; *** Requirements ***
; lib/util
; *** Code ***
; Parse the hex char at A and extract it's 0-15 numerical value. Put the result
; in A.
;
; On success, the carry flag is reset. On error, it is set.
parseHex:
; First, let's see if we have an easy 0-9 case
add a, 0xc6 ; maps '0'-'9' onto 0xf6-0xff
sub 0xf6 ; maps to 0-9 and carries if not a digit
ret nc
and 0xdf ; converts lowercase to uppercase
add a, 0xe9 ; map 0x11-x017 onto 0xFA - 0xFF
sub 0xfa ; map onto 0-6
ret c
; we have an A-F digit
add a, 10 ; C is clear, map back to 0xA-0xF
ret
; Parse string at (HL) as a decimal value and return value in DE.
; Reads as many digits as it can and stop when:
; 1 - A non-digit character is read
; 2 - The number overflows from 16-bit
; HL is advanced to the character following the last successfully read char.
; Error conditions are:
; 1 - There wasn't at least one character that could be read.
; 2 - Overflow.
; Sets Z on success, unset on error.
parseDecimal:
; First char is special: it has to succeed.
ld a, (hl)
; 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.
add a, 0xff-'9' ; maps '0'-'9' onto 0xf6-0xff
sub 0xff-9 ; maps to 0-9 and carries if not a digit
ret c ; Error. If it's C, it's also going to be NZ
; During this routine, we switch between HL and its shadow. On one side,
; we have HL the string pointer, and on the other side, we have HL the
; numerical result. We also use EXX to preserve BC, saving us a push.
parseDecimalSkip: ; enter here to skip parsing the first digit
exx ; HL as a result
ld h, 0
ld l, a ; load first digit in without multiplying
.loop:
exx ; HL as a string pointer
inc hl
ld a, (hl)
exx ; HL as a numerical result
; same as other above
add a, 0xff-'9'
sub 0xff-9
jr c, .end
ld b, a ; we can now use a for overflow checking
add hl, hl ; x2
sbc a, a ; a=0 if no overflow, a=0xFF otherwise
ld d, h
ld e, l ; de is x2
add hl, hl ; x4
rla
add hl, hl ; x8
rla
add hl, de ; x10
rla
ld d, a ; a is zero unless there's an overflow
ld e, b
add hl, de
adc a, a ; same as rla except affects Z
; Did we oveflow?
jr z, .loop ; No? continue
; error, NZ already set
exx ; HL is now string pointer, restore BC
; HL points to the char following the last success.
ret
.end:
push hl ; --> lvl 1, result
exx ; HL as a string pointer, restore BC
pop de ; <-- lvl 1, result
cp a ; ensure Z
ret
; Call parseDecimal and then check that HL points to a whitespace or a null.
parseDecimalC:
call parseDecimal
ret nz
ld a, (hl)
or a
ret z ; null? we're happy
jp isWS
; Parse string at (HL) as a hexadecimal value without the "0x" prefix and
; return value in DE.
; HL is advanced to the character following the last successfully read char.
; Sets Z on success.
parseHexadecimal:
ld a, (hl)
call parseHex ; before "ret c" is "sub 0xfa" in parseHex
; so carry implies not zero
ret c ; we need at least one char
push bc
ld de, 0
ld b, d
ld c, d
; The idea here is that the 4 hex digits of the result can be represented "bdce",
; where each register holds a single digit. Then the result is simply
; e = (c << 4) | e, d = (b << 4) | d
; However, the actual string may be of any length, so when loading in the most
; significant digit, we don't know which digit of the result it actually represents
; To solve this, after a digit is loaded into a (and is checked for validity),
; all digits are moved along, with e taking the latest digit.
.loop:
dec b
inc b ; b should be 0, else we've overflowed
jr nz, .end ; Z already unset if overflow
ld b, d
ld d, c
ld c, e
ld e, a
inc hl
ld a, (hl)
call parseHex
jr nc, .loop
ld a, b
add a, a \ add a, a \ add a, a \ add a, a
or d
ld d, a
ld a, c
add a, a \ add a, a \ add a, a \ add a, a
or e
ld e, a
xor a ; ensure z
.end:
pop bc
ret
; Parse string at (HL) as a binary value (010101) without the "0b" prefix and
; return value in E. D is always zero.
; HL is advanced to the character following the last successfully read char.
; Sets Z on success.
parseBinaryLiteral:
ld de, 0
.loop:
ld a, (hl)
add a, 0xff-'1'
sub 0xff-1
jr c, .end
rlc e ; sets carry if overflow, and affects Z
ret c ; Z unset if carry set, since bit 0 of e must be set
add a, e
ld e, a
inc hl
jr .loop
.end:
; HL is properly set
xor a ; ensure Z
ret
; Parses the string at (HL) and returns the 16-bit value in DE. The string
; can be a decimal literal (1234), a hexadecimal literal (0x1234) or a char
; literal ('X').
; HL is advanced to the character following the last successfully read char.
;
; 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:
ld de, 0 ; pre-fill
ld a, (hl)
cp 0x27 ; apostrophe
jr z, .char
; inline parseDecimalDigit
add a, 0xc6 ; maps '0'-'9' onto 0xf6-0xff
sub 0xf6 ; maps to 0-9 and carries if not a digit
ret c
; a already parsed so skip first few instructions of parseDecimal
jp nz, parseDecimalSkip
; maybe hex, maybe binary
inc hl
ld a, (hl)
inc hl ; already place it for hex or bin
cp 'x'
jr z, parseHexadecimal
cp 'b'
jr z, parseBinaryLiteral
; nope, just a regular decimal
dec hl \ dec hl
jp parseDecimal
; Parse string at (HL) and, if it is a char literal, sets Z and return
; corresponding value in E. D is always zero.
; HL is advanced to the character following the last successfully read char.
;
; 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.
.char:
inc hl
ld e, (hl) ; our result
inc hl
cp (hl)
; advance HL and return if good char
inc hl
ret z
; Z unset and there's an error
; In all error conditions, HL is advanced by 3. Rewind.
dec hl \ dec hl \ dec hl
; NZ already set
ret
; Returns whether A is a literal prefix, that is, a digit or an apostrophe.
isLiteralPrefix:
cp 0x27 ; apostrophe
ret z
; continue to isDigit
; Returns whether A is a digit
isDigit:
cp '0' ; carry implies not zero for cp
ret c
cp '9' ; zero unset for a > '9', but set for a='9'
ret nc
cp a ; ensure Z
ret