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https://github.com/cc65/cc65.git
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c7969a78b0
Fixed typo errors. Made the comments consistent across all those files.
490 lines
12 KiB
ArmAsm
490 lines
12 KiB
ArmAsm
;
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; Startup code for cc65 (CBM 600/700 version)
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;
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.export _exit, BRKVec
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.export __STARTUP__ : absolute = 1 ; Mark as startup
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.import callirq_y, initlib, donelib
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.import push0, callmain
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.import __BSS_RUN__, __BSS_SIZE__, __EXTZP_RUN__
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.import __INTERRUPTOR_COUNT__
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.import scnkey, UDTIM
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.include "zeropage.inc"
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.include "extzp.inc"
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.include "cbm610.inc"
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; ------------------------------------------------------------------------
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; The BASIC header and a small BASIC program. Since it isn't possible to start
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; programs in other banks using SYS, the BASIC program will write a small
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; machine code program into memory at $100; and, start that machine code
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; program. The machine code program will then start the machine language
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; code in bank 1, which will initialize the system by copying stuff from
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; the system bank, and start the application.
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;
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; Here's the BASIC program that's in the following lines:
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;
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; 10 for i=0 to 4
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; 20 read j
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; 30 poke 256+i,j
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; 40 next i
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; 50 sys 256
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; 60 data 120,169,1,133,0
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;
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; The machine program in the data lines is:
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;
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; sei
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; lda #$01
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; sta $00 <-- Switch to bank 1 after this command
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;
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; Initialization is complex not only because of the jumping from one bank
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; into another. but also because we want to save memory; and because of
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; that, we will use the system memory ($00-$3FF) for initialization stuff
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; that is overwritten later.
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;
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.segment "EXEHDR"
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.byte $03,$00,$11,$00,$0a,$00,$81,$20,$49,$b2,$30,$20,$a4,$20,$34,$00
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.byte $19,$00,$14,$00,$87,$20,$4a,$00,$27,$00,$1e,$00,$97,$20,$32,$35
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.byte $36,$aa,$49,$2c,$4a,$00,$2f,$00,$28,$00,$82,$20,$49,$00,$39,$00
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.byte $32,$00,$9e,$20,$32,$35,$36,$00,$4f,$00,$3c,$00,$83,$20,$31,$32
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.byte $30,$2c,$31,$36,$39,$2c,$31,$2c,$31,$33,$33,$2c,$30,$00,$00,$00
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;------------------------------------------------------------------------------
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; A table that contains values that must be transferred from the system zero-
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; page into our zero-page. Contains pairs of bytes, first one is the address
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; in the system ZP, second one is our ZP address. The table goes into page 2;
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; but, is declared here because it is needed earlier.
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.SEGMENT "PAGE2"
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; (We use .proc because we need both a label and a scope.)
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.proc transfer_table
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.byte $9F, DEVNUM
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.byte $CA, CURS_Y
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.byte $CB, CURS_X
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.byte $CC, graphmode
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.byte $D4, config
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.endproc
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;------------------------------------------------------------------------------
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; Page 3 data. This page contains the break vector and the bankswitch
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; subroutine that is copied into high memory on startup. The space occupied by
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; this routine will later be used for a copy of the bank 15 stack. It must be
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; saved since we're going to destroy it when calling bank 15.
