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zasm: add multiplication expressions
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@ -2,69 +2,85 @@
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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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call _parseExpr
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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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_parseExpr:
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ld a, '*'
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call _findAndSplit
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jp z, _applyMult
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ld a, '+'
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call _findAndSplit
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jp z, _applyPlus
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ld a, '-'
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call _findAndSplit
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jp z, _applyMinus
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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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; 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 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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; parse expression on the left (HL) and the right (DE) and put the results in
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; DE (left) and IX (right)
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_resolveLeftAndRight:
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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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; Now we have parsed everything to the left and we have its result in
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; IX. What we need to do now is the same thing on (DE) and then apply
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; the + operator. Let's save IX somewhere and parse this.
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ex hl, de ; right expr now in HL
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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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pop de ; numeric left expr result in DE
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jp parseExpr
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; Parse expr in (HL) and expr in (DE) and apply + operator to both sides.
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; Put result in IX.
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_applyPlus:
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call _resolveLeftAndRight
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ret nz
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; Good! let's do the math! IX has our right part, DE has our left one.
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add ix, de
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jr .end
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.sub:
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cp a ; ensure Z
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ret
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; Same as _applyPlus but with -
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_applyMinus:
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call _resolveLeftAndRight
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ret nz
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push ix
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pop hl
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ex de, 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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_applyMult:
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call _resolveLeftAndRight
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ret nz
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push ix \ pop bc
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call multDEBC
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push hl \ pop ix
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cp a ; ensure Z
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ret
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@ -210,3 +210,21 @@ findStringInList:
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pop de
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cp a ; ensure Z
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ret
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; DE * BC -> DE (high) and HL (low)
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multDEBC:
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ld hl, 0
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ld a, 0x10
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.loop:
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add hl, hl
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rl e
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rl d
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jr nc, .noinc
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add hl, bc
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jr nc, .noinc
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inc de
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.noinc:
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dec a
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jr nz, .loop
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ret
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@ -18,6 +18,7 @@ testNum: .db 1
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s1: .db "2+2", 0
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s2: .db "0x4001+0x22", 0
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s3: .db "FOO+BAR", 0
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s4: .db "3*3", 0
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sFOO: .db "FOO", 0
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sBAR: .db "BAR", 0
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@ -70,6 +71,17 @@ test:
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jp nz, fail
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call nexttest
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ld hl, s4
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call parseExpr
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jp nz, fail
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ld a, ixh
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or a
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jp nz, fail
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ld a, ixl
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cp 9
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jp nz, fail
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call nexttest
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; success
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xor a
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halt
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36
tools/tests/unit/test_util_z.asm
Normal file
36
tools/tests/unit/test_util_z.asm
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@ -0,0 +1,36 @@
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jp test
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#include "core.asm"
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#include "parse.asm"
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#include "zasm/util.asm"
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testNum: .db 1
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test:
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ld hl, 0xffff
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ld sp, hl
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ld de, 12
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ld bc, 4
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call multDEBC
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ld a, l
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cp 48
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jp nz, fail
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call nexttest
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; success
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xor a
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halt
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nexttest:
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ld a, (testNum)
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inc a
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ld (testNum), a
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ret
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fail:
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ld a, (testNum)
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halt
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@ -1,3 +1,4 @@
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; test expressions
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ld a, 'A'+3
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ld a, 'A'-0x20
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ld a, 8*5
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