* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * core_00.s * * All the subroutines * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * poly_fil: * This fills a polygon. * It consists of 2 parts: * part 1 - the the x-coordinates of all boundary points are entered in xbuf * part 2 - the holine routine fills the polygon from the values in xbuf * PART 1. Fill the buffer. * Regs: * a3: pointer to crds_in - coords. list (x1,y1,x2,y2,....x1,y1) * a2: pointer to xbuf * d0(x1),d1(y1),d2(x2),d3(y2),d4(vertex no)/(decision ver., * d5(lowest y),d6(highest y)/(the increment),d7(edge counter) * polygon vertices are ordered anticlockwise * Initialise all variables filxbuf: move.w no_in,d7 no. edges in polygon beq fil_end quit if none to do lea crds_in,a3 pointer to the coords. of vertices subq.w #1,d7 the counter move.w #199,d5 initial minimum y clr.w d6 initial maximum y filbuf1 lea xbuf,a2 init. buffer pointer addq.w #2,a2 point to ascending side (low word) move.w (a3)+,d0 next x1 move.w (a3)+,d1 next y1 move.w (a3)+,d2 next x2 move.w (a3)+,d3 next y2 subq.w #4,a3 point back to x2 * Find the lowest and highest y values: the filled range of xbuf cmp.w d5,d1 test(y1-miny) bge filbuf3 minimum y unchanged move.w d1,d5 minimum y is y1 filbuf3 cmp.w d1,d6 test(maxy-y1) bge filbuf5 unchanged move.w d1,d6 maximum y is y1 filbuf5 exg d5,a5 save minimum y exg d6,a6 save maximum y clr.w d4 init. decision var moveq #1,d6 init. increment * All lines fall into two catagories: [slope]<1, [slope]>1. * The only difference is whether x and y are increasing or decreasing. * See if line is ascending (slope > 0) or descending (slope < 0). cmp.w d1,d3 (y2-y1)=dy beq y_limits ignore horizontals altogether bgt ascend slope > 0 * It must be decending. Direct output to LHS of buffer. a2 must * be reduced and we have to reverse the order of the vertices. exg d0,d2 exchange x1 and x2 exg d1,d3 exchange y1 and y2 subq.w #2,a2 point to left hand buffer ascend sub.w d1,d3 now dy is +ve * Set up y1 as index to buffer lsl.w #2,d1 add.w d1,a2 * Check the sign of the slope sub.w d0,d2 (x2-x1)=dx beq vertical if it's vertical its a special case bgt pos_slope the slope is positive * It must have a negative slope but we deal with this by making the * increment negative neg.w d6 increment is decrement neg.w d2 and dx is positive * now decide whether the slope is high (>1) or low (<1) pos_slope: cmp.w d2,d3 test(dy-dx) bgt hislope slope is >1 * The slope is less than 1 so we want to increment x every time and then * check whether to also increment y. If so this value of x must be saved. * dx is the counter. Initial error D1=2dy-dx * If last D -ve, then x=x=inc, don't record x, D=D+err1 * If last D +ve, then x=x+inc,y=y+inc, record this x, D=D+err2 * err1=2dy; err2=2dy-2dx * dx in d2, dy in d3, incx in d6, x in d0 move.w d2,d5 subq.w #1,d5 dx-1 is the counter add.w d3,d3 2dy=err1 move.w d3,d4 2dy neg.w d2 -dx add.w d2,d4 2dy-dx = D1 add.w d4,d2 2dy-2dx=err2 move.w d0,(a2) save first x inc_x add.w d6,d0 x=x+incx tst.w d4 what is the decision? bmi no_stk don't inc y, don't record x add.w #4,a2 inc y so record x; find next buffer place move.w d0,(a2) save this x add.w d2,d4 update decision D=D+err2 bra.s next_x next one no_stk add.w d3,d4 D=D+err1 next_x dbra d5,inc_x increment x again bra y_limits * The slope is >1 so change the roles of dx and dy * This time we must increment y each time and record the value of x after * having done so. * Init error D1 = 2dx-dy * If last D -ve, then y=y+inc, D=D+err1, record x * If last D +ve, then x=x+inc, y=y+inc, D=D+err2, record x * err1=2dx, err2=2(dx-dy) * dx in d2, dy in d3, inc in d6, x in d0 hislope move.w d3,d5 subq.w #1,d5 dy-1 is counter add.w d2,d2 2dx=err1 move.w d2,d4 2dx neg.w d3 -dy add.w d3,d4 2dx-dy=D1 add.w d4,d3 2dx-2dy=err2 move.w d0,(a2) save 1st x inc_y addq.w #4,a2 next place in buffer (equivalent to incrementing y) tst.w d4 what is the decision? bmi same_x don't inc x add.w d6,d0 inc x add.w d3,d4 D=D+err2 bra.s next_y same_x add.w d2,d4 D=D+err1 next_y move.w d0,(a2) save the x value dbra d5,inc_y bra y_limits * the special case of a vertical line. x is constant. dy is the counter vertical: move.w d0,(a2) save next x addq.w #4,a2 next place in buffer dbra d3,vertical for all y * Restore the y limits y_limits: exg d5,a5 exg d6,a6 next_line: dbra d7,filbuf1 do all lines in this polygon next_poly: * This part ends with minimum y in d5 and maximum y in d6 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * PART 2 * set up the pointer lea xbuf,a1 base address sub.w d5,d6 no. lines to do-1 move.w d6,d7 is the counter beq fil_end quit if all sides are horizontal move.w d5,d3 minimum y is the start lsl.w #2,d5 4*minimum y = offset into xbuf add.w d5,a1 for the address to start move.w colour,d4 the colour move.l LineA_base,a0 Line A vars. structure clr.w fg_bp1(a0) reset clr.w fg_bp2(a0) colours clr.w fg_bp3(a0) clr.w fg_bp4(a0) lsr #1,d4 colour plane 0? bcc plne_1 move.w #1,fg_bp1(a0) plne_1 lsr #1,d4 colour plane 1? bcc plne_2 move.w #1,fg_bp2(a0) plne_2 lsr #1,d4 colour plane 2? bcc plne_3 move.w #1,fg_bp3(a0) plne_3 lsr #1,d4 colour plane 3? bcc pln_out move.w #1,fg_bp4(a0) pln_out clr.w wrt_mod(a0) overwrite lea fill,a2 move.l a2,patptr(a0) Here is the fill pattern move.w #0,patmsk(a0) consisting of 1 line move.w #0,multifil(a0) on all colour planes subq #1,d3 reduce initial y move.w d3,y1(a0) initial y-1 poly2 addq #1,y1(a0) next y move.w (a1)+,d2 next x1 move.w (a1)+,d1 next x2 cmp.w d2,d1 x2-x1 ble poly4 x1 must be =< x2 move.w d2,x1(a0) x1 move.w d1,x2(a0) x2 movem.l d0-d7/a0-a6,-(sp) save the registers dc.w hline Line A $a004 movem.l (sp)+,d0-d7/a0-a6 restore the registers poly4 dbra d7,poly2 repeat for all y values fil_end rts * Get the address of the Line A variables structure init_LineA: dc.w init move.l a0,LineA_base rts