collapseos/apps/zasm/util.asm

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2019-04-30 15:51:39 -04:00
; run RLA the number of times specified in B
rlaX:
; first, see if B == 0 to see if we need to bail out
inc b
dec b
ret z ; Z flag means we had B = 0
.loop: rla
djnz .loop
ret
callHL:
jp (hl)
ret
2019-05-09 21:21:08 -04:00
; Compare HL with DE and sets Z and C in the same way as a regular cp X where
; HL is A and DE is X.
cpHLDE:
ld a, h
cp d
ret nz ; if not equal, flags are correct
ld a, l
cp e
ret ; flags are correct
2019-05-12 22:07:21 -04:00
; HL - DE -> HL
subDEFromHL:
push af
ld a, l
sub e
ld l, a
ld a, h
sbc d
ld h, a
pop af
ret
2019-05-09 21:21:08 -04:00
; Returns length of string at (HL) in A.
strlen:
push bc
push hl
ld bc, 0
ld a, 0 ; look for null char
.loop:
cpi
jp z, .found
jr .loop
.found:
; How many char do we have? the (NEG BC)-1, which started at 0 and
; decreased at each CPI call. In this routine, we stay in the 8-bit
; realm, so C only.
ld a, c
neg
dec a
pop hl
pop bc
ret
2019-04-30 15:51:39 -04:00
; If string at (HL) starts with ( and ends with ), "enter" into the parens
; (advance HL and put a null char at the end of the string) and set Z.
; Otherwise, do nothing and reset Z.
enterParens:
ld a, (hl)
cp '('
ret nz ; nothing to do
push hl
ld a, 0 ; look for null char
; advance until we get null
.loop:
cpi
jp z, .found
jr .loop
.found:
dec hl ; cpi over-advances. go back to null-char
dec hl ; looking at the last char before null
ld a, (hl)
cp ')'
jr nz, .doNotEnter
; We have parens. While we're here, let's put a null
xor a
ld (hl), a
pop hl ; back at the beginning. Let's advance.
inc hl
cp a ; ensure Z
ret ; we're good!
.doNotEnter:
pop hl
call JUMP_UNSETZ
ret
2019-04-30 21:13:37 -04:00
; Find string (HL) in string list (DE) of size B. Each string is C bytes wide.
; Returns the index of the found string. Sets Z if found, unsets Z if not found.
findStringInList:
push de
push bc
.loop:
ld a, c
call JUMP_STRNCMP
ld a, c
call JUMP_ADDDE
jr z, .match
djnz .loop
; no match, Z is unset
pop bc
pop de
ret
.match:
; Now, we want the index of our string, which is equal to our initial B
; minus our current B. To get this, we have to play with our registers
; and stack a bit.
ld d, b
pop bc
ld a, b
sub d
pop de
cp a ; ensure Z
ret