C语言实现俄罗斯方块小游戏

这篇文章主要为大家详细介绍了Linux下C语言实现俄罗斯方块小游戏,文中示例代码介绍的非常详细,具有一定的参考价值,感兴趣的小伙伴们可以参考一下

C语言实现俄罗斯方块小游戏的制作代码,具体内容如下

 #include  #include  #include  #define TTY_PATH "/dev/tty" #define STTY_ON "stty raw -echo -F" #define STTY_OFF "stty -raw echo -F" int map[21][14]; char direct; int node[7][4][16]={ {{0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0},//长方形 {0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0}, {0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0}, {0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0}}, {{1,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0},//正方形 {1,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0}, {1,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0}, {1,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0}}, {{0,1,0,0,1,1,1,0,0,0,0,0,0,0,0,0},//3边加一中点 {0,1,0,0,0,1,1,0,0,1,0,0,0,0,0,0}, {0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0}, {0,1,0,0,1,1,0,0,0,1,0,0,0,0,0,0}}, {{0,1,1,0,0,1,0,0,0,1,0,0,0,0,0,0},//右锄头型 {0,0,0,0,1,1,1,0,0,0,1,0,0,0,0,0}, {0,1,0,0,0,1,0,0,1,1,0,0,0,0,0,0}, {1,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0}}, {{1,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0},//左锄头型 {0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,0}, {0,1,0,0,0,1,0,0,0,1,1,0,0,0,0,0}, {0,0,0,0,1,1,1,0,1,0,0,0,0,0,0,0}}, {{0,1,0,0,0,1,1,0,0,0,1,0,0,0,0,0},//右曲折型 {0,1,1,0,1,1,0,0,0,0,0,0,0,0,0,0}, {0,1,0,0,0,1,1,0,0,0,1,0,0,0,0,0}, {0,1,1,0,1,1,0,0,0,0,0,0,0,0,0,0}}, {{0,1,0,0,1,1,0,0,1,0,0,0,0,0,0,0},//左曲折型 {1,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0}, {0,1,0,0,1,1,0,0,1,0,0,0,0,0,0,0}, {1,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0}} }; typedef struct block { int x; int y; int blockType; int blockDirect; }Block; Block bl; void init_map()//初始化边框 { int i,j; for(i=0; i<21; i++) for(j=0; j<14; j++) { if(j==0 || j==13) map[i][j] = 200; else if(i==20) map[i][j] = 201; else map[i][j] = 0; } } void new_block()//生成随机的俄罗斯方块 { int blockType = rand()%7; int blockDirect = rand()%4; int x = 1; int y = 5; bl.x = x; bl.y = y; bl.blockType = blockType; bl.blockDirect = blockDirect; } void input()//将移动后的俄罗斯方块,导入地图中作标记 { int i, j; for(i=0; i<4; i++) for(j=0; j<4; j++) if(node[bl.blockType][bl.blockDirect][i*4+j]==1) { map[bl.x+i][bl.y+j] = 1; } } void output()//移动时,将之前俄罗斯方块在地图信息清空。 { int i, j; for(i=0; i<4; i++) for(j=0; j<4; j++) if(node[bl.blockType][bl.blockDirect][i*4+j]==1) { map[bl.x+i][bl.y+j] = 0; } } void change()//俄罗斯方格在碰撞后融入,固定 { int i, j; for(i=0; i<4; i++) for(j=0; j<4; j++) if(node[bl.blockType][bl.blockDirect][i*4+j]==1) { map[bl.x+i][bl.y+j] = 10; } for(j=1; j<13; j++) if(map[5][j] == 10) { system("clear"); printf("game over !!!!!!!!!\n"); exit(1); } } void print_map()//打印地图,显示信息 { int i,j; for(i=5; i<21; i++) { for(j=0; j<14; j++) { if(map[i][j]==200)//左右边界 printf("#"); else if(map[i][j]==201)//下边界 printf(" # "); else if(map[i][j]==0)//空白地 printf(" "); else if(map[i][j]==1)//移动的俄罗斯方块 printf(" * "); else if(map[i][j]==10)//固定的俄罗斯方块 printf(" @ "); } printf("\n"); } } void delLine(int n)//消行 { int i,j; for(j = 1; j<13; j++) map[n][j] = 0; for(i = n; i>5 ; i--) for(j = 1; j<13; j++) if(map[i-1][j] != 1) map[i][j] = map[i-1][j]; } void isFillLine()//是否满足消行条件 { int i,j; int fals; for(i=19; i>5; i--) { fals = 1; for(j=1; j<13; j++) { if(map[i][j] != 10) { fals = 0; continue; } } if(fals) { delLine(i); } } } void down()//下移 { int i, j; int fale = 1; for(i=3; i>=0; i--) for(j=0; j<4; j++) if(node[bl.blockType][bl.blockDirect][i*4+j] == 1) if(map[bl.x+i+1][bl.y+j] == 10 || map[bl.x+i+1][bl.y+j] == 201) { change(); fale = 0; new_block(); isFillLine(); return; } if(fale) { output(); bl.x += 1; input(); } } void right()//右移 { int i, j; int fale = 1; for(i=3; i>=0; i--) for(j=0; j<4; j++) if(node[bl.blockType][bl.blockDirect][i*4+j] == 1) if(map[bl.x+i][bl.y+j+1] == 10 || map[bl.x+i][bl.y+j+1] == 200) { fale = 0; return; } if(fale) { output(); bl.y += 1; input(); } } void left()//左移 { int i, j; int fale = 1; for(i=3; i>=0; i--) for(j=0; j<4; j++) if(node[bl.blockType][bl.blockDirect][i*4+j] == 1) if(map[bl.x+i][bl.y+j-1] == 10 || map[bl.x+i][bl.y+j-1] == 200) { fale = 0; return; } if(fale) { output(); bl.y -= 1; input(); } } void change_block()//俄罗斯方块变形 { int i,j; output(); int fals = 1; bl.blockDirect += 1; bl.blockDirect %= 4; for(i=0; i<4; i++) for(j=0; j<4; j++) if(node[bl.blockType][bl.blockDirect][i*4+j]==1) if(map[bl.x+i][bl.y+j] != 0 ) { fals = 0; break; } if(fals) { input(); }else { bl.blockDirect -= 1; input(); } } 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()//q 退出游戏,a,d 左右移动,空格变形 { srand(time(NULL)); init_map(); new_block(); input(); char ch; int num = 0; while(1) { usleep(500000); system(STTY_ON TTY_PATH); ch = in_direct(); system(STTY_OFF TTY_PATH); system("clear"); if(ch == 'a' && num <= 1) { left(); print_map(); num++; continue; }else if(ch == 'd' && num <= 1) { right(); print_map(); num++; continue; }else if(ch == ' ' && num <= 1 ) { change_block(); print_map(); num++; continue; }else if(ch == 'q') { system("clear"); printf("gave over!!!!!\n"); exit(0); } down(); print_map(); num = 0; } return 0; }

更多俄罗斯方块精彩文章请点击专题:俄罗斯方块游戏集合 进行学习。

更多有趣的经典小游戏实现专题,分享给大家:

C++经典小游戏汇总

python经典小游戏汇总

JavaScript经典游戏 玩不停

java经典小游戏汇总

javascript经典小游戏汇总

以上就是C语言实现俄罗斯方块小游戏的详细内容,更多请关注0133技术站其它相关文章!

赞(0) 打赏
未经允许不得转载:0133技术站首页 » C语言