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143 lines
3.0 KiB
NASM
143 lines
3.0 KiB
NASM
; Parse an expression yielding a truth value from (HL) and set A accordingly.
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; 0 for False, nonzero for True.
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; How it evaluates truth is that it looks for =, <, >, >= or <= in (HL) and,
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; if it finds it, evaluate left and right expressions separately. Then it
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; compares both sides and set A accordingly.
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; If comparison operators aren't found, the whole string is sent to parseExpr
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; and zero means False, nonzero means True.
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; **This routine mutates (HL).**
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; Z for success.
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parseTruth:
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push ix
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push de
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ld a, '='
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call .maybeFind
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jr z, .foundEQ
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ld a, '<'
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call .maybeFind
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jr z, .foundLT
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ld a, '>'
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call .maybeFind
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jr z, .foundGT
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jr .simple
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.success:
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cp a ; ensure Z
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.end:
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pop de
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pop ix
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ret
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.maybeFind:
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push hl ; --> lvl 1
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call findchar
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jr nz, .notFound
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; found! We want to keep new HL around. Let's pop old HL in DE
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pop de ; <-- lvl 1
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ret
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.notFound:
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; not found, restore HL
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pop hl ; <-- lvl 1
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ret
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.simple:
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call parseExpr
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jr nz, .end
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ld a, d
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or e
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jr .success
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.foundEQ:
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; we found an '=' char and HL is pointing to it. DE is pointing to the
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; beginning of our string. Let's separate those two strings.
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; But before we do that, to we have a '<' or a '>' at the left of (HL)?
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dec hl
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ld a, (hl)
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cp '<'
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jr z, .foundLTE
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cp '>'
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jr z, .foundGTE
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inc hl
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; Ok, we are a straight '='. Proceed.
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call .splitLR
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; HL now point to right-hand, DE to left-hand
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call .parseLeftRight
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jr nz, .end ; error, stop
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xor a ; clear carry and prepare value for False
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sbc hl, de
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jr nz, .success ; NZ? equality not met. A already 0, return.
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; Z? equality met, make A=1, set Z
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inc a
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jr .success
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.foundLTE:
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; Almost the same as '<', but we have two sep chars
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call .splitLR
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inc hl ; skip the '=' char
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call .parseLeftRight
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jr nz, .end
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ld a, 1 ; prepare for True
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sbc hl, de
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jr nc, .success ; Left <= Right, True
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; Left > Right, False
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dec a
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jr .success
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.foundGTE:
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; Almost the same as '<='
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call .splitLR
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inc hl ; skip the '=' char
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call .parseLeftRight
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jr nz, .end
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ld a, 1 ; prepare for True
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sbc hl, de
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jr z, .success ; Left == Right, True
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jr c, .success ; Left > Right, True
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; Left < Right, False
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dec a
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jr .success
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.foundLT:
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; Same thing as EQ, but for '<'
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call .splitLR
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call .parseLeftRight
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jr nz, .end
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xor a
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sbc hl, de
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jr z, .success ; Left == Right, False
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jr c, .success ; Left > Right, False
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; Left < Right, True
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inc a
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jr .success
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.foundGT:
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; Same thing as EQ, but for '>'
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call .splitLR
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call .parseLeftRight
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jr nz, .end
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xor a
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sbc hl, de
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jr nc, .success ; Left <= Right, False
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; Left > Right, True
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inc a
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jr .success
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.splitLR:
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xor a
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ld (hl), a
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inc hl
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ret
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; Given string pointers in (HL) and (DE), evaluate those two expressions and
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; place their corresponding values in HL and DE.
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.parseLeftRight:
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; let's start with HL
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push de ; --> lvl 1
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call parseExpr
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pop hl ; <-- lvl 1, orig DE
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ret nz
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push de ; --> lvl 1. save HL value in stack.
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; Now, for DE. (DE) is now in HL
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call parseExpr ; DE in place
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pop hl ; <-- lvl 1. restore saved HL
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ret
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