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74 lines
1.4 KiB
NASM
74 lines
1.4 KiB
NASM
; run RLA the number of times specified in B
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rlaX:
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; first, see if B == 0 to see if we need to bail out
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inc b
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dec b
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ret z ; Z flag means we had B = 0
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.loop: rla
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djnz .loop
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ret
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callHL:
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jp (hl)
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ret
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; If string at (HL) starts with ( and ends with ), "enter" into the parens
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; (advance HL and put a null char at the end of the string) and set Z.
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; Otherwise, do nothing and reset Z.
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enterParens:
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ld a, (hl)
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cp '('
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ret nz ; nothing to do
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push hl
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ld a, 0 ; look for null char
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; advance until we get null
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.loop:
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cpi
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jp z, .found
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jr .loop
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.found:
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dec hl ; cpi over-advances. go back to null-char
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dec hl ; looking at the last char before null
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ld a, (hl)
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cp ')'
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jr nz, .doNotEnter
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; We have parens. While we're here, let's put a null
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xor a
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ld (hl), a
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pop hl ; back at the beginning. Let's advance.
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inc hl
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cp a ; ensure Z
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ret ; we're good!
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.doNotEnter:
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pop hl
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call JUMP_UNSETZ
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ret
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; Find string (HL) in string list (DE) of size B. Each string is C bytes wide.
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; Returns the index of the found string. Sets Z if found, unsets Z if not found.
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findStringInList:
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push de
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push bc
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.loop:
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ld a, c
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call JUMP_STRNCMP
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ld a, c
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call JUMP_ADDDE
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jr z, .match
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djnz .loop
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; no match, Z is unset
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pop bc
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pop de
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ret
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.match:
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; Now, we want the index of our string, which is equal to our initial B
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; minus our current B. To get this, we have to play with our registers
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; and stack a bit.
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ld d, b
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pop bc
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ld a, b
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sub d
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pop de
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cp a ; ensure Z
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ret
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