如何从mouseX,mouseY到Processing中的矩形计算dist()

Question

如果它是一个点,那将是

dist(mouseX, mouseY, x, y)

对于

point(x,y)

但是如何从鼠标的当前位置计算dist()

rectMode(CORNERS);
rect(x1,y2,x2,y2);

谢谢

Answer 1

这样的事情应该这样做:

float distrect(float x, float y, float x1, float y1, float x2, float y2){
  float dx1 = x - x1;
  float dx2 = x - x2;
  float dy1 = y - y1;
  float dy2 = y - y2;

  if (dx1*dx2 < 0) { // x is between x1 and x2
    if (dy1*dy2 < 0) { // (x,y) is inside the rectangle
      return min(min(abs(dx1), abs(dx2)),min(abs(dy1),abs(dy2)));
    }
    return min(abs(dy1),abs(dy2));
  }
  if (dy1*dy2 < 0) { // y is between y1 and y2
    // we don't have to test for being inside the rectangle, it's already tested.
    return min(abs(dx1),abs(dx2));
  }
  return min(min(dist(x,y,x1,y1),dist(x,y,x2,y2)),min(dist(x,y,x1,y2),dist(x,y,x2,y1)));
}

基本上,你需要弄清楚关闭点是在一侧还是在角落里.这张图可能有所帮助,它显示了点与不同位置的矩形点的距离: