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2ddca57f3f
The goal was to be able to implement "(" in forth, but I realised that my
INTERPRET approach was wrong. Compiling the line beforehand is, after all,
not good. I'll have to change it again.
108 lines
2.0 KiB
NASM
108 lines
2.0 KiB
NASM
; Sets Z is A is ' ' or '\t' (whitespace)
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isWS:
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cp ' '
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ret z
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cp 0x09
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ret
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; Advance HL to next WS.
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; Set Z if WS found, unset if end-of-string.
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toWS:
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ld a, (hl)
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call isWS
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ret z
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cp 0x01 ; if a is null, carries and unsets z
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ret c
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inc hl
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jr toWS
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; Consume following whitespaces in HL until a non-WS is hit.
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; Set Z if non-WS found, unset if end-of-string.
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rdWS:
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ld a, (hl)
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cp 0x01 ; if a is null, carries and unsets z
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ret c
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call isWS
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jr nz, .ok
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inc hl
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jr rdWS
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.ok:
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cp a ; ensure Z
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ret
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; Copy string from (HL) in (DE), that is, copy bytes until a null char is
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; encountered. The null char is also copied.
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; HL and DE point to the char right after the null char.
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strcpyM:
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ld a, (hl)
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ld (de), a
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inc hl
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inc de
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or a
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jr nz, strcpyM
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ret
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; Like strcpyM, but preserve HL and DE
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strcpy:
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push hl
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push de
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call strcpyM
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pop de
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pop hl
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ret
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; Compares strings pointed to by HL and DE until one of them hits its null char.
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; If equal, Z is set. If not equal, Z is reset. C is set if HL > DE
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strcmp:
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push hl
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push de
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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 caller
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or a ; If our chars are null, stop the cmp
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inc hl
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inc de
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jr nz, .loop ; Z is carried through
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.end:
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pop de
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pop hl
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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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; Given a string at (HL), move HL until it points to the end of that string.
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strskip:
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push bc
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ex af, af'
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xor a ; look for null char
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ld b, a
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ld c, a
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cpir ; advances HL regardless of comparison, so goes one too far
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dec hl
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ex af, af'
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pop bc
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ret
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; Returns length of string at (HL) in A.
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; Doesn't include null termination.
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strlen:
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push bc
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xor a ; look for null char
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ld b, a
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ld c, a
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cpir ; advances HL to the char after the null
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.found:
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; How many char do we have? We have strlen=(NEG BC)-1, since BC started
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; at 0 and decreased at each CPIR loop. In this routine,
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; we stay in the 8-bit realm, so C only.
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add hl, bc
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sub c
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dec a
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pop bc
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ret
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