XFree86 4.3.0.1xf86-4_3_0_1PRE_xf86-4_3_0_1

1 files changed, 3 insertions, 587 deletions

diff --git a/mi/mizerline.c b/mi/mizerline.c
index 2279cb822..d1fbf631e 100644
--- a/mi/mizerline.c
+++ b/mi/mizerline.c
@@ -1,3 +1,4 @@
+/* $XFree86: xc/programs/Xserver/mi/mizerline.c,v 3.7 2001/12/14 20:00:29 dawes Exp $ */
 /***********************************************************
 
 Copyright 1987, 1998  The Open Group
@@ -55,591 +56,6 @@ SOFTWARE.
 #include "mi.h"
 #include "miline.h"
 
-/*
-
-The bresenham error equation used in the mi/mfb/cfb line routines is:
-
-	e = error
-	dx = difference in raw X coordinates
-	dy = difference in raw Y coordinates
-	M = # of steps in X direction
-	N = # of steps in Y direction
-	B = 0 to prefer diagonal steps in a given octant,
-	    1 to prefer axial steps in a given octant
-
-	For X major lines:
-		e = 2Mdy - 2Ndx - dx - B
-		-2dx <= e < 0
-
-	For Y major lines:
-		e = 2Ndx - 2Mdy - dy - B
-		-2dy <= e < 0
-
-At the start of the line, we have taken 0 X steps and 0 Y steps,
-so M = 0 and N = 0:
-
-	X major	e = 2Mdy - 2Ndx - dx - B
-		  = -dx - B
-
-	Y major	e = 2Ndx - 2Mdy - dy - B
-		  = -dy - B
-
-At the end of the line, we have taken dx X steps and dy Y steps,
-so M = dx and N = dy:
-
-	X major	e = 2Mdy - 2Ndx - dx - B
-		  = 2dxdy - 2dydx - dx - B
-		  = -dx - B
-	Y major e = 2Ndx - 2Mdy - dy - B
-		  = 2dydx - 2dxdy - dy - B
-		  = -dy - B
-
-Thus, the error term is the same at the start and end of the line.
-
-Let us consider clipping an X coordinate.  There are 4 cases which
-represent the two independent cases of clipping the start vs. the
-end of the line and an X major vs. a Y major line.  In any of these
-cases, we know the number of X steps (M) and we wish to find the
-number of Y steps (N).  Thus, we will solve our error term equation.
-If we are clipping the start of the line, we will find the smallest
-N that satisfies our error term inequality.  If we are clipping the
-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
-	2Ndx <= 2Mdy - dx - B + 2dx	2Ndx > 2Mdy - dx - B
-	2Ndx <= 2Mdy + dx - B		N > (2Mdy - dx - B) / 2dx
-	N <= (2Mdy + dx - B) / 2dx
-
-Since we are trying to find the smallest N that satisfies these
-equations, we should use the > inequality to find the smallest:
-
-	N = floor((2Mdy - dx - B) / 2dx) + 1
-	  = floor((2Mdy - dx - B + 2dx) / 2dx)
-	  = floor((2Mdy + dx - B) / 2dx)
-
-Case 1b: X major, ending X coordinate moved to M steps
-
-Same derivations as Case 1, but we want the largest N that satisfies
-the equations, so we use the <= inequality:
-
-	N = floor((2Mdy + dx - B) / 2dx)
-
-Case 2: X major, ending X coordinate moved by M steps
-
-		-2dx <= 2(dx - M)dy - 2(dy - N)dx - dx - B < 0
-		-2dx <= 2dxdy - 2Mdy - 2dxdy + 2Ndx - dx - B < 0
-		-2dx <= 2Ndx - 2Mdy - dx - B < 0
-	2Ndx >= 2Mdy + dx + B - 2dx	2Ndx < 2Mdy + dx + B
-	2Ndx >= 2Mdy - dx + B		N < (2Mdy + dx + B) / 2dx
-	N >= (2Mdy - dx + B) / 2dx
-
-Since we are trying to find the highest number of Y steps that
-satisfies these equations, we need to find the smallest N, so
-we should use the >= inequality to find the smallest:
-
