C语言实现2048游戏
时间:2021-03-17 09:39:14|栏目:C代码|点击: 次
本文实例为大家分享了C语言实现2048小游戏的具体代码,供大家参考,具体内容如下
具有以下特点:
1.linux下完成
2.非堵塞键盘读取
3.随机生成2和4
#include <stdio.h> #include <stdlib.h> #include <unistd.h> #define TTY_PATH "/dev/tty" #define STTY_ON "stty raw -echo -F" #define STTY_OFF "stty -raw echo -F" int map[4][4]; typedef struct node { int x; int y; int num; }Node; Node node; void init_map()//初始化全部方格 { int i, j; for(i=0; i<4; i++) for(j=0; j<4; j++) map[i][j] = 0; } void new_node()//新增方格,避免重复。 { int x = 0, y = 0 ,num = 0; do{ int fals = 1; int i,j; for(i=0; i<4; i++) for(j=0; j<4; j++) if(map[i][j] == 0) fals =0; if(fals)//新方格无法被填入,游戏失败 { system("clear"); printf("game over!!!!!!!!!!!\n"); exit(0); } x = rand() % 4; y = rand() % 4; num = rand() % 2; if(num == 0) num = 3; else if(num == 1) num = 5; }while(map[x][y] > 0); node.x = x; node.y = y; node.num = num; map[x][y] = node.num; } void show()//彩色打印数字 { int i, j; for(i=0; i<4; i++) { for(j=0; j<4; j++) { if(map[i][j]%2 == 1) { map[i][j] -= 1; printf("\33[31m%d\33[0m\t",map[i][j]); } else if(map[i][j] == 0) { printf("%d\t",map[i][j]); } else if(map[i][j] == 2) { printf("\33[32m%d\33[0m\t",map[i][j]); } else if(map[i][j] == 4) { printf("\33[33m%d\33[0m\t",map[i][j]); } else if(map[i][j] == 8) { printf("\33[34m%d\33[0m\t",map[i][j]); } else if(map[i][j] == 16) { printf("\33[35m%d\33[0m\t",map[i][j]); } else if(map[i][j] == 32) { printf("\33[36m%d\33[0m\t",map[i][j]); } else { printf("\33[44m%d\33[0m\t",map[i][j]); } } printf("\n"); } } void left() { int i, j, z, tmp; for(i=0; i<4; i++)//全体方格左移 for(j=0; j<4; j++) if(map[i][j] == 0 ) for(z = j + 1; z<4; z++) if(map[i][z] > 0) { tmp = map[i][j]; map[i][j] = map[i][z]; map[i][z] = tmp; break; } for(i=0; i<4; i++)//如果方格数字相同则,相加,通过主函数多次调用,排列好 for(j=0; j<4; j++) if(map[i][j] > 0 ) for(z = j + 1; z<4; z++) if(map[i][z] > 0) if(map[i][z] == map[i][j]) { map[i][j] *= 2; map[i][z] = 0; }else break; else break; else break; } void right() { int i, j, z, tmp; for(i=0; i<4; i++) for(j=3; j>=0; j--) if(map[i][j] == 0 ) for(z = j-1; z>=0; z--) if(map[i][z] > 0) { tmp = map[i][j]; map[i][j] = map[i][z]; map[i][z] = tmp; break; } for(i=0; i<4; i++) for(j=3; j>=0; j--) if(map[i][j] > 0 ) for(z = j-1; z>=0; z--) if(map[i][z] > 0) if(map[i][z] == map[i][j]) { map[i][j] *= 2; map[i][z] = 0; }else break; else break; else break; } void up() { int i, j, z, tmp; for(i=0; i<4; i++) for(j=0; j<4; j++) if(map[j][i] == 0 ) for(z = j+1; z<4; z++) if(map[z][i] > 0) { tmp = map[j][i]; map[j][i] = map[z][i]; map[z][i] = tmp; break; } for(i=0; i<4; i++) for(j=0; j<4; j++) if(map[j][i] > 0 ) for(z = j+1; z<4; z++) if(map[z][i] > 0) if(map[z][i] == map[j][i]) { map[j][i] *= 2; map[z][i] = 0; }else break; else break; else break; } void down() { int i, j, z, tmp; for(i=0; i<4; i++) for(j=3; j>=0; j--) if(map[j][i] == 0 ) for(z = j-1; z>=0; z--) if(map[z][i] > 0) { tmp = map[j][i]; map[j][i] = map[z][i]; map[z][i] = tmp; break; } for(i=0; i<4; i++) for(j=3; j>=0; j--) if(map[j][i] > 0 ) for(z = j-1; z>=0; z--) if(map[z][i] > 0) if(map[z][i] == map[j][i]) { map[j][i] *= 2; map[z][i] = 0; }else break; else break; else break; } void move(char ch) { switch(ch) { case 'a': left(); break; case 'd': right(); break; case 'w': up(); break; case 's': down(); break; } } char in_direct()//非堵塞输入 { fd_set fd; struct timeval tv; char ch; FD_ZERO(&fd); FD_SET(0, &fd); tv.tv_sec = 0; tv.tv_usec = 10; if(select(1, &fd ,NULL, NULL, &tv) > 0) { ch = getchar(); } return ch; } int main() { srand(time(NULL)); init_map(); new_node(); show(); char ch; int i=0; while(1) { system(STTY_ON TTY_PATH); ch = in_direct(); system(STTY_OFF TTY_PATH); if(ch=='a'||ch=='d'||ch=='s'||ch=='w') { system("clear"); for(i=0;i<3;i++)//重复多次才能排序好 move(ch); new_node(); show(); } if(ch=='q')//退出游戏 { system("clear"); printf("game over!!!!!!!!\n"); break; } usleep(500000); } return 0; }
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