x + 1;
   }
  break;
  case 'a': // 向左
   if(NPC(deriction_left()) == 0){
    role.newP.x = role.newP.x - 1;
   }
  break;
   case 'w': // 向上
    if(NPC(deriction_up()) == 0){
    role.newP.y = role.newP.y - 1;
    }
  break;
 default: ;
 }
  system("cls");
}

char npc_info[4][80] = {
 "我：你就是传说中的NPC?\n",
 "NPC：你这是看不起NPC嘛！！！\n",
 "我：不不不，我是来接任务的！\n",
 "NPC：我有一头猪不见了，你帮我找回来吧！\n",
};


int NPC(int npc){
 printf("%d",npc);
 if(npc == 4){//遇到NPC
  system("cls");
  gameinfo();
  int i = 0,j=0;
  for(j=0;j < 4; j++){
   for(i=0; i < sizeof(npc_info[j]); i++){
    printf("%c",npc_info[j][i]);
    _sleep(10);
   }
   printf("\n");
  }
  printf("是否接收任务？\n");
  printf("1. 是  2. 否\n");
  char key = getchar();
  if(key == '1'){
   display[10][9] = 5;
   return 0;
  } else{
   return 4;
  } 
  
  return 4;
 }else if(npc == 5){//找到了猪
  system("cls");
  gameinfo();
  printf("恭喜你，任务完成！\n");
  system("pause");
  return 0;
 }
 return npc;
}



/* 向下 */
int deriction_down(){
 return display[role.newP.y + 1][role.newP.x];
}
/* 向上 */
int deriction_up(){
 return display[role.newP.y - 1][role.newP.x];
}
/* 向左 */
int deriction_left(){
 return display[role.newP.y][role.newP.x - 1];  
}
/* 向右 */
int deriction_right(){
 return display[role.newP.y][role.newP.x + 1];
}

/*
* 角色向上走
*/
void role_display(){
  display[role.oldP.y][role.oldP.x] = 0;
 display[role.newP.y][role.newP.x] = 1;
}

试试其它关键字

　RPG游戏　

同语言下

可能有用的

舞维天下贡献的其它代码(4)