-	N = ceiling((2Mdy - dx + B) / 2dx)
-	  = floor((2Mdy - dx + B + 2dx - 1) / 2dx)
-	  = floor((2Mdy + dx + B - 1) / 2dx)
-
-Case 2b: X major, starting X coordinate moved to M steps from end
-
-Same derivations as Case 2, but we want the smallest number of Y
-steps, so we want the highest N, so we use the < inequality:
-
-	N = ceiling((2Mdy + dx + B) / 2dx) - 1
-	  = 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
-	2Ndx >= 2Mdy + dy + B - 2dy	2Ndx < 2Mdy + dy + B
-	2Ndx >= 2Mdy - dy + B		N < (2Mdy + dy + B) / 2dx
-	N >= (2Mdy - dy + B) / 2dx
-
-Since we are trying to find the smallest N that satisfies these
-equations, we should use the >= inequality to find the smallest:
-
-	N = ceiling((2Mdy - dy + B) / 2dx)
-	  = floor((2Mdy - dy + B + 2dx - 1) / 2dx)
-	  = floor((2Mdy - dy + B - 1) / 2dx) + 1
-
-Case 3b: Y major, ending X coordinate moved to M steps
-
-Same derivations as Case 3, but we want the largest N that satisfies
-the equations, so we use the < inequality:
-
-	N = ceiling((2Mdy + dy + B) / 2dx) - 1
-	  = floor((2Mdy + dy + B + 2dx - 1) / 2dx) - 1
-	  = floor((2Mdy + dy + B + 2dx - 1 - 2dx) / 2dx)
-	  = floor((2Mdy + dy + B - 1) / 2dx)
-
-Case 4: Y major, ending X coordinate moved by M steps
-
-		-2dy <= 2(dy - N)dx - 2(dx - M)dy - dy - B < 0
-		-2dy <= 2dxdy - 2Ndx - 2dxdy + 2Mdy - dy - B < 0
-		-2dy <= 2Mdy - 2Ndx - dy - B < 0
-	2Ndx <= 2Mdy - dy - B + 2dy	2Ndx > 2Mdy - dy - B
-	2Ndx <= 2Mdy + dy - B		N > (2Mdy - dy - B) / 2dx
-	N <= (2Mdy + dy - B) / 2dx
-
-Since we are trying to find the highest number of Y steps that
-satisfies these equations, we need to find the smallest N, so
-we should use the > inequality to find the smallest:
-
-	N = floor((2Mdy - dy - B) / 2dx) + 1
-
-Case 4b: Y major, starting X coordinate moved to M steps from end
-
-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
-
-		-2dx <= 2Mdy - 2Ndx - dx - B < 0
-	2Mdy >= 2Ndx + dx + B - 2dx	2Mdy < 2Ndx + dx + B
-	2Mdy >= 2Ndx - dx + B		M < (2Ndx + dx + B) / 2dy
-	M >= (2Ndx - dx + B) / 2dy
-
-Since we are trying to find the smallest M, we use the >= inequality:
-
-	M = ceiling((2Ndx - dx + B) / 2dy)
-	  = floor((2Ndx - dx + B + 2dy - 1) / 2dy)
-	  = floor((2Ndx - dx + B - 1) / 2dy) + 1
-
-Case 5b: X major, ending Y coordinate moved to N steps
-
-Same derivations as Case 5, but we want the largest M that satisfies
-the equations, so we use the < inequality:
-
-	M = ceiling((2Ndx + dx + B) / 2dy) - 1
-	  = floor((2Ndx + dx + B + 2dy - 1) / 2dy) - 1
-	  = floor((2Ndx + dx + B + 2dy - 1 - 2dy) / 2dy)
-	  = floor((2Ndx + dx + B - 1) / 2dy)
-
-Case 6: X major, ending Y coordinate moved by N steps
-
-		-2dx <= 2(dx - M)dy - 2(dy - N)dx - dx - B < 0
-		-2dx <= 2dxdy - 2Mdy - 2dxdy + 2Ndx - dx - B < 0
-		-2dx <= 2Ndx - 2Mdy - dx - B < 0
-	2Mdy <= 2Ndx - dx - B + 2dx	2Mdy > 2Ndx - dx - B
-	2Mdy <= 2Ndx + dx - B		M > (2Ndx - dx - B) / 2dy
-	M <= (2Ndx + dx - B) / 2dy
-
-Largest # of X steps means smallest M, so use the > inequality:
-
-	M = floor((2Ndx - dx - B) / 2dy) + 1
-
-Case 6b: X major, starting Y coordinate moved to N steps from end
-
-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
