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+/* $XFree86: xc/programs/Xserver/mi/mizerclip.c,v 1.2 2001/08/06 20:51:20 dawes Exp $ */
+/***********************************************************
+
+Copyright 1987, 1998 The Open Group
+
+Permission to use, copy, modify, distribute, and sell this software and its
+documentation for any purpose is hereby granted without fee, provided that
+the above copyright notice appear in all copies and that both that
+copyright notice and this permission notice appear in supporting
+documentation.
+
+The above copyright notice and this permission notice shall be included in
+all copies or substantial portions of the Software.
+
+THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
+IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
+FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
+OPEN GROUP BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN
+AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
+CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
+
+Except as contained in this notice, the name of The Open Group shall not be
+used in advertising or otherwise to promote the sale, use or other dealings
+in this Software without prior written authorization from The Open Group.
+
+
+Copyright 1987 by Digital Equipment Corporation, Maynard, Massachusetts.
+
+ All Rights Reserved
+
+Permission to use, copy, modify, and distribute this software and its
+documentation for any purpose and without fee is hereby granted,
+provided that the above copyright notice appear in all copies and that
+both that copyright notice and this permission notice appear in
+supporting documentation, and that the name of Digital not be
+used in advertising or publicity pertaining to distribution of the
+software without specific, written prior permission.
+
+DIGITAL DISCLAIMS ALL WARRANTIES WITH REGARD TO THIS SOFTWARE, INCLUDING
+ALL IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS, IN NO EVENT SHALL
+DIGITAL BE LIABLE FOR ANY SPECIAL, INDIRECT OR CONSEQUENTIAL DAMAGES OR
+ANY DAMAGES WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS,
+WHETHER IN AN ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION,
+ARISING OUT OF OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS
+SOFTWARE.
+
+******************************************************************/
+#ifdef HAVE_DIX_CONFIG_H
+#include <dix-config.h>
+#endif
+
+#include <X11/X.h>
+
+#include "misc.h"
+#include "scrnintstr.h"
+#include "gcstruct.h"
+#include "windowstr.h"
+#include "pixmap.h"
+#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.
+
+*/
+
+/* 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 = 0;
+ int clip1, clip2;
+ int x1, y1, x2, y2;
+ int x1_orig, y1_orig, x2_orig, y2_orig;
+ int xmajor;
+ int negslope = 0, anchorval = 0;
+ unsigned int eqn = 0;
+
+ 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;
+}
generated by cgit v1.2.3 (git 2.25.1) at 2024-11-06 07:39:22 +0000