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zasm: add support for numerical constants

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Virgil Dupras 5 年之前
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共有 1 個檔案被更改,包括 147 行新增10 行删除
  1. +147
    -10
      apps/zasm/zasm.asm

+ 147
- 10
apps/zasm/zasm.asm 查看文件

@@ -33,6 +33,64 @@ rlaX:
djnz .loop
ret

; Parse the decimal char at A and extract it's 0-9 numerical value. Put the
; result in A.
;
; On success, the carry flag is reset. On error, it is set.
parseDecimal:
; First, let's see if we have an easy 0-9 case
cp '0'
ret c ; if < '0', we have a problem
cp '9'+1
; We are in the 0-9 range
sub a, '0' ; C is clear
ret

; Parses the string at (HL) and returns the 16-bit value in IX.
; As soon as the number doesn't fit 16-bit any more, parsing stops and the
; number is invalid. If the number is valid, Z is set, otherwise, unset.
parseNumber:
push hl
push de

ld ix, 0
.loop:
ld a, (hl)
cp 0
jr z, .end ; success!
call parseDecimal
jr c, .error

; Now, let's add A to IX. First, multiply by 10.
ld d, ixh ; we need a copy of the initial copy for later
ld e, ixl
add ix, ix ; x2
add ix, ix ; x4
add ix, ix ; x8
add ix, de ; x9
add ix, de ; x10
add a, ixl
jr nc, .nocarry
inc ixh
.nocarry:
ld ixl, a

; We didn't bother checking for the C flag at each step because we
; check for overflow afterwards. If ixh < d, we overflowed
ld a, ixh
cp d
jr c, .error ; carry is set? overflow

inc hl
jr .loop

.error:
call unsetZ
.end:
pop de
pop hl
ret

; Sets Z is A is ';', CR, LF, or null.
isLineEnd:
cp ';'
@@ -112,14 +170,24 @@ toWord:
; HL is advanced to the next word. Z is set if there's a next word.
readArg:
push de
ld de, tmpVal
ld de, tmpBuf
call readWord
push hl
ld hl, tmpVal
ld hl, tmpBuf
call parseArg
pop hl
pop de
ld (de), a
; When A is a number, IX is set with the value of that number. Because
; We don't use the space allocated to store those numbers in any other
; occasion, we store IX there unconditonally, LSB first.
inc de
ld a, ixl
ld (de), a
inc de
ld a, ixh
ld (de), a

call toWord
ret

@@ -168,6 +236,9 @@ strlen:
; Return value 0xff holds a special meaning: arg is not empty, but doesn't match
; any argspec (A == 0 means arg is empty). A return value of 0xff means an
; error.
;
; If the parsed argument is a number constant, 'N' is returned and IX contains
; the value of that constant.
parseArg:
call strlen
cp 0
@@ -190,8 +261,14 @@ parseArg:
ld a, 5
call JUMP_ADDDE
djnz .loop1
; exhausted? we have a problem os specifying a wrong argspec. This is
; an internal consistency error.

; We exhausted the argspecs. Let's see if it's a number
call parseNumber
jr nz, .notanumber
; Alright, we have a parsed number in IX. We're done.
ld a, 'N'
jr .end
.notanumber:
ld a, 0xff
jr .end
.found:
@@ -291,11 +368,24 @@ findInGroup:

; Compare argspec from instruction table in A with argument in (HL).
; For constant args, it's easy: if A == (HL), it's a success.
; If it's not this, then we check if it's a numerical arg.
; If A is a group ID, we do something else: we check that (HL) exists in the
; groupspec (argGrpTbl)
matchArg:
cp a, (hl)
ret z
; not an exact match, let's check for numerical constants.
cp 'N'
jr z, .expectsNumber
cp 'n'
jr z, .expectsNumber
jr .notNumber
.expectsNumber:
ld a, (hl)
cp 'N' ; In parsed arg, we don't have 'n', only 'N'
ret ; whether we match or not, the result of Z is the good
; one
.notNumber:
; A bit of a delicate situation here: we want A to go in H but also
; (HL) to go in A. If not careful, we overwrite each other. EXX is
; necessary to avoid invoving other registers.
@@ -336,6 +426,8 @@ matchPrimaryRow:

; Parse line at (HL) and write resulting opcode(s) in (DE). Returns the number
; of bytes written in A.
;
; Overwrites IX
parseLine:
call readLine
; Check whether we have errors. We don't do any parsing if we do.
@@ -368,7 +460,6 @@ parseLine:
; We have our matching instruction row. We're getting pretty near our
; goal here!
; First, let's go in IX mode. It's easier to deal with offsets here.
push ix
ld ixh, d
ld ixl, e
; First, let's see if we're dealing with a group here
@@ -409,18 +500,63 @@ parseLine:
pop hl

; Success!
jr .end
jr .writeFirstOpcode
.notgroup:
; not a group? easy as pie: we return the opcode directly.
ld a, (ix+7) ; upcode is on 8th byte
.end:
.writeFirstOpcode:
; At the end, we have our final opcode in A!
pop ix
pop de
ld (de), a

; Good, we are probably finished here for many primary opcodes. However,
; some primary opcodes take 8 or 16 bit constants as an argument and
; if that's the case here, we need to write it too.
; We still have our instruction row in IX. Let's revisit it.
push hl ; we use HL to point to the currently read arg
ld a, (ix+4) ; first argspec
ld hl, curArg1
cp 'N'
jr z, .withWord
cp 'n'
jr z, .withByte
ld a, (ix+5) ; second argspec
ld hl, curArg2
cp 'N'
jr z, .withWord
cp 'n'
jr z, .withByte
; nope, no number, aright, only one opcode
ld a, 1
jr .end
.withByte:
; verify that the MSB in argument is zero
inc hl
inc hl ; MSB is 2nd byte
ld a, (hl)
dec hl ; HL now points to LSB
cp 0
jr nz, .numberTruncated
; Clear to proceed. HL already points to our number
inc de
ldi
ld a, 2
jr .end

.withWord:
inc hl ; HL now points to LSB
ldi ; LSB written, we point to MSB now
ldi ; MSB written
ld a, 3
jr .end
.numberTruncated:
; problem: not zero, so value is truncated. error
xor a
.end:
pop hl
ret


; In instruction metadata below, argument types arge indicated with a single
; char mnemonic that is called "argspec". This is the table of correspondance.
; Single letters are represented by themselves, so we don't need as much
@@ -523,6 +659,7 @@ instrTBlPrimary:
.db "LD",0,0, 's', 'h', 0, 0x0a ; LD SP, HL
.db "LD",0,0, 'l', 0xb, 0, 0b01110000 ; LD (HL), r
.db "LD",0,0, 0xb, 'l', 3, 0b01000110 ; LD r, (HL)
.db "LD",0,0, 'l', 'n', 0, 0x36 ; LD (HL), n
.db "NOP", 0, 0, 0, 0, 0x00 ; NOP
.db "OR",0,0, 'l', 0, 0, 0xb6 ; OR (HL)
.db "OR",0,0, 0xb, 0, 0, 0b10110000 ; OR r
@@ -556,6 +693,6 @@ curArg2:
.db 0, 0, 0

; space for tmp stuff
tmpVal:
.db 0, 0, 0, 0, 0
tmpBuf:
.fill 0x20


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