如何在Processing中计算从mouseX、mouseY到矩形的dist()

问题描述

如果它是某个点的距离，那就是

If it was the dist to a point it would be

dist(mouseX, mouseY, x, y)

为了

point(x,y)

但是我怎样才能从鼠标的当前位置计算 dist() 到

but how can I calculate dist() from the mouse's current position to

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

谢谢

推荐答案

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，您需要确定闭合点是在一侧还是在角落.这张图片可能会有所帮助，它显示了一个点与该点不同位置的矩形的距离:

Basically, you need to figure out if the closes point is on one of the sides, or in the corner. This picture may help, it shows the distance of a point from a rectangle for different positions of the point:

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