636 lines
20 KiB
C
636 lines
20 KiB
C

/***********************************************************






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 <dixconfig.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 x2M floor((2Mdy + dx + B  1) / 2dx)



2b: X major move x1 to x2M 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 x2M floor((2Mdy  dy  B) / 2dx) + 1



4b: Y major move x1 to x2M 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 y2N floor((2Ndx  dx  B) / 2dy) + 1



6b: X major move y1 to y2N 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 y2N floor((2Ndx + dy + B  1) / 2dy)



8b: Y major move y1 to y2N 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 (B1) 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 (B1) 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



*



*/



_X_EXPORT 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;



}
