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lib/expr: make recursion process a bit more orderly
Instead of going left and right, finding operators chars and replacing them with nulls, we parse expressions in a more orderly manner, one chunk at a time. I think it qualifies as "recursive descent", but I'm not sure. This allows us to preserve the string we parse and should also make the implementation of parens much easier.
This commit is contained in:
Virgil Dupras
parent
7410891ad1
commit
213614af33
@ -1,7 +1,5 @@
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; *** Requirements ***
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; findchar
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; multDEBC
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; callIXI
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; ari
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;
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; *** Defines ***
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;
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@ -11,167 +9,132 @@
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;
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; *** Code ***
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;
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; Parse expression in string at (HL) and returns the result in DE.
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; **This routine mutates (HL).**
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; We expect (HL) to be disposable: we mutate it to avoid having to make a copy.
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; This routine needs to be able to mutate (HL), but it takes care of restoring
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; the string to its original value before returning.
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; Sets Z on success, unset on error.
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parseExpr:
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push iy
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push ix
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push hl
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call _parseExpr
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push ix \ pop de
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call _parseAddSubst
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pop hl
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pop ix
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pop iy
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ret
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_parseExpr:
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ld de, exprTbl
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.loop:
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ld a, (de)
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or a
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jp z, EXPR_PARSE ; no operator, just parse the literal
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push de ; --> lvl 1. save operator row
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call _findAndSplit
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jr z, .found
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pop de ; <-- lvl 1
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inc de \ inc de \ inc de
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jr .loop
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.found:
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; Operator found, string splitted. Left in (HL), right in (DE)
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call _resolveLeftAndRight
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; Whether _resolveLeftAndRight was a success, we pop our lvl 1 stack
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; out, which contains our operator row. We pop it in IX.
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; L-R numbers are parsed in HL (left) and DE (right).
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pop ix ; <-- lvl 1
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; *** Op signature ***
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; The signature of "operators routines" (.plus, .mult, etc) below is this:
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; Combine HL and DE with an operator (+, -, *, etc) and put the result in DE.
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; Destroys HL and A. Never fails. Yes, that's a problem for division by zero.
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; Don't divide by zero. All other registers are protected.
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; Given a running result in DE, a rest-of-expression in (HL), a parse routine
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; in IY and an apply "operator routine" in IX, (HL/DE --> DE)
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; With that, parse the rest of (HL) and apply the operation on it, then place
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; HL at the end of the parsed string, with A containing the last char of it,
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; which can be either an operator or a null char.
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; Z for success.
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;
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_parseApply:
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push de ; --> lvl 1, left result
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push ix ; --> lvl 2, routine to apply
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inc hl ; after op char
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call callIY ; --> DE
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pop ix ; <-- lvl 2, routine to apply
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; Here we do some stack kung fu. We have, in HL, a string pointer we
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; want to keep. We have, in (SP), our left result we want to use.
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ex (sp), hl ; <-> lvl 1
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ret nz
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; Resolving left and right succeeded, proceed!
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inc ix ; point to routine pointer
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call callIXI
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push de \ pop ix
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cp a ; ensure Z
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push af ; --> lvl 2, save ending operator
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call callIX
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pop af ; <-- lvl 2, restore operator.
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pop hl ; <-- lvl 1, restore str pointer
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ret
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; Given a string in (HL) and a separator char in A, return a splitted string,
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; that is, the same (HL) string but with the found A char replaced by a null
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; char. DE points to the second part of the split.
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; Sets Z if found, unset if not found.
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_findAndSplit:
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push hl
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call .skipCharLiteral
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call findchar
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jr nz, .end ; nothing found
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; Alright, we have our char and we're pointing at it. Let's replace it
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; with a null char.
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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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ex de, hl ; DE now points to the second part of the split
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cp a ; ensure Z
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.end:
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pop hl ; HL is back to the start
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ret
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.skipCharLiteral:
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; special case: if our first char is ', skip the first 3 characters
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; so that we don't mistake a literal for an iterator
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push af
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ld a, (hl)
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cp 0x27 ; '
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jr nz, .skipCharLiteralEnd ; not a '
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xor a ; check for null char during skipping
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; skip 3
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inc hl
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cp (hl)
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jr z, .skipCharLiteralEnd
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inc hl
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cp (hl)
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jr z, .skipCharLiteralEnd
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inc hl
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.skipCharLiteralEnd:
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pop af
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ret
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.find:
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; parse expression on the left (HL) and the right (DE) and put the results in
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; HL (left) and DE (right)
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_resolveLeftAndRight:
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ld a, (hl)
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; Unless there's an error, this routine completely resolves any valid expression
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; from (HL) and puts the result in DE.
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; Destroys HL
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; Z for success.
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_parseAddSubst:
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call _parseMultDiv
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ret nz
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.loop:
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; do we have an operator?
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or a
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jr z, .noleft
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; Parse left operand in (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 ; return immediately if error
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.parseright:
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; Now we have parsed everything to the left and we have its result in
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; DE. What we need to do now is the same thing on (DE) and then apply
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; the + operator. Let's save DE somewhere and parse this.
