1
0
mirror of https://github.com/cc65/cc65.git synced 2024-11-18 00:07:21 +00:00
cc65/libsrc/cbm610/crt0.s

490 lines
12 KiB
ArmAsm
Raw Normal View History

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