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-rw-r--r--mi/mizerline.c590
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;
generated by cgit v1.2.3 (git 2.39.1) at 2024-11-24 06:34:23 +0000