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.segment "PAGE3"
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BRKVec: .addr _exit ; BRK indirect vector
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.proc callbank15
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excrts := $FF05 ; In bank 15 ROM
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.org $FECB
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entry: php
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pha
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lda #$0F ; Bank 15
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sta IndReg
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txa
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pha
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tya
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pha
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sei
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ldy #$FF
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lda (sysp1),y
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tay
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lda ExecReg
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sta (sysp1),y
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dey
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lda #.hibyte(excrts-1)
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sta (sysp1),y
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dey
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lda #.lobyte(excrts-1)
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sta (sysp1),y
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tya
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sec
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sbc #7
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sta $1FF ; Save new sp
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tay
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tsx
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pla
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iny
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sta (sysp1),y
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pla
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iny
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sta (sysp1),y
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pla
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iny
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sta (sysp1),y
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pla
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iny
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sta (sysp1),y
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lda $105,x
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sec
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sbc #3
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iny
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sta (sysp1),y
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lda $106,x
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sbc #0
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iny
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sta (sysp1),y
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ldy $1FF ; Restore sp in bank 15
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lda #.hibyte(expull-1)
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sta (sysp1),y
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dey
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lda #.lobyte(expull-1)
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sta (sysp1),y
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dey
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pla
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pla
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tsx
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stx $1FF
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tya
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tax
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txs
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lda IndReg
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jmp $FFF6
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expull: pla
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tay
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pla
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tax
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pla
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plp
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rts
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.if (expull <> $FF2E)
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.error "Symbol expull must be aligned with Kernal in bank 15"
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.endif
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.reloc
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.endproc
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;------------------------------------------------------------------------------
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; The code in the target bank when switching back will be put at the bottom
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; of the stack. We will jump here to switch segments. The range $F2..$FF is
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; not used by any Kernal routine.
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.segment "STARTUP"
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Back: sta ExecReg
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; We are at $100 now. The following snippet is a copy of the code that is poked
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; in the system bank memory by the BASIC header program; it's only for
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; documentation, and not actually used here:
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sei
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lda #$01
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sta ExecReg
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; This is the actual starting point of our code after switching banks for
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; startup. Beware: The following code will get overwritten as soon as we
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; use the stack (since it's in page 1)! We jump to another location since
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; we need some space for subroutines that aren't used later.
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jmp Origin
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; Hardware vectors, copied to $FFF6
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.proc vectors
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sta ExecReg
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rts
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nop
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.word nmi ; NMI vector
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.word 0 ; Reset -- not used
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.word irq ; IRQ vector
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.endproc
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; Initializers for the extended zero-page. See "extzp.s".
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.proc extzp
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.word $0100 ; sysp1
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.word $0300 ; sysp3
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.word $d800 ; crtc
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.word $da00 ; sid
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.word $db00 ; ipccia
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.word $dc00 ; cia
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.word $dd00 ; acia
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.word $de00 ; tpi1
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.word $df00 ; tpi2
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.word $ea29 ; ktab1
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.word $ea89 ; ktab2
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.word $eae9 ; ktab3
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.word $eb49 ; ktab4
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.endproc
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; Switch the indirect segment to the system bank.
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Origin: lda #$0F
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sta IndReg
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; Initialize the extended zero-page.
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ldx #.sizeof(extzp)-1
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L1: lda extzp,x
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sta <__EXTZP_RUN__,x
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dex
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bpl L1
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; Save the old stack pointer from the system bank; and, set up our hw sp.
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tsx
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txa
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ldy #$FF
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sta (sysp1),y ; Save system stack point into $F:$1FF
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ldx #$FE ; Leave $1FF untouched for cross-bank calls
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txs ; Set up our own stack
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; Copy stuff from the system zero-page to ours.
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lda #.sizeof(transfer_table)
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sta ktmp
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L2: ldx ktmp
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ldy transfer_table-2,x
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lda transfer_table-1,x
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tax
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lda (sysp0),y
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sta $00,x
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dec ktmp
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dec ktmp
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bne L2
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; Set the interrupt, NMI, and other vectors.
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ldx #.sizeof(vectors)-1
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L3: lda vectors,x
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sta $10000 - .sizeof(vectors),x
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dex
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bpl L3
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; Set up the C stack.
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lda #.lobyte(callbank15::entry)
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sta sp
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lda #.hibyte(callbank15::entry)
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sta sp+1
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; Set up the subroutine and jump vector table that redirects Kernal calls to
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; the system bank.
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ldy #.sizeof(callbank15)
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@L1: lda callbank15-1,y
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sta callbank15::entry-1,y
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dey
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bne @L1
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; Set up the jump vector table. Y is zero on entry.
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ldx #45-1 ; Number of vectors
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@L2: lda #$20 ; JSR opcode
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sta $FF6F,y
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iny
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lda #.lobyte(callbank15::entry)
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sta $FF6F,y
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iny
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lda #.hibyte(callbank15::entry)
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sta $FF6F,y
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iny
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dex
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bpl @L2
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; Set the indirect segment to the bank that we're executing in.
