.proc circle_help ; This is because jmp cant jump that long! .include "circle.inc" ;; WARNING expects C=1 before and C =0 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ;;We have named the parts of the circle as such. ;; * | ;; qbb | qab ;; | ;; qba | qaa ;;---------------X-----------------> X ;; qca | qda ;; | ;; qcb | qdb ;; | * ;; v Y ;; The q stands for quarter, whe have 4 quarter, and each quarter is split into 2 ;; We first calculate all btp_mem_pos for the inverted half quarters! calculate: ;; qab = 2*center - qcb ;; qca = 2*center - qaa ;; qbq = 2*center - qda ;; a = 2*center - b comes from that a = center -(b-center) LDA btp_mem_pos_center_two SBC btp_mem_pos_qcb STA btp_mem_pos_qab LDA btp_mem_pos_center_two + 1 SBC btp_mem_pos_qcb + 1 STA btp_mem_pos_qab + 1 LDA btp_mem_pos_center_two SBC btp_mem_pos STA btp_mem_pos_qca LDA btp_mem_pos_center_two + 1 SBC btp_mem_pos + 1 STA btp_mem_pos_qca + 1 LDA btp_mem_pos_center_two SBC btp_mem_pos_qda STA btp_mem_pos_qba LDA btp_mem_pos_center_two + 1 SBC btp_mem_pos_qda + 1 STA btp_mem_pos_qba + 1 end_calculation: ;; Lets draw all half-quatrons of the circle. This draws only 8 pixels per iteration. draw_qaa: LDA byte_to_paint ;A byte containing a single 1. Coresponds to X position in the chunk. ORA (btp_mem_pos), Y STA (btp_mem_pos), Y draw_qcb:; xy swaped STY temp LDA byte_to_paint STA temp__ ;;modify Y_pos LDY #$00 jmp goto_loop loop: INY goto_loop: ASL byte_to_paint; lsr asl BCC loop STY temp___ revert_byte_to_paint: LDA temp__ STA byte_to_paint ;;modify X_pos LDX temp LDA binary_factor, X; (see END.s) LDY temp___ ORA (btp_mem_pos_qcb), Y STA (btp_mem_pos_qcb), Y; draw_qdb:; xy swaped and y is inverted. ;;modify X LDX temp LDA binary_factor, X; (see END.s) TAX LDA inverse_factor_value, X;; (see END.s) ;;Uses modifyed Y from above ORA (btp_mem_pos_qdb), Y STA (btp_mem_pos_qdb), Y; draw_qab:; xy swoped + mirroring ;SEC LDA #$07 SBC temp___ TAY LDA #$07 SBC temp TAX LDA binary_factor, X; (see END.s) ORA (btp_mem_pos_qab), Y STA (btp_mem_pos_qab), Y LDY temp draw_qda:; y is inverted ;; invert Y, this is shared with qca LDA #$07 SBC temp TAY LDA byte_to_paint ORA (btp_mem_pos_qda), Y STA (btp_mem_pos_qda), Y draw_qca: ;;mirror technique ;; calculate byte_to_paint_inv 00000001 --> 10000000, 00000010 --> 01000000 ... etc ;; uses a table! STX temp_ LDX byte_to_paint LDA inverse_factor_value, X;; (see END.s) TAX ;; A is saved to x because qba use this as well ; A = byte_to_paint_inv ORA (btp_mem_pos_qca), Y STA (btp_mem_pos_qca), Y ;;Recover the Y value (we changed it because evrything is inverted) LDY temp draw_qba: ;;mirror_technique TXA ORA (btp_mem_pos_qba), Y STA (btp_mem_pos_qba), Y increment_y_pos: INC Y_rel ; y++ DEY BPL increment_y_pos_end move_8px_down: LDY #$07 ;; Switch to chunk bellow ; So we subtract #$0140 ; C = 1 because branching! Sub_16 btp_mem_pos, btp_mem_pos + 1, #$40, #$01, ! ;-320 ;; Y is inverted Add_16 btp_mem_pos_qda, btp_mem_pos_qda + 1, #$3f, #$01,! ;+320 ;; X and Y has swopped Sub_16 btp_mem_pos_qcb, btp_mem_pos_qcb + 1, #$07, #$00,! ;-8 ;; X and Y has swoped and Y has inverted Add_16 btp_mem_pos_qdb, btp_mem_pos_qdb +1, #$07, #$00,! ;+8 increment_y_pos_end: RTS .endproc