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71 lines
1.4 KiB
NASM
71 lines
1.4 KiB
NASM
; Parse expression in string at (HL) and returns the result in IX.
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; We expect (HL) to be disposable: we mutate it to avoid having to make a copy.
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; Sets Z on success, unset on error.
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parseExpr:
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push bc
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push de
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push hl
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ld a, '+'
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call findchar
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jr z, .hasExpr
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pop hl
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push hl
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ld a, '-'
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call findchar
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jr nz, .noExpr
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ld c, '-'
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jr .hasExpr
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.hasPlus:
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ld c, '+'
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jr .hasExpr
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.hasExpr:
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; Alright, we have a +/ and we're pointing at it. Let's advance HL and
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; recurse. But first, let's change this + into a null char. It will be
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; handy later.
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xor a
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ld (hl), a ; + changed to \0
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inc hl
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pop de ; we pop out the HL we pushed earlier into DE
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; That's our original beginning of string.
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call _applyExprToHL
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pop de
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pop bc
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ret
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.noExpr:
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pop hl
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pop de
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pop bc
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jp parseNumberOrSymbol
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; Parse number or symbol in (DE) and expression in (HL) and apply operator
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; specified in C to them.
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_applyExprToHL:
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call parseExpr
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ret nz ; return immediately if error
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; Now we have parsed everything to the right and we have its result in
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; IX. What we need to do now is parseNumberOrSymbol on (DE) and apply
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; operator. Let's save IX somewhere and parse this.
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ex hl, de
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push ix
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pop de
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call parseNumberOrSymbol
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ret nz ; error
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; Good! let's do the math! IX has our left part, DE has our right one.
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ld a, c ; restore operator
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cp '-'
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jr z, .sub
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; addition
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add ix, de
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jr .end
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.sub:
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push ix
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pop hl
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sbc hl, de
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push hl
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pop ix
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.end:
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cp a ; ensure Z
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ret
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