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draw 1/2 of circle by circle-help-subroutine

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hugova 2025-06-29 13:12:01 +02:00
parent 219f8c155d
commit c4add9a62c
3 changed files with 111 additions and 41 deletions
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8
wip-hugo/routines/circle/circle.inc
View file

@ -9,5 +9,9 @@
Y_rel = $E3
temp = $E4
temp_ = $E7
btp_mem_pos_center = $E5 ;; 16bit <fix>
btp_mem_pos_inv = $EE ;; 16bit <fix>
; E8 - EA is used by pixel.inc
btp_mem_pos_center = $E5 ; 16bit value (uses E6)
btp_mem_pos_center_two = btp_mem_pos_center
btp_mem_pos_inv_y =$EB ; 16bit value (uses EC)
Y_inv_x = $EC
btp_mem_pos_inv = $EE ; 16bit value (uses EF)

57
wip-hugo/routines/circle/circle.s
View file

@ -3,7 +3,7 @@
.proc circle
.include "circle.inc"
;; We use the algorithm jerkos method
;; https://schwarzers.com/algorithms/
;; git
;; X_rel = radius (share the same address)
@ -27,8 +27,9 @@ draw_center_px_in_circle:
LDA btp_mem_pos + 1
STA btp_mem_pos_center + 1
;; btp_mem_pos_center_two = 2*btp_mem_pos_center | used later for calculating btp_mem_pos_inv
btp_mem_pos_center_two = btp_mem_pos_center
Mult_16 btp_mem_pos_center, btp_mem_pos_center + 1
;; fix offset of 8 bytes, idk why this is needed
Sub_16 btp_mem_pos_center, btp_mem_pos_center + 1, #$40, #$01 ;-320
draw_right_px_in_circle:
@ -41,44 +42,19 @@ draw_right_px_in_circle:
JSR pixel_draw
;; This sets byte_to_paint, btp_mem_pos and Y
LDA btp_mem_pos
STA btp_mem_pos_inv_y
LDA btp_mem_pos + 1
STA btp_mem_pos_inv_y + 1
;; fix offset of 8 bytes, idk why this is needed
Sub_16 btp_mem_pos_inv_y, btp_mem_pos_inv_y + 1, #$40, #$01 ;-320
SEC ;; See draw_pixel_inv
while_x_bigger_then_y:
draw_pixel:
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_pixel_inv:
;;C = 1 beacause the branching to while_x_bigger_then_y and SEC on first ittteration.
LDA btp_mem_pos_center_two
SBC btp_mem_pos
STA btp_mem_pos_inv
LDA btp_mem_pos_center_two + 1
SBC btp_mem_pos + 1
STA btp_mem_pos_inv + 1
;; calculate the inverted y value Y = 7-Y
STY temp
LDA #$07
;; C = 1 because arithmatic above
SBC temp
TAY
;; 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)
LDX temp_
; A = byte_to_paint_inv
ORA (btp_mem_pos_inv), Y
STA (btp_mem_pos_inv), Y
;;Recover the Y value (we changed it because evrything is inverted)
LDY temp
;;Draw pixels
JSR circle_help
increment_y_pos:
INC Y_rel ; y++
@ -87,9 +63,10 @@ increment_y_pos:
move_8px_down:
LDY #$07
;; Switch to chunk bellow
; So we subtract #$4001
; So we subtract #$0140
; C = 1 because branching!
Sub_16 btp_mem_pos, btp_mem_pos + 1, #$40, #$01, ! ;-320
Add_16 btp_mem_pos_inv_y, btp_mem_pos_inv_y + 1, #$40, #$01 ;+320
increment_y_pos_end:
;;t1 += y
@ -115,6 +92,7 @@ move_8px_left:
;; -8.
;; C = 1 because branching
Sub_16 btp_mem_pos, btp_mem_pos + 1, #$08, #$00, !
Sub_16 btp_mem_pos_inv_y, btp_mem_pos_inv_y + 1, #$08, #$00, !
;; Restores byte to paint
LDA #%00000001
STA byte_to_paint
@ -128,4 +106,5 @@ endif:
BCS while_x_bigger_then_y
RTS
.include "routines/circle/circle_help.s"
.endproc

87
wip-hugo/routines/circle/circle_help.s Normal file
View file

@ -0,0 +1,87 @@
.proc circle_help ;; This is because jmp cant jump that long!
.include "circle.inc"
;;We have named the parts of the circle as such.
;; |
;; q22 | q12
;; |
;; q21 | q11
;;---------------X-----------------> X
;; q31 | q41
;; |
;; q32 | q42
;; |
;; v Y
;; The q stands for quarter, whe have 4 quarter, and each quarter is split into 2
draw_pixel:;;q11
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_pixel_inv_y:;;q41
STY temp
LDA #$07
SEC
SBC temp
TAY
LDA byte_to_paint
ORA (btp_mem_pos_inv_y), Y
STA (btp_mem_pos_inv_y), Y
;hihi:
;jmp hihi
LDY temp
draw_pixel_inv_y_mirror:
SEC
LDA btp_mem_pos_center_two
SBC btp_mem_pos_inv_y
STA btp_mem_pos_inv
LDA btp_mem_pos_center_two + 1
SBC btp_mem_pos_inv_y + 1
STA btp_mem_pos_inv + 1
STX temp_
LDX byte_to_paint
LDA inverse_factor_value, X;; (see END.s)
LDX temp_
; A = byte_to_paint_inv
ORA (btp_mem_pos_inv), Y
STA (btp_mem_pos_inv), Y
draw_pixel_mirror:;;q31
SEC
;;C = 1 beacause the branching to while_x_bigger_then_y and SEC on first ittteration.
LDA btp_mem_pos_center_two
SBC btp_mem_pos
STA btp_mem_pos_inv
LDA btp_mem_pos_center_two + 1
SBC btp_mem_pos + 1
STA btp_mem_pos_inv + 1
;; calculate the inverted y value Y = 7-Y
STY temp
LDA #$07
;; C = 1 because arithmatic above
SBC temp
TAY
;; 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)
LDX temp_
; A = byte_to_paint_inv
ORA (btp_mem_pos_inv), Y
STA (btp_mem_pos_inv), Y
;;Recover the Y value (we changed it because evrything is inverted)
LDY temp
RTS
.endproc