Remove bodge macro in circledraw and use jmp in other addressing mode
This commit is contained in:
hugova
parent
f54ecb75b9
commit
590d7533d9
3 changed files with 162 additions and 161 deletions
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@ -9,7 +9,9 @@
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Y_rel = $E2
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Y_copy = $E3
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temp_ = $E4
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temo__ = $E5
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temp__ = $E5
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temp___ = $ED
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jmp_location_pointer = $EE ;16 bit value (uses EF)
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byte_to_paint_qaa = byte_to_paint
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byte_to_paint_qcb = $EB ;16bit value (uses EC)
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@ -17,15 +19,13 @@
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btp_mem_pos_center = $E6 ; 16bit value (uses E7)
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btp_mem_pos_center_two = btp_mem_pos_center
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temp__ = $D0
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temp___ = $D1
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btp_mem_pos_qaa = btp_mem_pos
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btp_mem_pos_qcb = $D2 ; 16bit value (uses D3)
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btp_mem_pos_qdb = $D4 ;16bit value (uses D5)
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btp_mem_pos_qda =$D6 ; 16bit value (uses D7)
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btp_mem_pos_qcb = $D0 ; 16bit value (uses D1)
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btp_mem_pos_qdb = $D2 ;16bit value (uses D3)
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btp_mem_pos_qda =$D4 ; 16bit value (uses D5)
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;;mirrord
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btp_mem_pos_qab = $D8
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btp_mem_pos_qca = $DA
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btp_mem_pos_qba = $DC
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btp_mem_pos_qbb = $DE
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btp_mem_pos_qab = $D6
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btp_mem_pos_qca = $D8
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btp_mem_pos_qba = $DA
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btp_mem_pos_qbb = $DC
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@ -3,8 +3,13 @@
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.proc circle
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.include "circle.inc"
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;; X_rel = radius (share the same address)
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;; Because loop is so big, We need to save position in pointer
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LDA #<while_x_bigger_then_y
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STA jmp_location_pointer
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LDA #>while_x_bigger_then_y
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STA jmp_location_pointer + 1
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;; X_rel = radius (share the same address)
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;;Y_rel =0
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LDA #$00
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STA Y_rel
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@ -70,7 +75,149 @@ while_x_bigger_then_y:
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SEC
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;;Draw pixels and does the ypos incrementation logic
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;; WARNING expects C=1 before and C =0 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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JSR circle_help
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;; WARNING expects C=1 before and C =0 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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;;We have named the parts of the circle as such.
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;; |
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;; qbb (7) | qab (8)
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;; |
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;; qba (2) | qaa (1)
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;;---------------X-----------------> X
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;; qca (4) | qda (3)
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;; |
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;; qcb (5) | qdb (6)
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;; |
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;; v Y
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;; The q stands for quarter, whe have 4 quarter, and each quarter is split into 2
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;; We first calculate all btp_mem_pos for the inverted half quarters!
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calculate:
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;; qab = 2*center - qcb
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;; qca = 2*center - qaa
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;; qba = 2*center - qda
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;; qbb = 2*center - qdb
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;; a = 2*center - b comes from that a = center -(b-center)
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LDA btp_mem_pos_center_two
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SBC btp_mem_pos_qcb
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STA btp_mem_pos_qab
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LDA btp_mem_pos_center_two + 1
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SBC btp_mem_pos_qcb + 1
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STA btp_mem_pos_qab + 1
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LDA btp_mem_pos_center_two
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SBC btp_mem_pos
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STA btp_mem_pos_qca
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LDA btp_mem_pos_center_two + 1
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SBC btp_mem_pos + 1
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STA btp_mem_pos_qca + 1
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LDA btp_mem_pos_center_two
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SBC btp_mem_pos_qda
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STA btp_mem_pos_qba
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LDA btp_mem_pos_center_two + 1
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SBC btp_mem_pos_qda + 1
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STA btp_mem_pos_qba + 1
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LDA btp_mem_pos_center_two
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SBC btp_mem_pos_qdb
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STA btp_mem_pos_qbb
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LDA btp_mem_pos_center_two + 1
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SBC btp_mem_pos_qdb + 1
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STA btp_mem_pos_qbb + 1
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end_calculation:
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;; Lets draw all half-quatrons of the circle. This draws only 8 pixels per iteration.
