update to xserver 21.1.12

xserver/mi/mipoly.h

@@ -122,7 +122,7 @@ typedef struct _ScanLineListBlock {
  * to scanlines() :  Must be an even number
  */
 #define NUMPTSTOBUFFER 200
-
+
 /*
  *
  *     a few macros for the inner loops of the fill code where

xserver/mi/miscanfill.h

@@ -50,7 +50,7 @@ from The Open Group.
  *     overhead is out of the question.
  *     See the author for a derivation if needed.
  */
-
+
 /*
  *  In scan converting polygons, we want to choose those pixels
  *  which are inside the polygon.  Thus, we add .5 to the starting
@@ -95,7 +95,7 @@ from The Open Group.
         } \
     } \
 }
-
+
 #define BRESINCRPGON(d, minval, m, m1, incr1, incr2) { \
     if (m1 > 0) { \
         if (d > 0) { \
@@ -117,7 +117,7 @@ from The Open Group.
         } \
     } \
 }
-
+
 /*
  *     This structure contains all of the information needed
  *     to run the bresenham algorithm.

xserver/mi/mizerclip.c

@@ -109,7 +109,7 @@ end of the line, we will find the largest number of Y steps that
 satisfies the inequality.  In that case, since we are representing
 the Y steps as (dy - N), we will actually want to solve for the
 smallest N in that equation.
-
+
 Case 1:  X major, starting X coordinate moved by M steps
 
 		-2dx <= 2Mdy - 2Ndx - dx - B < 0
@@ -157,7 +157,7 @@ steps, so we want the highest N, so we use the < inequality:
 	  = floor((2Mdy + dx + B + 2dx - 1) / 2dx) - 1
 	  = floor((2Mdy + dx + B + 2dx - 1 - 2dx) / 2dx)
 	  = floor((2Mdy + dx + B - 1) / 2dx)
-
+
 Case 3: Y major, starting X coordinate moved by M steps
 
 		-2dy <= 2Ndx - 2Mdy - dy - B < 0
@@ -203,7 +203,7 @@ Same analysis as Case 4, but we want the smallest number of Y steps
 which means the largest N, so we use the <= inequality:
 
 	N = floor((2Mdy + dy - B) / 2dx)
-
+
 Now let's try the Y coordinates, we have the same 4 cases.
 
 Case 5: X major, starting Y coordinate moved by N steps
@@ -248,7 +248,7 @@ Same derivations as Case 6, but we want the smallest # of X steps
 which means the largest M, so use the <= inequality:
 
 	M = floor((2Ndx + dx - B) / 2dy)
-
+
 Case 7: Y major, starting Y coordinate moved by N steps
 
 		-2dy <= 2Ndx - 2Mdy - dy - B < 0
@@ -293,7 +293,7 @@ steps which means the largest M, so we use the < inequality:
 	  = floor((2Ndx + dy + B + 2dy - 1) / 2dy) - 1
 	  = floor((2Ndx + dy + B + 2dy - 1 - 2dy) / 2dy)
 	  = floor((2Ndx + dy + B - 1) / 2dy)
-
+
 So, our equations are:
 
 	1:  X major move x1 to x1+M	floor((2Mdy + dx - B) / 2dx)