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bobobobo
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Nobody described the algorithm here, but the Graphics Gems algorithm is simply:

  1. Using your ray's direction vector, determine which 3 of the 6 candidate planes would be hit first. If your (unnormalized) ray direction vector is (-1, 1, -1), then the 3 planes that are possible to be hit are +x, -y, and +z.

  2. Of the 3 candidate planes, do find the t-value for the intersection for each. Accept the plane that gets the largest t value as being the plane that got hit, and check that the hit is within the box. The diagram in the text makes this clear:

enter image description here

My implementation:

bool AABB::intersects( const Ray& ray )
{
  // EZ cases: if the ray starts inside the box, or ends inside
  // the box, then it definitely hits the box.
  // I'm using this code for ray tracing with an octree,
  // so I needed rays that start and end within an
  // octree node to COUNT as hits.
  // You could modify this test to (ray starts inside and ends outside)
  // to qualify as a hit if you wanted to NOT count totally internal rays
  if( containsIn( ray.startPos ) || containsIn( ray.getEndPoint() ) )
    return true ; 

  // the algorithm says, find 3 t's,
  Vector t ;

  // LARGEST t is the only one we need to test if it's on the face.
  for( int i = 0 ; i < 3 ; i++ )
  {
    if( ray.direction.e[i] > 0 ) // CULL BACK FACE
      t.e[i] = ( min.e[i] - ray.startPos.e[i] ) / ray.direction.e[i] ;
    else
      t.e[i] = ( max.e[i] - ray.startPos.e[i] ) / ray.direction.e[i] ;
  }

  int mi = t.maxIndex() ;
  if( BetweenIn( t.e[mi], 0, ray.length ) )
  {
    Vector pt = ray.at( t.e[mi] ) ;

    // check it's in the box in other 2 dimensions
    int o1 = ( mi + 1 ) % 3 ; // i=0: o1=1, o2=2, i=1: o1=2,o2=0 etc.
    int o2 = ( mi + 2 ) % 3 ;

    return BetweenIn( pt.e[o1], min.e[o1], max.e[o1] ) &&
           BetweenIn( pt.e[o2], min.e[o2], max.e[o2] ) ;
  }

  return false ; // the ray did not hit the box.
}
bobobobo
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