2048游戏粤嵌嵌入式系统代码
时间: 2024-04-07 22:25:49 浏览: 12
2048游戏是一款益智类游戏,玩家需要通过合并相同数字的方块,最终得到一个数字为2048的方块。下面是一个简单的2048游戏的嵌入式系统代码示例:
```c
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
#define SIZE 4
int board[SIZE][SIZE];
void init_board() {
for (int i = 0; i < SIZE; i++) {
for ( j = 0; j < SIZE; j++) {
board[i][j] = 0;
}
}
}
void print_board() {
system("cls");
printf("2048 Game\n");
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
printf("%4d", board[i][j]);
}
printf("\n");
}
}
void generate_number() {
int empty_count = 0;
int empty_pos[SIZE * SIZE][2];
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
if (board[i][j] == 0) {
empty_pos[empty_count][0] = i;
empty_pos[empty_count][1] = j;
empty_count++;
}
}
}
if (empty_count == 0) {
return;
}
int index = rand() % empty_count;
int x = empty_pos[index][0];
int y = empty_pos[index][1];
board[x][y] = (rand() % 2 + 1) * 2;
}
void move_left() {
for (int i = 0; i < SIZE; i++) {
int current = 0;
int last_merged = -1;
for (int j = 0; j < SIZE; j++) {
if (board[i][j] != 0) {
if (current == 0) {
current = board[i][j];
} else {
if (board[i][j] == current && last_merged != j - 1) {
board[i][j] *= 2;
board[i][last_merged + 1] = 0;
last_merged = j;
current = 0;
} else {
current = board[i][j];
}
}
}
}
}
}
// 其他移动方向的函数(move_right, move_up, move_down)类似,这里省略
int main() {
init_board();
generate_number();
generate_number();
while (1) {
print_board();
char ch = getch();
if (ch == 'a') {
move_left();
} else if (ch == 'd') {
move_right();
} else if (ch == 'w') {
move_up();
} else if (ch == 's') {
move_down();
} else if (ch == 'q') {
break;
}
generate_number();
}
return 0;
}
```
这段代码实现了一个简单的2048游戏,使用了一个4x4的二维数组来表示游戏面板,通过键盘输入控制方块的移动和合并。你可以根据需要进行修改和扩展。