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;; Note that I have the draw_qxx in listed pairs. Each pair chair the same Y-register :)
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STY Y_copy
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draw_qaa:
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LDA byte_to_paint ;A byte containing a single 1. Coresponds to X position in the chunk.
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ORA (btp_mem_pos), Y
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STA (btp_mem_pos), Y
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draw_qba: ;;mirror_technique
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LDX byte_to_paint
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LDA inverse_factor_value, X;; (see END.s)
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TAX
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ORA (btp_mem_pos_qba), Y
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STA (btp_mem_pos_qba), Y
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draw_qda:; y is inverted
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;; invert Y, this is shared with qca
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LDA #$07
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SBC Y_copy
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TAY
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LDA byte_to_paint
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ORA (btp_mem_pos_qda), Y
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STA (btp_mem_pos_qda), Y
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draw_qca: ;;mirror technique
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TXA
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ORA (btp_mem_pos_qca), Y
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STA (btp_mem_pos_qca), Y
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draw_qcb:; xy swoped
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LDA log, X
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TAY
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;;modify X_pos
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LDX Y_copy
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LDA binary_factor, X; (see END.s)
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TAX
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ORA (btp_mem_pos_qcb), Y
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STA (btp_mem_pos_qcb), Y
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draw_qdb:; xy swaped and y is inverted.
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LDA inverse_factor_value, X
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STA temp__
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;;Uses modifyed Y from above
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ORA (btp_mem_pos_qdb), Y
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STA (btp_mem_pos_qdb), Y
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draw_qbb:
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STY temp_
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LDA #$07
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SBC temp_
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TAY
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TXA
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ORA (btp_mem_pos_qbb), Y
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STA (btp_mem_pos_qbb), Y
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draw_qab:; xy swoped + mirroring
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LDA temp__
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ORA (btp_mem_pos_qab), Y
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STA (btp_mem_pos_qab), Y
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;;Recover the Y value (we changed it because evrything is inverted)
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LDY Y_copy
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increment_y_pos:
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INC Y_rel ; y++
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DEY
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BPL increment_y_pos_end
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move_8px_down:
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LDY #$07
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;; Switch to chunk bellow
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; So we subtract #$0140
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; C = 1 because branching!
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Sub_16 btp_mem_pos, btp_mem_pos + 1, #$40, #$01, ! ;-320
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;; Y is inverted
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Add_16 btp_mem_pos_qda, btp_mem_pos_qda + 1, #$3f, #$01,! ;+320
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;; X and Y has swopped
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Sub_16 btp_mem_pos_qcb, btp_mem_pos_qcb + 1, #$07, #$00,! ;-8
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;; X and Y has swoped and Y has inverted
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Add_16 btp_mem_pos_qdb, btp_mem_pos_qdb +1, #$07, #$00,! ;+8
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increment_y_pos_end:
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;;t1 += y
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@ -114,7 +261,8 @@ endif:
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LDA X_rel
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CMP Y_rel
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BCS while_x_bigger_then_y
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BCC end
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JMP (jmp_location_pointer)
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end:
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RTS
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.include "routines/circle/circle_help.s"
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.endproc
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@ -1,147 +0,0 @@
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.proc circle_help ; This is because jmp cant jump that long!
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.include "circle.inc"
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;; WARNING expects C=1 before and C =0 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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;;We have named the parts of the circle as such.
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;; |
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;; qbb (7) | qab (8)
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;; |
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;; qba (2) | qaa (1)
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;;---------------X-----------------> X
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;; qca (4) | qda (3)
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;; |
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;; qcb (5) | qdb (6)
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;; |
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;; v Y
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;; The q stands for quarter, whe have 4 quarter, and each quarter is split into 2
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;; We first calculate all btp_mem_pos for the inverted half quarters!