-	2Mdy <= 2Ndx - dy - B + 2dy	2Mdy > 2Ndx - dy - B
-	2Mdy <= 2Ndx + dy - B		M > (2Ndx - dy - B) / 2dy
-	M <= (2Ndx + dy - B) / 2dy
-
-To find the smallest M, use the > inequality:
-
-	M = floor((2Ndx - dy - B) / 2dy) + 1
-	  = floor((2Ndx - dy - B + 2dy) / 2dy)
-	  = floor((2Ndx + dy - B) / 2dy)
-
-Case 7b: Y major, ending Y coordinate moved to N steps
-
-Same derivations as Case 7, but we want the largest M that satisfies
-the equations, so use the <= inequality:
-
-	M = floor((2Ndx + dy - B) / 2dy)
-
-Case 8: Y major, ending Y coordinate moved by N steps
-
-		-2dy <= 2(dy - N)dx - 2(dx - M)dy - dy - B < 0
-		-2dy <= 2dxdy - 2Ndx - 2dxdy + 2Mdy - dy - B < 0
-		-2dy <= 2Mdy - 2Ndx - dy - B < 0
-	2Mdy >= 2Ndx + dy + B - 2dy	2Mdy < 2Ndx + dy + B
-	2Mdy >= 2Ndx - dy + B		M < (2Ndx + dy + B) / 2dy
-	M >= (2Ndx - dy + B) / 2dy
-
-To find the highest X steps, find the smallest M, use the >= inequality:
-
-	M = ceiling((2Ndx - dy + B) / 2dy)
-	  = floor((2Ndx - dy + B + 2dy - 1) / 2dy)
-	  = floor((2Ndx + dy + B - 1) / 2dy)
-
-Case 8b: Y major, starting Y coordinate moved to N steps from the end
-
-Same derivations as Case 8, but we want to find the smallest # of X
-steps which means the largest M, so we use the < inequality:
-
-	M = ceiling((2Ndx + dy + B) / 2dy) - 1
-	  = 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)
-	1b: X major move x2 to x1+M	floor((2Mdy + dx - B) / 2dx)
-	2:  X major move x2 to x2-M	floor((2Mdy + dx + B - 1) / 2dx)
-	2b: X major move x1 to x2-M	floor((2Mdy + dx + B - 1) / 2dx)
-
-	3:  Y major move x1 to x1+M	floor((2Mdy - dy + B - 1) / 2dx) + 1
-	3b: Y major move x2 to x1+M	floor((2Mdy + dy + B - 1) / 2dx)
-	4:  Y major move x2 to x2-M	floor((2Mdy - dy - B) / 2dx) + 1
-	4b: Y major move x1 to x2-M	floor((2Mdy + dy - B) / 2dx)
-
-	5:  X major move y1 to y1+N	floor((2Ndx - dx + B - 1) / 2dy) + 1
-	5b: X major move y2 to y1+N	floor((2Ndx + dx + B - 1) / 2dy)
-	6:  X major move y2 to y2-N	floor((2Ndx - dx - B) / 2dy) + 1
-	6b: X major move y1 to y2-N	floor((2Ndx + dx - B) / 2dy)
-
-	7:  Y major move y1 to y1+N	floor((2Ndx + dy - B) / 2dy)
-	7b: Y major move y2 to y1+N	floor((2Ndx + dy - B) / 2dy)
-	8:  Y major move y2 to y2-N	floor((2Ndx + dy + B - 1) / 2dy)
-	8b: Y major move y1 to y2-N	floor((2Ndx + dy + B - 1) / 2dy)
-
-We have the following constraints on all of the above terms:
-
-	0 < M,N <= 2^15		 2^15 can be imposed by miZeroClipLine
-	0 <= dx/dy <= 2^16 - 1
-	0 <= B <= 1
-
-The floor in all of the above equations can be accomplished with a
-simple C divide operation provided that both numerator and denominator
-are positive.
-
-Since dx,dy >= 0 and since moving an X coordinate implies that dx != 0
-and moving a Y coordinate implies dy != 0, we know that the denominators
-are all > 0.
-
-For all lines, (-B) and (B-1) are both either 0 or -1, depending on the
-bias.  Thus, we have to show that the 2MNdxy +/- dxy terms are all >= 1
-or > 0 to prove that the numerators are positive (or zero).
-
-For X Major lines we know that dx > 0 and since 2Mdy is >= 0 due to the
-constraints, the first four equations all have numerators >= 0.
-
-For the second four equations, M > 0, so 2Mdy >= 2dy so (2Mdy - dy) >= dy
-So (2Mdy - dy) > 0, since they are Y major lines.  Also, (2Mdy + dy) >= 3dy
-or (2Mdy + dy) > 0.  So all of their numerators are >= 0.
-
-For the third set of four equations, N > 0, so 2Ndx >= 2dx so (2Ndx - dx)
->= dx > 0.  Similarly (2Ndx + dx) >= 3dx > 0.  So all numerators >= 0.
-
-For the fourth set of equations, dy > 0 and 2Ndx >= 0, so all numerators
-are > 0.
-
-To consider overflow, consider the case of 2 * M,N * dx,dy + dx,dy.  This
-is bounded <= 2 * 2^15 * (2^16 - 1) + (2^16 - 1)
-	   <= 2^16 * (2^16 - 1) + (2^16 - 1)
-	   <= 2^32 - 2^16 + 2^16 - 1
-	   <= 2^32 - 1
-Since the (-B) and (B-1) terms are all 0 or -1, the maximum value of
-the numerator is therefore (2^32 - 1), which does not overflow an unsigned
-32 bit variable.
-
-*/
-
-#define MIOUTCODES(outcode, x, y, xmin, ymin, xmax, ymax) \
-{\
-     if (x < xmin) outcode |= OUT_LEFT;\
-     if (x > xmax) outcode |= OUT_RIGHT;\
-     if (y < ymin) outcode |= OUT_ABOVE;\
-     if (y > ymax) outcode |= OUT_BELOW;\
-}
-
-/* Bit codes for the terms of the 16 clipping equations defined below. */
-
-#define T_2NDX		(1 << 0)
-#define T_2MDY		(0)				/* implicit term */
-#define T_DXNOTY	(1 << 1)
-#define T_DYNOTX	(0)				/* implicit term */
-#define T_SUBDXORY	(1 << 2)
-#define T_ADDDX		(T_DXNOTY)			/* composite term */
-#define T_SUBDX		(T_DXNOTY | T_SUBDXORY)		/* composite term */
-#define T_ADDDY		(T_DYNOTX)			/* composite term */
-#define T_SUBDY		(T_DYNOTX | T_SUBDXORY)		/* composite term */
-#define T_BIASSUBONE	(1 << 3)
-#define T_SUBBIAS	(0)				/* implicit term */
-#define T_DIV2DX	(1 << 4)
-#define T_DIV2DY	(0)				/* implicit term */
-#define T_ADDONE	(1 << 5)
-
-/* Bit masks defining the 16 equations used in miZeroClipLine. */
-
-#define EQN1	(T_2MDY | T_ADDDX | T_SUBBIAS    | T_DIV2DX)
-#define EQN1B	(T_2MDY | T_ADDDX | T_SUBBIAS    | T_DIV2DX)
-#define EQN2	(T_2MDY | T_ADDDX | T_BIASSUBONE | T_DIV2DX)
-#define EQN2B	(T_2MDY | T_ADDDX | T_BIASSUBONE | T_DIV2DX)
-
-#define EQN3	(T_2MDY | T_SUBDY | T_BIASSUBONE | T_DIV2DX | T_ADDONE)
-#define EQN3B	(T_2MDY | T_ADDDY | T_BIASSUBONE | T_DIV2DX)
-#define EQN4	(T_2MDY | T_SUBDY | T_SUBBIAS    | T_DIV2DX | T_ADDONE)
-#define EQN4B	(T_2MDY | T_ADDDY | T_SUBBIAS    | T_DIV2DX)
-
-#define EQN5	(T_2NDX | T_SUBDX | T_BIASSUBONE | T_DIV2DY | T_ADDONE)
-#define EQN5B	(T_2NDX | T_ADDDX | T_BIASSUBONE | T_DIV2DY)
-#define EQN6	(T_2NDX | T_SUBDX | T_SUBBIAS    | T_DIV2DY | T_ADDONE)
-#define EQN6B	(T_2NDX | T_ADDDX | T_SUBBIAS    | T_DIV2DY)
-
-#define EQN7	(T_2NDX | T_ADDDY | T_SUBBIAS    | T_DIV2DY)
-#define EQN7B	(T_2NDX | T_ADDDY | T_SUBBIAS    | T_DIV2DY)
-#define EQN8	(T_2NDX | T_ADDDY | T_BIASSUBONE | T_DIV2DY)
-#define EQN8B	(T_2NDX | T_ADDDY | T_BIASSUBONE | T_DIV2DY)
-
-/* miZeroClipLine
- *
- * returns:  1 for partially clipped line
- *          -1 for completely clipped line
- *
- */
-int
-miZeroClipLine(xmin, ymin, xmax, ymax,
-	       new_x1, new_y1, new_x2, new_y2,
-	       adx, ady,
-	       pt1_clipped, pt2_clipped, octant, bias, oc1, oc2)
