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collapseos/apps/zasm/parse.asm
Virgil Dupras 6dbbfa837d apps/ed: add (dummy) line number processing
Starting to feel interactive...
2019-07-13 11:53:30 -04:00

204 lines
3.8 KiB
NASM

; Parse string at (HL) as a hexadecimal value and return value in IX under the
; same conditions as parseLiteral.
parseHexadecimal:
call hasHexPrefix
ret nz
push hl
push de
ld d, 0
inc hl ; get rid of "0x"
inc hl
call strlen
cp 3
jr c, .single
cp 4
jr c, .doubleShort ; 0x123
cp 5
jr c, .double ; 0x1234
; too long, error
jr .error
.double:
call parseHexPair
jr c, .error
inc hl ; now HL is on first char of next pair
ld d, a
jr .single
.doubleShort:
ld a, (hl)
call parseHex
jr c, .error
inc hl ; now HL is on first char of next pair
ld d, a
.single:
call parseHexPair
jr c, .error
ld e, a
cp a ; ensure Z
jr .end
.error:
call unsetZ
.end:
push de \ pop ix
pop de
pop hl
ret
; Sets Z if (HL) has a '0x' prefix.
hasHexPrefix:
ld a, (hl)
cp '0'
ret nz
push hl
inc hl
ld a, (hl)
cp 'x'
pop hl
ret
; Parse string at (HL) as a binary value (0b010101) and return value in IX.
; High IX byte is always clear.
; Sets Z on success.
parseBinaryLiteral:
call hasBinPrefix
ret nz
push bc
push hl
push de
ld d, 0
inc hl ; get rid of "0b"
inc hl
call strlen
or a
jr z, .error ; empty, error
cp 9
jr nc, .error ; >= 9, too long
; We have a string of 8 or less chars. What we'll do is that for each
; char, we rotate left and set the LSB according to whether we have '0'
; or '1'. Error out on anything else. C is our stored result.
ld b, a ; we loop for "strlen" times
ld c, 0 ; our stored result
.loop:
rlc c
ld a, (hl)
inc hl
cp '0'
jr z, .nobit ; no bit to set
cp '1'
jr nz, .error ; not 0 or 1
; We have a bit to set
inc c
.nobit:
djnz .loop
ld e, c
cp a ; ensure Z
jr .end
.error:
call unsetZ
.end:
push de \ pop ix
pop de
pop hl
pop bc
ret
; Sets Z if (HL) has a '0b' prefix.
hasBinPrefix:
ld a, (hl)
cp '0'
ret nz
push hl
inc hl
ld a, (hl)
cp 'b'
pop hl
ret
; Parse string at (HL) and, if it is a char literal, sets Z and return
; corresponding value in IX. High IX byte is always clear.
;
; A valid char literal starts with ', ends with ' and has one character in the
; middle. No escape sequence are accepted, but ''' will return the apostrophe
; character.
parseCharLiteral:
ld a, 0x27 ; apostrophe (') char
cp (hl)
ret nz
push hl
push de
inc hl
inc hl
cp (hl)
jr nz, .end ; not ending with an apostrophe
inc hl
ld a, (hl)
or a ; cp 0
jr nz, .end ; string has to end there
; Valid char, good
ld d, a ; A is zero, take advantage of that
dec hl
dec hl
ld a, (hl)
ld e, a
cp a ; ensure Z
.end:
push de \ pop ix
pop de
pop hl
ret
; Parses the string at (HL) and returns the 16-bit value in IX. The string
; can be a decimal literal (1234), a hexadecimal literal (0x1234) or a char
; literal ('X').
;
; 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.
parseLiteral:
call parseCharLiteral
ret z
call parseHexadecimal
ret z
call parseBinaryLiteral
ret z
jp parseDecimal
; Parse string in (HL) and return its numerical value whether its a number
; literal or a symbol. Returns value in IX.
; Sets Z if number or symbol is valid, unset otherwise.
parseNumberOrSymbol:
call parseLiteral
ret z
; Not a number. Try PC
push de
ld de, .sDollar
call strcmp
pop de
jr z, .returnPC
; Not PC either, try symbol
call symSelect
call symFind
jr nz, .notfound
; Found! let's fetch value
push de
call symGetVal
; value in DE. We need it in IX
push de \ pop ix
pop de
cp a ; ensure Z
ret
.notfound:
; If not found, check if we're in first pass. If we are, it doesn't
; matter that we didn't find our symbol. Return success anyhow.
; Otherwise return error. Z is already unset, so in fact, this is the
; same as jumping to zasmIsFirstPass
jp zasmIsFirstPass
.returnPC:
push hl
call zasmGetPC
push hl \ pop ix
pop hl
ret
.sDollar:
.db '$', 0