問(wèn)題描述

我正在嘗試制作一種紙牌散開(kāi)的紙牌游戲.現(xiàn)在使用具有功能的 Allegro API 來(lái)顯示它:

I'm trying to make a card game where the cards fan out. Right now to display it Im using the Allegro API which has a function:

al_draw_rotated_bitmap(OBJECT_TO_ROTATE,CENTER_X,CENTER_Y,X
        ,Y,DEGREES_TO_ROTATE_IN_RADIANS);

所以有了這個(gè)，我可以輕松地制作我的粉絲效果.問(wèn)題是知道哪張卡在鼠標(biāo)下面.為此，我想到了進(jìn)行多邊形碰撞測(cè)試.我只是不確定如何旋轉(zhuǎn)卡片上的 4 個(gè)點(diǎn)來(lái)組成多邊形.我基本上需要做和Allegro一樣的操作.

so with this I can make my fan effect easily. The problem is then knowing which card is under the mouse. To do this I thought of doing a polygon collision test. I'm just not sure how to rotate the 4 points on the card to make up the polygon. I basically need to do the same operation as Allegro.

比如卡片的4點(diǎn)是:

card.x

card.y

card.x + card.width

card.y + card.height

我需要一個(gè)類似的功能:

I would need a function like:

POINT rotate_point(float cx,float cy,float angle,POINT p)
{
}

謝謝

推薦答案

先減去樞軸點(diǎn)(cx,cy)，然后旋轉(zhuǎn)，再添加點(diǎn).

First subtract the pivot point (cx,cy), then rotate it, then add the point again.

未經(jīng)測(cè)試:

POINT rotate_point(float cx,float cy,float angle,POINT p)
{
  float s = sin(angle);
  float c = cos(angle);

  // translate point back to origin:
  p.x -= cx;
  p.y -= cy;

  // rotate point
  float xnew = p.x * c - p.y * s;
  float ynew = p.x * s + p.y * c;

  // translate point back:
  p.x = xnew + cx;
  p.y = ynew + cy;
  return p;
}

這篇關(guān)于圍繞另一個(gè)點(diǎn)旋轉(zhuǎn)一個(gè)點(diǎn) (2D)的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！