-    int xmin, ymin, xmax, ymax;
-    int *new_x1, *new_y1, *new_x2, *new_y2;
-    int *pt1_clipped, *pt2_clipped;
-    unsigned int adx, ady;
-    int octant;
-    unsigned int bias;
-    int oc1, oc2;
-{
-    int swapped = 0;
-    int clipDone = 0;
-    CARD32 utmp;
-    int clip1, clip2;
-    int x1, y1, x2, y2;
-    int x1_orig, y1_orig, x2_orig, y2_orig;
-    int xmajor;
-    int negslope, anchorval;
-    unsigned int eqn;
-
-    x1 = x1_orig = *new_x1;
-    y1 = y1_orig = *new_y1;
-    x2 = x2_orig = *new_x2;
-    y2 = y2_orig = *new_y2;
-
-    clip1 = 0;
-    clip2 = 0;
-
-    xmajor = IsXMajorOctant(octant);
-    bias = ((bias >> octant) & 1);
-
-    while (1)
-    {
-        if ((oc1 & oc2) != 0)			/* trivial reject */
-	{
-	    clipDone = -1;
-	    clip1 = oc1;
-	    clip2 = oc2;
-	    break;
-	}
-        else if ((oc1 | oc2) == 0)		/* trivial accept */
-        {
-	    clipDone = 1;
-	    if (swapped)
-	    {
-	        SWAPINT_PAIR(x1, y1, x2, y2);
-	        SWAPINT(clip1, clip2);
-	    }
-	    break;
-        }
-        else			/* have to clip */
-        {
-	    /* only clip one point at a time */
-	    if (oc1 == 0)
-	    {
-	        SWAPINT_PAIR(x1, y1, x2, y2);
-	        SWAPINT_PAIR(x1_orig, y1_orig, x2_orig, y2_orig);
-	        SWAPINT(oc1, oc2);
-	        SWAPINT(clip1, clip2);
-	        swapped = !swapped;
-	    }
-    
-	    clip1 |= oc1;
-	    if (oc1 & OUT_LEFT)
-	    {
-		negslope = IsYDecreasingOctant(octant);
-		utmp = xmin - x1_orig;
-		if (utmp <= 32767)		/* clip based on near endpt */
-		{
-		    if (xmajor)
-			eqn = (swapped) ? EQN2 : EQN1;
-		    else
-			eqn = (swapped) ? EQN4 : EQN3;
-		    anchorval = y1_orig;
-		}
-		else				/* clip based on far endpt */
-		{
-		    utmp = x2_orig - xmin;
-		    if (xmajor)
-			eqn = (swapped) ? EQN1B : EQN2B;
-		    else
-			eqn = (swapped) ? EQN3B : EQN4B;
-		    anchorval = y2_orig;
-		    negslope = !negslope;
-		}
-		x1 = xmin;
-	    }
-	    else if (oc1 & OUT_ABOVE)
-	    {
-		negslope = IsXDecreasingOctant(octant);
-		utmp = ymin - y1_orig;
-		if (utmp <= 32767)		/* clip based on near endpt */
-		{
-		    if (xmajor)
-			eqn = (swapped) ? EQN6 : EQN5;
-		    else
-			eqn = (swapped) ? EQN8 : EQN7;
-		    anchorval = x1_orig;
-		}
-		else				/* clip based on far endpt */
-		{
-		    utmp = y2_orig - ymin;
-		    if (xmajor)
-			eqn = (swapped) ? EQN5B : EQN6B;
-		    else
-			eqn = (swapped) ? EQN7B : EQN8B;
-		    anchorval = x2_orig;
-		    negslope = !negslope;
-		}
-		y1 = ymin;
-	    }
-	    else if (oc1 & OUT_RIGHT)
-	    {
-		negslope = IsYDecreasingOctant(octant);
-		utmp = x1_orig - xmax;
-		if (utmp <= 32767)		/* clip based on near endpt */
-		{
-		    if (xmajor)
-			eqn = (swapped) ? EQN2 : EQN1;
-		    else
-			eqn = (swapped) ? EQN4 : EQN3;
-		    anchorval = y1_orig;
-		}
-		else				/* clip based on far endpt */
-		{
-		    /*
-		     * Technically since the equations can handle
-		     * utmp == 32768, this overflow code isn't
-		     * needed since X11 protocol can't generate
-		     * a line which goes more than 32768 pixels
-		     * to the right of a clip rectangle.