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calculate:
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;; qab = 2*center - qcb
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;; qca = 2*center - qaa
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;; qba = 2*center - qda
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;; qbb = 2*center - qdb
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;; a = 2*center - b comes from that a = center -(b-center)
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LDA btp_mem_pos_center_two
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SBC btp_mem_pos_qcb
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STA btp_mem_pos_qab
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LDA btp_mem_pos_center_two + 1
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SBC btp_mem_pos_qcb + 1
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STA btp_mem_pos_qab + 1
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LDA btp_mem_pos_center_two
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SBC btp_mem_pos
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STA btp_mem_pos_qca
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LDA btp_mem_pos_center_two + 1
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SBC btp_mem_pos + 1
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STA btp_mem_pos_qca + 1
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LDA btp_mem_pos_center_two
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SBC btp_mem_pos_qda
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STA btp_mem_pos_qba
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LDA btp_mem_pos_center_two + 1
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SBC btp_mem_pos_qda + 1
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STA btp_mem_pos_qba + 1
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LDA btp_mem_pos_center_two
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SBC btp_mem_pos_qdb
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STA btp_mem_pos_qbb
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LDA btp_mem_pos_center_two + 1
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SBC btp_mem_pos_qdb + 1
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STA btp_mem_pos_qbb + 1
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end_calculation:
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;; Lets draw all half-quatrons of the circle. This draws only 8 pixels per iteration.
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;; Note that I have the draw_qxx in listed pairs. Each pair chair the same Y-register :)
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STY Y_copy
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draw_qaa:
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LDA byte_to_paint ;A byte containing a single 1. Coresponds to X position in the chunk.
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ORA (btp_mem_pos), Y
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STA (btp_mem_pos), Y
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draw_qba: ;;mirror_technique
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LDX byte_to_paint
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LDA inverse_factor_value, X;; (see END.s)
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TAX
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ORA (btp_mem_pos_qba), Y
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STA (btp_mem_pos_qba), Y
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draw_qda:; y is inverted
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;; invert Y, this is shared with qca
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LDA #$07
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SBC Y_copy
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TAY
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LDA byte_to_paint
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ORA (btp_mem_pos_qda), Y
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STA (btp_mem_pos_qda), Y
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draw_qca: ;;mirror technique
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TXA
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ORA (btp_mem_pos_qca), Y
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STA (btp_mem_pos_qca), Y
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draw_qcb:; xy swoped
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LDA log, X
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TAY
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;;modify X_pos
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LDX Y_copy
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LDA binary_factor, X; (see END.s)
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TAX
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ORA (btp_mem_pos_qcb), Y
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STA (btp_mem_pos_qcb), Y
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draw_qdb:; xy swaped and y is inverted.
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LDA inverse_factor_value, X
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STA temp__
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;;Uses modifyed Y from above
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ORA (btp_mem_pos_qdb), Y
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STA (btp_mem_pos_qdb), Y
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draw_qbb:
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STY temp_
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LDA #$07
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SBC temp_
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TAY
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TXA
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ORA (btp_mem_pos_qbb), Y
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STA (btp_mem_pos_qbb), Y
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draw_qab:; xy swoped + mirroring
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LDA temp__
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ORA (btp_mem_pos_qab), Y
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STA (btp_mem_pos_qab), Y
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;;Recover the Y value (we changed it because evrything is inverted)
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LDY Y_copy
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increment_y_pos:
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INC Y_rel ; y++
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DEY
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BPL increment_y_pos_end
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move_8px_down:
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LDY #$07
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;; Switch to chunk bellow
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; So we subtract #$0140
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; C = 1 because branching!
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Sub_16 btp_mem_pos, btp_mem_pos + 1, #$40, #$01, ! ;-320
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;; Y is inverted
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Add_16 btp_mem_pos_qda, btp_mem_pos_qda + 1, #$3f, #$01,! ;+320
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;; X and Y has swopped
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Sub_16 btp_mem_pos_qcb, btp_mem_pos_qcb + 1, #$07, #$00,! ;-8
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;; X and Y has swoped and Y has inverted
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Add_16 btp_mem_pos_qdb, btp_mem_pos_qdb +1, #$07, #$00,! ;+8
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increment_y_pos_end:
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RTS
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.endproc
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