cca3157c66
* addHL and subHL affect flags, and are smaller Most importantly, addHL and subHL now affect the flags as you would expect from a 16 bit addition/subtraction. This seems like it'd be preferred behaviour, however I realise any code relying on it not affecting flags would break. One byte saved in addHL, and two bytes saved in subHL. Due to the branching nature of the original code, it's difficult to compare speeds, subHL is either 1 or 6 cycles faster depending on branching, and addHL is between -1 and 3 cycles faster. If the chance of a carry is 50%, addHL is expected to be a cycle faster, but for a chance of carry below 25% (so a < 0x40) this will be up to a cycle slower. * Update core.asm * Reworked one use of addHL By essentially inlining both addHL and cpHLDE, 100 cycles are saved, but due to the registers not needing preserving, a byte is saved too. * Corrected spelling error in comment * Reworked second use of addHL 43 cycles saved, and no more addHL in critical loops. No bytes saved or used. * Fixed tabs and spacing, and made a comment clearer. * Clearer comments * Adopted push/pop notation
244 lines
4.4 KiB
NASM
244 lines
4.4 KiB
NASM
; core
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;
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; Routines used by pretty much all parts. You will want to include it first
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; in your glue file.
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; *** CONSTS ***
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.equ ASCII_BS 0x08
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.equ ASCII_CR 0x0d
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.equ ASCII_LF 0x0a
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.equ ASCII_DEL 0x7f
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; *** DATA ***
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; Useful data to point to, when a pointer is needed.
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P_NULL: .db 0
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; *** REGISTER FIDDLING ***
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; add the value of A into DE
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addDE:
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push af
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add a, e
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jr nc, .end ; no carry? skip inc
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inc d
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.end:
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ld e, a
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pop af
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noop: ; piggy backing on the first "ret" we have
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ret
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; copy (HL) into DE, then exchange the two, utilising the optimised HL instructions.
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; ld must be done little endian, so least significant byte first.
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intoHL:
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push de
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ld e, (hl)
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inc hl
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ld d, (hl)
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ex de, hl
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pop de
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ret
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intoDE:
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ex de, hl
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call intoHL
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ex de, hl ; de preserved by intoHL, so no push/pop needed
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ret
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intoIX:
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push ix
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ex (sp), hl ;swap hl with ix, on the stack
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call intoHL
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ex (sp), hl ;restore hl from stack
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pop ix
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ret
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; add the value of A into HL
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; affects carry flag according to the 16-bit addition, Z, S and P untouched.
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addHL:
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push de
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ld d, 0
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ld e, a
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add hl, de
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pop de
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ret
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; subtract the value of A from HL
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; affects flags according to the 16-bit subtraction.
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subHL:
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push de
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ld d, 0
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ld e, a
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or a ;reset carry flag
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sbc hl, de ;There is no 'sub hl, de', so we must use sbc
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pop de
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ret
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; Compare HL with DE and sets Z and C in the same way as a regular cp X where
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; HL is A and DE is X.
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cpHLDE:
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push hl
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or a ;reset carry flag
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sbc hl, de ;There is no 'sub hl, de', so we must use sbc
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pop hl
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ret
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; Write the contents of HL in (DE)
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; de and hl are preserved, so no pushing/popping necessary
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writeHLinDE:
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ex de, hl
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ld (hl), e
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inc hl
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ld (hl), d
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dec hl
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ex de, hl
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ret
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; Call the method (IX) is a pointer to. In other words, call intoIX before
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; callIX
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callIXI:
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push ix
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call intoIX
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call callIX
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pop ix
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ret
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; jump to the location pointed to by IX. This allows us to call IX instead of
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; just jumping it. We use IX because we seldom use this for arguments.
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callIX:
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jp (ix)
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callIY:
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jp (iy)
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; Ensures that Z is unset (more complicated than it sounds...)
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unsetZ:
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push bc
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ld b, a
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inc b
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cp b
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pop bc
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ret
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; *** STRINGS ***
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; Fill B bytes at (HL) with A
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fill:
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push bc
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push hl
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.loop:
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ld (hl), a
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inc hl
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djnz .loop
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pop hl
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pop bc
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ret
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; Increase HL until the memory address it points to is equal to A for a maximum
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; of 0xff bytes. Returns the new HL value as well as the number of bytes
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; iterated in A.
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; If a null char is encountered before we find A, processing is stopped in the
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; same way as if we found our char (so, we look for A *or* 0)
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; Set Z if the character is found. Unsets it if not
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findchar:
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push bc
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ld c, a ; let's use C as our cp target
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ld a, 0xff
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ld b, a
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.loop: ld a, (hl)
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cp c
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jr z, .match
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or a ; cp 0
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jr z, .nomatch
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inc hl
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djnz .loop
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.nomatch:
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call unsetZ
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jr .end
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.match:
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; We ran 0xff-B loops. That's the result that goes in A.
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ld a, 0xff
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sub b
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cp a ; ensure Z
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.end:
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pop bc
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ret
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; Format the lower nibble of A into a hex char and stores the result in A.
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fmtHex:
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and 0xf
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cp 10
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jr nc, .alpha ; if >= 10, we have alpha
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add a, '0'
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ret
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.alpha:
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add a, 'A'-10
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ret
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; Formats value in A into a string hex pair. Stores it in the memory location
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; that HL points to. Does *not* add a null char at the end.
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fmtHexPair:
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push af
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; let's start with the rightmost char
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inc hl
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call fmtHex
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ld (hl), a
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; and now with the leftmost
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dec hl
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pop af
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push af
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and 0xf0
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rra \ rra \ rra \ rra
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call fmtHex
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ld (hl), a
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pop af
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ret
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; Compares strings pointed to by HL and DE up to A count of characters. If
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; equal, Z is set. If not equal, Z is reset.
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strncmp:
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push bc
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push hl
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push de
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ld b, a
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.loop:
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ld a, (de)
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cp (hl)
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jr nz, .end ; not equal? break early. NZ is carried out
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; to the called
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cp 0 ; If our chars are null, stop the cmp
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jr z, .end ; The positive result will be carried to the
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; caller
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inc hl
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inc de
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djnz .loop
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; We went through all chars with success, but our current Z flag is
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; unset because of the cp 0. Let's do a dummy CP to set the Z flag.
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cp a
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.end:
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pop de
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pop hl
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pop bc
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; Because we don't call anything else than CP that modify the Z flag,
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; our Z value will be that of the last cp (reset if we broke the loop
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; early, set otherwise)
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ret
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; Transforms the character in A, if it's in the a-z range, into its upcase
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; version.
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upcase:
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cp 'a'
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ret c ; A < 'a'. nothing to do
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cp 'z'+1
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ret nc ; A >= 'z'+1. nothing to do
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; 'a' - 'A' == 0x20
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sub 0x20
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ret
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