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push de ; --> lvl 1
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; right expr in (HL)
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call parseExpr ; DE is set
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pop hl ; <-- lvl 1. left value
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ret ; Z is parseExpr's result
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.noleft:
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; special case: is (HL) zero? If yes, it means that our left operand
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; is empty. consider it as 0
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ex de, hl ; (DE) goes in (HL) for .parseright
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ld de, 0
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jr .parseright
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; Routines in here all have the same signature: they take two numbers, DE (left)
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; and IX (right), apply the operator and put the resulting number in DE.
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; The table has 3 bytes per row: 1 byte for operator and 2 bytes for routine
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; pointer.
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exprTbl:
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.db '+'
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.dw .plus
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.db '-'
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.dw .minus
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.db '*'
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.dw .mult
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.db '/'
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.dw .div
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.db '%'
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.dw .mod
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.db '&'
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.dw .and
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.db 0x7c ; '|'
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.dw .or
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.db '^'
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.dw .xor
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.db '}'
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.dw .rshift
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.db '{'
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.dw .lshift
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.db 0 ; end of table
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ret z ; null char, we're done
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; We have an operator. Resolve the rest of the expr then apply it.
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ld ix, .plus
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cp '+'
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jr z, .found
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ld ix, .minus
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cp '-'
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ret nz ; unknown char, error
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.found:
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ld iy, _parseMultDiv
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call _parseApply
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ret nz
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jr .loop
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.plus:
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add hl, de
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ex de, hl
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ret
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.minus:
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or a ; clear carry
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sbc hl, de
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ex de, hl
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ret
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; Parse (HL) as far as it can, that is, resolving expressions at its level or
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; lower (anything but + and -).
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; A is set to the last op it encountered. Unless there's an error, this can only
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; be +, - or null. Null if we're done parsing, + and - if there's still work to
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; do.
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; (HL) points to last op encountered.
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; DE is set to the numerical value of everything that was parsed left of (HL).
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_parseMultDiv:
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call _parseBitShift
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ret nz
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.loop:
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; do we have an operator?
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or a
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ret z ; null char, we're done
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; We have an operator. Resolve the rest of the expr then apply it.
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ld ix, .mult
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cp '*'
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jr z, .found
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ld ix, .div
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cp '/'
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jr z, .found
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ld ix, .mod
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cp '%'
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jr z, .found
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; might not be an error, return success
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cp a
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ret
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.found:
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ld iy, _parseBitShift
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call _parseApply
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ret nz
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jr .loop
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.mult:
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push bc ; --> lvl 1
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ld b, h
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ld c, l
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call multDEBC ; --> HL
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pop bc ; <-- lvl 1
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ex de, hl
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ret
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.div:
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; divide takes HL/DE
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push bc
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ld a, l
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push bc ; --> lvl 1
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call divide
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ld e, c
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ld d, b
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pop bc
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pop bc ; <-- lvl 1
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ret
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.mod:
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@ -179,6 +142,39 @@ exprTbl:
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ex de, hl
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ret
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; Same as _parseMultDiv, but a layer lower.
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_parseBitShift:
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call _parseNumber
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ret nz
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.loop:
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; do we have an operator?
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or a
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ret z ; null char, we're done
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; We have an operator. Resolve the rest of the expr then apply it.
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ld ix, .and
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cp '&'
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jr z, .found
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ld ix, .or
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cp 0x7c ; '|'
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jr z, .found
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ld ix, .xor
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cp '^'
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jr z, .found
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ld ix, .rshift
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cp '}'
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jr z, .found
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ld ix, .lshift
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cp '{'
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jr z, .found
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; might not be an error, return success
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cp a
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ret
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.found:
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ld iy, _parseNumber
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call _parseApply
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ret nz
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jr .loop
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.and:
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ld a, h
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and d
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@ -209,26 +205,130 @@ exprTbl:
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ld a, e
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and 0xf
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ret z
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push bc
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push bc ; --> lvl 1
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ld b, a
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.rshiftLoop:
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srl h
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rr l
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djnz .rshiftLoop
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ex de, hl
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pop bc
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pop bc ; <-- lvl 1
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ret
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.lshift:
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ld a, e
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and 0xf
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ret z
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push bc
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push bc ; --> lvl 1
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ld b, a
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.lshiftLoop:
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sla l
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rl h
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djnz .lshiftLoop
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ex de, hl
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pop bc
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pop bc ; <-- lvl 1
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ret
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; Parse first number of expression at (HL). A valid number is anything that can
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; be parsed by EXPR_PARSE and is followed either by a null char or by any of the
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; operator chars. This routines takes care of replacing an operator char with
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; the null char before calling EXPR_PARSE and then replace the operator back
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; afterwards.
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; HL is moved to the char following the number having been parsed.
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; DE contains the numerical result.
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; A contains the operator char following the number (or null). Only on success.
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; Z for success.