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lda ExecReg
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sta IndReg
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; Zero the BSS segment. We will do that here instead of calling the routine
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; in the common library, since we have the memory anyway; and this way,
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; it's reused later.
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lda #<__BSS_RUN__
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sta ptr1
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lda #>__BSS_RUN__
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sta ptr1+1
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lda #0
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tay
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; Clear full pages.
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ldx #>__BSS_SIZE__
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beq Z2
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Z1: sta (ptr1),y
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iny
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bne Z1
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inc ptr1+1 ; Next page
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dex
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bne Z1
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; Clear the remaining page.
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Z2: ldx #<__BSS_SIZE__
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beq Z4
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Z3: sta (ptr1),y
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iny
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dex
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bne Z3
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Z4: jmp Init
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; ------------------------------------------------------------------------
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; We are at $200 now. We may now start calling subroutines safely since
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; the code we execute is no longer in the stack page.
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.segment "PAGE2"
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; Activate the chained interrupt handlers; then, enable interrupts.
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Init: lda #.lobyte(__INTERRUPTOR_COUNT__*2)
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sta irqcount
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cli
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; Call module constructors.
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jsr initlib
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; Push the command-line arguments; and, call main().
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jsr callmain
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; Call the module destructors. This is also the exit() entry and the default entry
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; point for the break vector.
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_exit: pha ; Save the return code
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jsr donelib ; Run module destructors
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lda #$00
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sta irqcount ; Disable custom irq handlers
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; Address the system bank.
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lda #$0F
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sta IndReg
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; Copy stuff back from our zero-page to the system's.
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.if 0
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lda #.sizeof(transfer_table)
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sta ktmp
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@L0: ldx ktmp
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ldy transfer_table-2,x
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lda transfer_table-1,x
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tax
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lda $00,x
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sta (sysp0),y
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dec ktmp
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dec ktmp
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bne @L0
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.endif
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; Place the program return code into BASIC's status variable.
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pla
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ldy #$9C ; ST
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sta (sysp0),y
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; Set up the welcome code at the stack bottom in the system bank.
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ldy #$FF
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lda (sysp1),y ; Load system bank sp
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tax
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iny ; Y = 0
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lda #$58 ; CLI opcode
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sta (sysp1),y
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iny
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lda #$60 ; RTS opcode
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sta (sysp1),y
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lda IndReg
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sei
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txs
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jmp Back
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; -------------------------------------------------------------------------
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; The IRQ handler goes into PAGE2. For performance reasons, and to allow
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; easier chaining, we do handle the IRQs in the execution bank (instead of
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; passing them to the system bank).
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; This is the mapping of the active IRQ register of the 6525 (tpi1):
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;
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; Bit 7 6 5 4 3 2 1 0
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; | | | | ^ 50 Hz.
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; | | | ^ SRQ IEEE 488
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; | | ^ CIA
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; | ^ IRQB ext. Port
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; ^ ACIA
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irq: pha
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txa
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pha
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tya
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pha
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lda IndReg
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pha
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lda ExecReg
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sta IndReg ; Be sure to address our segment
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tsx
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lda $105,x ; Get the flags from the stack
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and #$10 ; Test break flag
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bne dobrk
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; It's an IRQ.
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cld
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; Call the chained IRQ handlers.
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ldy irqcount
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beq irqskip
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jsr callirq_y ; Call the functions
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; Done with the chained IRQ handlers; check the TPI for IRQs, and handle them.
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irqskip:lda #$0F
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sta IndReg
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ldy #TPI::AIR
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lda (tpi1),y ; Interrupt Register 6525
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beq noirq
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; 50/60Hz. interrupt
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cmp #%00000001 ; ticker IRQ?
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bne irqend
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jsr scnkey ; Poll the keyboard
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jsr UDTIM ; Bump the time
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; Done.
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irqend: ldy #TPI::AIR
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sta (tpi1),y ; Clear interrupt
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noirq: pla
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sta IndReg
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pla
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tay
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pla
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tax
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pla
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nmi: rti
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dobrk: jmp (BRKVec)
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; -------------------------------------------------------------------------
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; Data area
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.bss
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irqcount: .byte 0
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