僅供參考,這裏是鏈接到我的任務(當然是一個C語言課程):Snake不知道我的蛇遊戲的邏輯有什麼問題嗎?
口被認爲是「<」,「>」,「V」或「^」取決於方向,身體部分應該是「*」。所以首先,我的上下箭頭鍵控制被切換,但它打印正確的字符(即,當它向下移動時打印「v」,向上移動時打印「^」),並且我不明白爲什麼。
此外,隨着頭部的移動,它剛脫離身體並且不會打印新的身體片段(請參閱Screencap)。
下面是我的自定義結構,主,scene_render和scene_update功能(我們是不允許修改此分配的主要功能):
struct Point{
int x;
int y;
};
struct Snake {
struct Point segments[MAX_SEGMENTS];
int num_segments;
int dir; //0=up, 1=down, 2=left, 3=right
};
struct Scene {
// TODO: add fields
struct Snake snake;
struct Point fruit;
};
主要
int main(void) {
// Important: do NOT modify the main function
struct Scene scene;
scene_init(&scene);
int keep_going = 1;
while (keep_going == 1) {
scene_render(&scene);
cons_update();
scene_delay(&scene);
keep_going = scene_update(&scene);
}
return 0;
}
scene_render
void scene_render(const struct Scene *s) {
// TODO: add your code
int num=s->snake.num_segments;
cons_clear_screen();
cons_move_cursor(s->snake.segments[0].y, s->snake.segments[0].x);
if(s->snake.dir==0)
{
cons_printw("v");
}
else if(s->snake.dir==1)
{
cons_printw("^");
}
else if(s->snake.dir==2)
{
cons_printw(">");
}
else
{
cons_printw("<");
}
for(int i=1; i <num; i++)
{
cons_move_cursor(s->snake.segments[i].y, s->snake.segments[i].x);
cons_printw("*");
}
cons_move_cursor(23, 79); //just to get blinking cursor out of the way
}
scene_update
int scene_update(struct Scene *s) {
// This function should return 0 if the player presses 'q',
// 1 otherwise.
struct Scene temp=*s;
int num=temp.snake.num_segments;
int key = cons_get_keypress();
// TODO: add your code
if(key=='q')
{
return 0;
}
if(key==UP_ARROW && temp.snake.dir!=1)
{
temp.snake.dir=0;
temp.snake.segments[0].y=(temp.snake.segments[0].y)+1;
for(int i=1; i<num; i++)
{
temp.snake.segments[i].y=s->snake.segments[i].y;
temp.snake.segments[i].x=s->snake.segments[i].x;
}
}
else if(key==DOWN_ARROW && temp.snake.dir!=0)
{
temp.snake.dir=1;
temp.snake.segments[0].y=(temp.snake.segments[0].y)-1;
for(int i=1; i<num; i++)
{
temp.snake.segments[i].y=s->snake.segments[i].y;
temp.snake.segments[i].x=s->snake.segments[i].x;
}
}
else if(key==LEFT_ARROW && temp.snake.dir!=3)
{
temp.snake.dir=2;
temp.snake.segments[0].x=(temp.snake.segments[0].x)-1;
for(int i=1; i<num; i++)
{
temp.snake.segments[i].y=s->snake.segments[i].y;
temp.snake.segments[i].x=s->snake.segments[i].x;
}
}
else if(key==RIGHT_ARROW && temp.snake.dir!=2)
{
temp.snake.dir=3;
temp.snake.segments[0].x=(temp.snake.segments[0].x)+1;
for(int i=1; i<num; i++)
{
temp.snake.segments[i].y=s->snake.segments[i].y;
temp.snake.segments[i].x=s->snake.segments[i].x;
}
}
else if(key==-1)
{
if(temp.snake.dir==0)
{
temp.snake.segments[0].y=(temp.snake.segments[0].y)+1;
for(int i=1; i<num; i++)
{
temp.snake.segments[i].y=s->snake.segments[i].y;
temp.snake.segments[i].x=s->snake.segments[i].x;
}
}
else if(temp.snake.dir==1)
{
temp.snake.segments[0].y=(temp.snake.segments[0].y)-1;
for(int i=1; i<num; i++)
{
temp.snake.segments[i].y=s->snake.segments[i].y;
temp.snake.segments[i].x=s->snake.segments[i].x;
}
}
else if(temp.snake.dir==2)
{
temp.snake.segments[0].x=(temp.snake.segments[0].x)-1;
for(int i=1; i<num; i++)
{
temp.snake.segments[i].y=s->snake.segments[i].y;
temp.snake.segments[i].x=s->snake.segments[i].x;
}
}
else
{
temp.snake.segments[0].x=(temp.snake.segments[0].x)+1;
for(int i=1; i<num; i++)
{
temp.snake.segments[i].y=s->snake.segments[i].y;
temp.snake.segments[i].x=s->snake.segments[i].x;
}
}
}
temp.fruit.x=s->fruit.x;
temp.fruit.y=s->fruit.y;
*s=temp;
return 1;
}
這解決了我的問題,並且我能夠修復上下控件,謝謝! –
什麼是附加段的最簡單方法?我有if語句,說如果蛇頭的座標等於水果的座標,那麼做(某些事情) –
@ mjones.udri我會在與當前尾部相同的位置添加一個段(在另一個頂部有兩個段)。蛇移動一次後,新的尾巴將保持原位,然後蛇再次移動後,尾巴將開始追蹤它前面的棋子。 – JS1