-		     */
-		    utmp = xmax - x2_orig;
-		    if (xmajor)
-			eqn = (swapped) ? EQN1B : EQN2B;
-		    else
-			eqn = (swapped) ? EQN3B : EQN4B;
-		    anchorval = y2_orig;
-		    negslope = !negslope;
-		}
-		x1 = xmax;
-	    }
-	    else if (oc1 & OUT_BELOW)
-	    {
-		negslope = IsXDecreasingOctant(octant);
-		utmp = y1_orig - ymax;
-		if (utmp <= 32767)		/* clip based on near endpt */
-		{
-		    if (xmajor)
-			eqn = (swapped) ? EQN6 : EQN5;
-		    else
-			eqn = (swapped) ? EQN8 : EQN7;
-		    anchorval = x1_orig;
-		}
-		else				/* clip based on far endpt */
-		{
-		    /*
-		     * Technically since the equations can handle
-		     * utmp == 32768, this overflow code isn't
-		     * needed since X11 protocol can't generate
-		     * a line which goes more than 32768 pixels
-		     * below the bottom of a clip rectangle.
-		     */
-		    utmp = ymax - y2_orig;
-		    if (xmajor)
-			eqn = (swapped) ? EQN5B : EQN6B;
-		    else
-			eqn = (swapped) ? EQN7B : EQN8B;
-		    anchorval = x2_orig;
-		    negslope = !negslope;
-		}
-		y1 = ymax;
-	    }
-
-	    if (swapped)
-		negslope = !negslope;
-
-	    utmp <<= 1;			/* utmp = 2N or 2M */
-	    if (eqn & T_2NDX)
-		utmp = (utmp * adx);
-	    else /* (eqn & T_2MDY) */
-		utmp = (utmp * ady);
-	    if (eqn & T_DXNOTY)
-		if (eqn & T_SUBDXORY)
-		    utmp -= adx;
-		else
-		    utmp += adx;
-	    else /* (eqn & T_DYNOTX) */
-		if (eqn & T_SUBDXORY)
-		    utmp -= ady;
-		else
-		    utmp += ady;
-	    if (eqn & T_BIASSUBONE)
-		utmp += bias - 1;
-	    else /* (eqn & T_SUBBIAS) */
-		utmp -= bias;
-	    if (eqn & T_DIV2DX)
-		utmp /= (adx << 1);
-	    else /* (eqn & T_DIV2DY) */
-		utmp /= (ady << 1);
-	    if (eqn & T_ADDONE)
-		utmp++;
-
-	    if (negslope)
-		utmp = -utmp;
-
-	    if (eqn & T_2NDX)	/* We are calculating X steps */
-		x1 = anchorval + utmp;
-	    else		/* else, Y steps */
-		y1 = anchorval + utmp;
-
-	    oc1 = 0;
-	    MIOUTCODES(oc1, x1, y1, xmin, ymin, xmax, ymax);
-        }
-    }
-
-    *new_x1 = x1;
-    *new_y1 = y1;
-    *new_x2 = x2;
-    *new_y2 = y2;
-    
-    *pt1_clipped = clip1;
-    *pt2_clipped = clip2;
-
-    return clipDone;
-}
-
-
 /* Draw lineSolid, fillStyle-independent zero width lines.
  *
  * Must keep X and Y coordinates in "ints" at least until after they're
@@ -687,13 +103,13 @@ miZeroLine(pDraw, pGC, mode, npt, pptInit)
     int		npt;		/* number of points */
     DDXPointPtr pptInit;
 {
-    int Nspans, current_y;
+    int Nspans, current_y = 0;
     DDXPointPtr ppt; 
     DDXPointPtr pspanInit, spans;
     int *pwidthInit, *widths, list_len;
     int xleft, ytop, xright, ybottom;
     int new_x1, new_y1, new_x2, new_y2;
-    int x, y, x1, y1, x2, y2, xstart, ystart;
+    int x = 0, y = 0, x1, y1, x2, y2, xstart, ystart;
     int oc1, oc2;
     int result;
     int pt1_clipped, pt2_clipped = 0;