* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * core_01.s * * Program core for Part 2. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * A version of the Sutherland-Hodgman clipping algorithm. * It goes round the polygon clipping it against one boundary at * a time; it goes round four times in all. * regs: * a0(crds_in),a1(crds_out).a2(no_out),a3((saved) crds_out) * d1(x1),d2(y1),d3(x2),d4(y2),d5(saved x2), d6(saved y2) * d0(current limit) clip: * first clip against xmin bsr clip_ld1 set up pointers tst.w d7 any sides to clip? beq clip_end quit if none * do 1st point as a special case move.w (a0)+,d5 1st x move.w (a0)+,d6 1st y move.w xmin,d0 this limit cmp.w d0,d5 test(x1-xmin) bge xmin_save inside limit bra xmin_update outside limit * do successive vertices in turn xmin_next: move.w (a0)+,d3 x2 move.w (a0)+,d4 y2 move d3,d5 save x2 move d4,d6 save y2 * now test for position sub.w d0,d3 x2-xmin bge xmin_x2in x2 is in * x2 is outside, where is x1? sub.w d0,d1 x1-xmin blt xmin_update both x2 and x1 are out * x2 is out but x1 is in so find intersection, * needs dx1(+ve) in d1, dx2(-ve) in d3, y1 in d2 and y2 in d4 * finds the y-intercept and save it bsr y_intercept * but because its out, don't save x2 bra xmin_update xmin_x2in: * x2 is in but where is x1? sub.w d0,d1 x1-xmin bge xmin_save both x1 and x2 are in * x2 is in but x1 is out so find intercept * but must have the -ve one in d3, so switch exg d1,d3 exg d2,d4 bsr y_intercept xmin_save: move.w d5,(a1)+ save x move.w d6,(a1)+ save y addq.w #1,(a2) inc count xmin_update: move d5,d1 x1:=x2 move d6,d2 y1:=y2 dbf d7,xmin_next * the last point must be the same as the first movea.l a3,a4 pointer to first x subq #4,a1 point to last x cmpm.l (a4)+,(a1)+ check 1st and last x and y beq xmin_dec already the same move.l (a3),(a1) move first to last bra clip_xmax xmin_dec: tst.w (a2) if count beq clip_xmax is not already zero subq.w #1,(a2) reduce it clip_xmax: * Now clip against xmax. Essentially the same as above except that * the order of subtraction is reversed so that the same subroutine * can be used to find the intercept. bsr clip_ld2 set up pointers tst.w d7 any to do? beq clip_ymin no * do 1st point as a special case move.w (a0)+,d5 1st x move.w (a0)+,d6 1st y move.w xmax,d0 cmp.w d5,d0 test(xmax-x1) bge xmax_save inside limit bra xmax_update outside limit * do successive vertices in turn xmax_next: move.w (a0)+,d3 x2 move.w (a0)+,d4 y2 move d3,d5 save x2 move d4,d6 save y2 * now test for position sub.w d0,d3 neg.w d3 xmax-x2 bge xmax_x2in x2 is in * x2 is outside, where is x1? sub.w d0,d1 neg.w d1 xmax-x1 blt xmax_update both x2 and x1 are out * x2 is out but x1 is in so find intersection * needs dx1(+ve) in d1, dx2(-ve) in d3, y1 in d2 and y2 in d4 * find the intercept and save it bsr y_intercept * but because its out, don't save x2 bra xmax_update xmax_x2in: * x2 is in but where is x1? sub.w d0,d1 neg.w d1 xmax-x1 bge xmax_save both x1 and x2 are in * x2 is in but x1 is out so find intercept * but must have the -ve one in d3, so switch exg d1,d3 exg d2,d4 bsr y_intercept xmax_save: move.w d5,(a1)+ save x move.w d6,(a1)+ save y addq.w #1,(a2) inc count xmax_update: move d5,d1 x1:=x2 move d6,d2 y1:=y2 dbf d7,xmax_next * the last point must be the same as the first movea.l a3,a4 pointer to first x subq #4,a1 point to last x cmpm.l (a4)+,(a1)+ check 1st and last x and y beq xmax_dec already the same move.l (a3),(a1) move first to last bra clip_ymin xmax_dec: tst.w (a2) if count beq clip_ymin is not already zero subq.w #1,(a2) reduce it clip_ymin: * clip against ymin bsr clip_ld1 set up pointers tst.w d7 any to do? beq clip_ymax no * do 1st point as a special case move.w (a0)+,d5 1st x move.w (a0)+,d6 1st y move.w ymin,d0 this limit cmp.w d0,d6 test(y1-ymin) bge ymin_save inside limit bra ymin_update outside limit * do successive vertices in turn ymin_next: move.w (a0)+,d3 x2 move.w (a0)+,d4 y2 move d3,d5 save x2 move d4,d6 save x1 * now test for position sub.w d0,d4 y2-xmin bge ymin_y2in y2 is in * y2 is outside, where is y1? sub.w d0,d2 y1-xmin blt ymin_update both y2 and y1 are out * y2 is out but y1 is in so find intersection * needs x1 in d1, x2 in d3, dy1 in d2 and dy2 in d4 * find the intercept and save it bsr x_intercept * but because its out, don't save y2 bra ymin_update ymin_y2in: * y2 is in but where is y1? sub.w d0,d2 y1-ymin bge ymin_save both y1 and y2 are in * y2 is in but y1 is out so find intercept * but must have the -ve one in d4, so switch exg d1,d3 exg d2,d4 bsr x_intercept ymin_save: move.w d5,(a1)+ save x move.w d6,(a1)+ save y addq.w #1,(a2) inc no ymin_update: move d5,d1 x1:=x2 move d6,d2 y1:=y2 dbf d7,ymin_next * the last point must be the same as the first movea.l a3,a4 pointer to first x subq #4,a1 point to last x cmpm.l (a4)+,(a1)+ check 1st and last x and y beq ymin_dec already the same move.l (a3),(a1) move first to last bra clip_ymax ymin_dec: tst.w (a2) if count beq clip_ymax is not already zero subq.w #1,(a2) reduce it clip_ymax: * Now clip against ymax. Essentially the same as above except * the order of subtraction has been reversed so that the * same subroutine can be used. bsr clip_ld2 set up pointers tst.w d7 any to do? beq clip_end no * do 1st point as a special case move.w (a0)+,d5 1st x move.w (a0)+,d6 1st y move.w ymax,d0 cmp.w d6,d0 test(ymax-y1) bge ymax_save inside limit bra ymax_update outside limit * do successive vertices in turn ymax_next: move.w (a0)+,d3 x2 move.w (a0)+,d4 y2 move d3,d5 save x2 move d4,d6 save y2 * now test for position sub.w d0,d4 neg.w d4 ymax-y2 bge ymax_y2in y2 is in * y2 is outside, where is y1? sub.w d0,d2 neg.w d2 ymax-y1 blt ymax_update both x2 and x1 are out * y2 is out but y1 is in so find intersection * needs x1 in d1, x2 in d3, dy1(+ve) in d3 and dy2(-ve) in d4 * find the intercept and save it bsr x_intercept * but because its out, don't save y2 bra ymax_update ymax_y2in: * y2 is in but where is y1 sub.w d0,d2 neg.w d2 ymax-y1 bge ymax_save both y1 and y2 are in * y2 is in but y1 is out so find intercept * but must have the -ve one in d4, so switch exg d1,d3 exg d2,d4 bsr x_intercept ymax_save: move.w d5,(a1)+ save x move.w d6,(a1)+ save y addq.w #1,(a2) inc no ymax_update: move d5,d1 x1:=x2 move d6,d2 y1:=y2 dbf d7,ymax_next * the last point must be the same as the first movea.l a3,a4 pointer to first x subq #4,a1 point to last x cmpm.l (a4)+,(a1)+ check 1st and last x and y beq ymax_dec already the same move.l (a3),(a1) move first to last bra clip_end ymax_dec: tst.w (a2) if count beq clip_end is not alredy zero subq.w #1,(a2) reduce it clip_end: rts clip_ld1: * first set up the pointers for the first and third passes lea crds_in,a0 pointer to vertex coords. before clip lea crds_out,a1 and after the this clip move.l a1,a3 saved move.w no_in,d7 this many sides before lea no_out,a2 where the number after is stored clr.w no_out rts clip_ld2: * set up the pointers for the second and fourth passes * ensures the final output is at the same place as initial input lea crds_out,a0 pointer to vertex coords before clip lea crds_in,a1 and after this clip move.l a1,a3 saved move.w no_out,d7 this many sides before lea no_in,a2 where the number after is stored clr.w no_in rts y_intercept: * Find the y-intercept on the clipping boundary x = k of the * line joining p1(x1,y1) to p2(x2,y2). * entry: * d1: (x1-k) - a positive number * d3: (x2-k) - a negative number * d2: y1, d4: y2 tst.w d1 point on boundary beq yint_out already saved tst.w d3 point on boundary beq yint_out will be saved movem d5/d6,-(sp) save x2, y2 yint_in move.w d2,d6 y1 add.w d4,d6 y1+y2 asr.w #1,d6 (y1+y2)/2 = , a possible intercept move d1,d5 dx1 add.w d3,d5 dx1+dx2 asr.w #1,d5 ()/2 = beq yint_end if /2=0, boundary reached bgt yint_loop if not loop again move d5,d3 unless is -ve, and becomes new dx2 move d6,d4 and is new y2 bra yint_in and try again yint_loop: move d5,d1 is new dx1 move d6,d2 is new y1 bra yint_in yint_end: move.w d0,(a1)+ store x boundary move.w d6,(a1)+ and as the coords of a new vertex addq.w #1,(a2) and increment the vertex count movem (sp)+,d5/d6 restore regs yint_out: rts x_intercept: * Finds the x-intercept on the clipping boundary y = k of the * line joining p1(x1,y1) to p2(x2,y2) * entry: * d1: x1, d3: x2 * d2: (y1-k) - a positive number * d4: (y2-k) - a negative number * tst.w d2 point on boundary beq xint_out already saved tst.w d4 point on boundary beq xint_out wil be saved movem d5/d6,-(sp) save x2, y2 xint_in move d1,d5 x1 add.w d3,d5 x1+x2 asr.w #1,d5 ()/2 = a possible intercept move d2,d6 dy1 add.w d4,d6 dy1+dy2 asr.w #1,d6 (dy1+dy2)/2 = beq xint_end if = 0, boundary reached bgt xint_loop if not loop again move d6,d4 unless is -ve and becomes dy2 move d5,d3 and becomes x2 bra xint_in and try again xint_loop: move d5,d1 is new dx1 move d6,d2 and is new dy1 bra xint_in xint_end: move.w d5,(a1)+ store intercept move.w d0,(a1)+ and the boundary y as new vertex coords addq.w #1,(a2) and increment the vertex count movem (sp)+,d5/d6 restore regs xint_out rts next vertex * leaves with: * a list of vertex coordinates at coords_in * the number of polygon sides at no_in. include core_00.s add on the previous core