2D line intersection point with a bounding box

What would be an efficient algorithm for calculating the intersection point of a line starting a (x,y), with an angle of θ, and a bounding rectangle with co-orindates (0,0 to w,h)? Assuming the line origin is within the box.

In basic ascii art (for a line starting at 2,6 and θ approx 9π/8):

 h---------------...h,w
 .               ...
 .               ... 
 .               ...
 7               ...
 6   x           ...
 5  /            ...
 4 /             ...
 3/    开发者_运维问答          ...
 2               ...
/1               ...
 0 1 2 3 4 5 6 7 ... w

All variables, except θ, are integers.

Assume we parametrize the line with a variable t, i.e. each point on the line can be written as

[x(t),y(t)] = [x,y] + t * [dx,dy]

with constants

dx = cos(theta)
dy = sin(theta)

Then you can first calculate the cut through the vertical boundaries:

if dx<0 then        // line going to the left -> intersect with x==0
  t1 = -x/dx
  x1 = 0
  y1 = y + t1*dy
else if dx>0 then   // line going to the right -> intersect with x==w
  t1 = (w-x)/dx
  x1 = w
  y1 = y + t1*dy
else                // line going straight up or down -> no intersection possible
  t1 = infinity
end

where t1 is the distance to the intersection point and [x1,y1] the point itself. Second you do the same for the upper and lower boundaries:

if dy<0 then        // line going downwards -> intersect with y==0
  t2 = -y/dy
  x2 = x + t2*dx
  y2 = 0
else if dy>0 then   // line going upwards -> intersect with y==h
  t2 = (h-y)/dy
  x2 = x + t2*dx
  y2 = h
else
  t2 = infinity
end

Now you just have to select the point with less distance from your origin [x,y].

if t1 < t2 then
  xb = x1
  yb = y1
else
  xb = x2
  yb = y2
end

Note that this algorithm only works if the starting point is inside the bounding box, i.e. 0 <= x <= w and 0 <= y <= h.