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use unsafe flag in macros to make circle draw faster

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hugova 2025-06-30 23:09:14 +02:00
parent bdca0cd82a
commit ba2db967c4
2 changed files with 55 additions and 59 deletions
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24
wip-hugo/routines/circle/circle.s
View file

@ -26,15 +26,16 @@ draw_center_px_in_circle:
STA btp_mem_pos_center
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
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_lower_px_in_circle:
CLC
LDA Y_pos
STA temp
CLC
ADC radius
STA Y_pos
@ -53,7 +54,7 @@ draw_lower_px_in_circle:
LDA temp
STA Y_pos
draw_right_px_in_circle:
;; C = 0, because Mult_16
CLC
LDA X_pos
ADC radius
STA X_pos
@ -70,22 +71,19 @@ draw_right_px_in_circle:
;; fix offset of 8 bytes, idk why this is needed
Sub_16 btp_mem_pos_qda, btp_mem_pos_qda + 1, #$40, #$01 ;-320
SEC ;; circle_help expect this
while_x_bigger_then_y:
;;Draw pixels and does the ypos incrementation logic
;; WARNING expects C=1 before and C =0 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
JSR circle_help
;;t1 += y
CLC
; C==0 from circle_help
LDA t1
ADC Y_rel
STA t1
;; t2 = t1 - x
SEC
LDA t1
SBC X_rel
STA t2
;; if t2 < 0 then skip to endif
@ -100,18 +98,18 @@ move_8px_left:
;; Next chunk is 8 addresses away. Look in pixel_draw for more detail.
;; -8.
;; C = 1 because branching
Sub_16 btp_mem_pos, btp_mem_pos + 1, #$08, #$00, !
Sub_16 btp_mem_pos, btp_mem_pos + 1, #$08, #$00, ! ;+8
;; Y is inverted
Sub_16 btp_mem_pos_qda, btp_mem_pos_qda + 1, #$08, #$00, !
Sub_16 btp_mem_pos_qda, btp_mem_pos_qda + 1, #$08, #$00, ! ;+8
;; X and Y has swoped
Sub_16 btp_mem_pos_qcb, btp_mem_pos_qcb + 1, #$40, #$01 ;-320
Sub_16 btp_mem_pos_qcb, btp_mem_pos_qcb + 1, #$40, #$01, ! ;-320
;; X and Y has swoped and Y has inverted
Sub_16 btp_mem_pos_qdb, btp_mem_pos_qdb +1, #$40, #$01 ;+320
Sub_16 btp_mem_pos_qdb, btp_mem_pos_qdb +1, #$40, #$01, ! ;+320
;; Restores byte to paint
LDA #%00000001
STA byte_to_paint
LDX #%00000001
STX byte_to_paint
decrement_x_pos_end:
LDA t2
STA t1 ; t1 = t2

90
wip-hugo/routines/circle/circle_help.s
View file

@ -1,47 +1,52 @@
.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_inverted_bmp:
;; qab = 2*center - qcb
;; qca = 2*center - qaa
;; qbq = 2*center - qda
;;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
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
;; 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
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_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_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
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.
@ -76,8 +81,6 @@ revert_byte_to_paint:
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)
@ -90,8 +93,7 @@ draw_qdb:; xy swaped and y is inverted.
STA (btp_mem_pos_qdb), Y;
draw_qab:; xy swoped + mirroring
SEC
;SEC
LDA #$07
SBC temp___
TAY
@ -138,9 +140,6 @@ draw_qba: ;;mirror_technique
ORA (btp_mem_pos_qba), Y
STA (btp_mem_pos_qba), Y
increment_y_pos:
INC Y_rel ; y++
DEY
@ -151,13 +150,12 @@ move_8px_down:
; 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, #$40, #$01 ;+320
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, #$08, #$00
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, #$08, #$00
Add_16 btp_mem_pos_qdb, btp_mem_pos_qdb +1, #$07, #$00,! ;+8
increment_y_pos_end:
RTS
.endproc