-// fast PointInfrontOfTriangle
-// subtracts v1 from v0 and v2, combined into a crossproduct, combined with a
-// dotproduct of the light location relative to the first point of the
-// triangle (any point works, since any triangle is obviously flat), and
-// finally a comparison to determine if the light is infront of the triangle
-// (the goal of this statement) we do not need to normalize the surface
-// normal because both sides of the comparison use it, therefore they are
-// both multiplied the same amount... furthermore the subtract can be done
-// on the vectors, saving a little bit of math in the dotproducts
-#define PointInfrontOfTriangle(p,a,b,c) (((p)[0] - (a)[0]) * (((a)[1] - (b)[1]) * ((c)[2] - (b)[2]) - ((a)[2] - (b)[2]) * ((c)[1] - (b)[1])) + ((p)[1] - (a)[1]) * (((a)[2] - (b)[2]) * ((c)[0] - (b)[0]) - ((a)[0] - (b)[0]) * ((c)[2] - (b)[2])) + ((p)[2] - (a)[2]) * (((a)[0] - (b)[0]) * ((c)[1] - (b)[1]) - ((a)[1] - (b)[1]) * ((c)[0] - (b)[0])) > 0)
+/*! Fast PointInfrontOfTriangle.
+ * subtracts v1 from v0 and v2, combined into a crossproduct, combined with a
+ * dotproduct of the light location relative to the first point of the
+ * triangle (any point works, since any triangle is obviously flat), and
+ * finally a comparison to determine if the light is infront of the triangle
+ * (the goal of this statement) we do not need to normalize the surface
+ * normal because both sides of the comparison use it, therefore they are
+ * both multiplied the same amount... furthermore a subtract can be done on
+ * the point to eliminate one dotproduct
+ * this is ((p - a) * cross(a-b,c-b))
+ */
+#define PointInfrontOfTriangle(p,a,b,c) \
+( ((p)[0] - (a)[0]) * (((a)[1] - (b)[1]) * ((c)[2] - (b)[2]) - ((a)[2] - (b)[2]) * ((c)[1] - (b)[1])) \
++ ((p)[1] - (a)[1]) * (((a)[2] - (b)[2]) * ((c)[0] - (b)[0]) - ((a)[0] - (b)[0]) * ((c)[2] - (b)[2])) \
++ ((p)[2] - (a)[2]) * (((a)[0] - (b)[0]) * ((c)[1] - (b)[1]) - ((a)[1] - (b)[1]) * ((c)[0] - (b)[0])) > 0)
+