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_parseNumber:
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; Special case 1: number starts with '-'
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ld a, (hl)
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cp '-'
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jr nz, .skip1
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; We have a negative number. Parse normally, then subst from zero
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inc hl
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call _parseNumber
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push hl ; --> lvl 1
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ex af, af' ; preserve flags
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or a ; clear carry
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ld hl, 0
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sbc hl, de
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ex de, hl
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ex af, af' ; restore flags
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pop hl ; <-- lvl 1
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ret
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.skip1:
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; End of special case 1
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push ix
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; Copy beginning of string to DE, we'll need it later
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ld d, h
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ld e, l
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; Special case 2: we have a char literal. If we have a char literal, we
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; don't want to go through the "_isOp" loop below because if that char
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; is one of our operators, we're messing up our processing. So, set
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; ourselves 3 chars further and continue from there. EXPR_PARSE will
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; take care of validating those 3 chars.
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cp 0x27 ; apostrophe (') char
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jr nz, .skip2
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; "'". advance HL by 3
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inc hl \ inc hl \ inc hl
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; End of special case 2
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.skip2:
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dec hl ; offset "inc-hl-before" in loop
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.loop:
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inc hl
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ld a, (hl)
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call _isOp
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jr nz, .loop
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; (HL) and A is an op or a null
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push af ; --> lvl 1 save op
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push hl ; --> lvl 2 save end of string
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; temporarily put a null char instead of the op
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xor a
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ld (hl), a
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ex de, hl ; rewind to beginning of number
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call EXPR_PARSE
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ex af, af' ; keep result flags away while we restore (HL)
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push ix \ pop de ; result in DE
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pop hl ; <-- lvl 2, end of string
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pop af ; <-- lvl 1, saved op
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ld (hl), a
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ex af, af' ; restore Z from EXPR_PARSE
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jr nz, .end
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; HL is currently at the end of the number's string
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; On success, have A be the operator char following the number
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ex af, af'
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.end:
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pop ix
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ret
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; Sets Z if A contains a valid operator char or a null char.
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_isOp:
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or a
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ret z
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push hl ; --> lvl 1
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; Set A' to zero for quick end-of-table checks
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ex af, af'
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xor a
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ex af, af'
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ld hl, .exprChars
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.loop:
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cp (hl)
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jr z, .found
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ex af, af'
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cp (hl)
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jr z, .notFound ; end of table
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ex af, af'
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inc hl ; next char
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jr .loop
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.notFound:
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ex af, af' ; restore orig A
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inc a ; unset Z
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.found:
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; Z already set
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pop hl ; <-- lvl 1
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ret
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.exprChars:
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.db "+-*/%&|^{}", 0
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@ -51,6 +51,24 @@ assertEQW:
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.msg:
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.db "HL != DE", CR, LF, 0
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; Given a list of pointer to test data structures in HL and a pointer to a test
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; routine in IX, call (IX) with HL pointing to the test structure until the list
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; points to a zero. See testParseExpr in test_expr for an example usage.
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testList:
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push hl ; --> lvl 1
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call intoHL
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ld a, h
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or l
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jr z, .end
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call callIX
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call nexttest
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pop hl ; <-- lvl 1
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inc hl \ inc hl
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jr testList
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.end:
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pop hl ; <-- lvl 1
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ret
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nexttest:
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ld a, (testNum)
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inc a
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@ -109,35 +109,18 @@ test:
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halt
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testParseExpr:
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ld iy, .t1
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call .testEQ
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ld iy, .t2
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call .testEQ
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ld iy, .t3
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call .testEQ
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ld iy, .t4
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call .testEQ
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ld iy, .t5
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call .testEQ
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ld iy, .t6
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call .testEQ
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ld iy, .t7
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call .testEQ
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ld iy, .t8
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call .testEQ
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ld iy, .t9
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call .testEQ
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ret
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ld hl, .alltests
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ld ix, .test
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jp testList
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.testEQ:
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push iy \ pop hl
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.test:
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push hl \ pop iy
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inc hl \ inc hl
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call parseExpr
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call assertZ
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ld l, (iy)
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ld h, (iy+1)
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call assertEQW
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jp nexttest
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jp assertEQW
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.t1:
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.dw 7
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@ -166,3 +149,13 @@ testParseExpr:
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.t9:
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.dw 10
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.db "2*3+4", 0
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; There was this untested regression during the replacement of find-and-subst
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; parseExpr to the recursive descent one. It was time consuming to find. Here
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; it goes, here it stays.
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.t10:
|
||||
.dw '-'+1
|
||||
.db "'-'+1", 0
|
||||
|
||||
.alltests:
|
||||
.dw .t1, .t2, .t3, .t4, .t5, .t6, .t7, .t8, .t9, .t10, 0
|
||||
|
||||
@ -13,9 +13,9 @@ cmpas() {
|
||||
echo ok
|
||||
else
|
||||
echo actual
|
||||
echo $ACTUAL
|
||||
echo "$ACTUAL"
|
||||
echo expected
|
||||
echo $EXPECTED
|
||||
echo "$EXPECTED"
|
||||
exit 1
|
||||
fi
|
||||
}
|
||||
|
||||
Loading…
Reference in New Issue
Block a user