fe15bafeca
During expression parsing, if a local label was parsed, it would select the local registry and keep that selection, making subsequent global labels register in the wrong place.
202 lines
3.8 KiB
NASM
202 lines
3.8 KiB
NASM
; Parse string at (HL) as a hexadecimal value and return value in IX under the
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; same conditions as parseLiteral.
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parseHexadecimal:
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call hasHexPrefix
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ret nz
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push hl
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push de
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ld d, 0
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inc hl ; get rid of "0x"
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inc hl
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call strlen
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cp 3
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jr c, .single
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cp 4
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jr c, .doubleShort ; 0x123
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cp 5
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jr c, .double ; 0x1234
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; too long, error
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jr .error
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.double:
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call parseHexPair
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jr c, .error
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inc hl ; now HL is on first char of next pair
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ld d, a
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jr .single
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.doubleShort:
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ld a, (hl)
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call parseHex
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jr c, .error
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inc hl ; now HL is on first char of next pair
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ld d, a
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.single:
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call parseHexPair
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jr c, .error
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ld e, a
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cp a ; ensure Z
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jr .end
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.error:
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call unsetZ
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.end:
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push de \ pop ix
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pop de
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pop hl
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ret
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; Sets Z if (HL) has a '0x' prefix.
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hasHexPrefix:
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ld a, (hl)
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cp '0'
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ret nz
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push hl
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inc hl
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ld a, (hl)
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cp 'x'
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pop hl
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ret
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; Parse string at (HL) as a binary value (0b010101) and return value in IX.
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; High IX byte is always clear.
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; Sets Z on success.
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parseBinaryLiteral:
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call hasBinPrefix
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ret nz
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push bc
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push hl
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push de
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ld d, 0
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inc hl ; get rid of "0b"
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inc hl
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call strlen
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or a
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jr z, .error ; empty, error
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cp 9
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jr nc, .error ; >= 9, too long
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; We have a string of 8 or less chars. What we'll do is that for each
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; char, we rotate left and set the LSB according to whether we have '0'
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; or '1'. Error out on anything else. C is our stored result.
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ld b, a ; we loop for "strlen" times
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ld c, 0 ; our stored result
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.loop:
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rlc c
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ld a, (hl)
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inc hl
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cp '0'
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jr z, .nobit ; no bit to set
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cp '1'
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jr nz, .error ; not 0 or 1
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; We have a bit to set
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inc c
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.nobit:
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djnz .loop
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ld e, c
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cp a ; ensure Z
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jr .end
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.error:
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call unsetZ
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.end:
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push de \ pop ix
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pop de
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pop hl
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pop bc
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ret
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; Sets Z if (HL) has a '0b' prefix.
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hasBinPrefix:
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ld a, (hl)
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cp '0'
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ret nz
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push hl
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inc hl
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ld a, (hl)
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cp 'b'
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pop hl
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ret
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; Parse string at (HL) and, if it is a char literal, sets Z and return
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; corresponding value in IX. High IX byte is always clear.
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;
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; A valid char literal starts with ', ends with ' and has one character in the
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; middle. No escape sequence are accepted, but ''' will return the apostrophe
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; character.
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parseCharLiteral:
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ld a, 0x27 ; apostrophe (') char
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cp (hl)
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ret nz
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push hl
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push de
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inc hl
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inc hl
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cp (hl)
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jr nz, .end ; not ending with an apostrophe
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inc hl
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ld a, (hl)
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or a ; cp 0
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jr nz, .end ; string has to end there
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; Valid char, good
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ld d, a ; A is zero, take advantage of that
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dec hl
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dec hl
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ld a, (hl)
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ld e, a
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cp a ; ensure Z
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.end:
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push de \ pop ix
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pop de
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pop hl
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ret
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; Parses the string at (HL) and returns the 16-bit value in IX. The string
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; can be a decimal literal (1234), a hexadecimal literal (0x1234) or a char
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; literal ('X').
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;
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; As soon as the number doesn't fit 16-bit any more, parsing stops and the
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; number is invalid. If the number is valid, Z is set, otherwise, unset.
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parseLiteral:
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call parseCharLiteral
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ret z
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call parseHexadecimal
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ret z
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call parseBinaryLiteral
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ret z
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jp parseDecimal
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; Parse string in (HL) and return its numerical value whether its a number
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; literal or a symbol. Returns value in IX.
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; Sets Z if number or symbol is valid, unset otherwise.
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parseNumberOrSymbol:
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call parseLiteral
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ret z
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; Not a number. Try PC
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push de ; --> lvl 1
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ld de, .sDollar
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call strcmp
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pop de ; <-- lvl 1
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jr z, .returnPC
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; Not PC either, try symbol
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push de ; --> lvl 1
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call symFindVal ; --> DE
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jr nz, .notfound
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; value in DE. We need it in IX
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push de \ pop ix
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pop de ; <-- lvl 1
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cp a ; ensure Z
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ret
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.notfound:
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pop de ; <-- lvl 1
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; If not found, check if we're in first pass. If we are, it doesn't
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; matter that we didn't find our symbol. Return success anyhow.
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; Otherwise return error. Z is already unset, so in fact, this is the
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; same as jumping to zasmIsFirstPass
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jp zasmIsFirstPass
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.returnPC:
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push hl
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call zasmGetPC
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push hl \ pop ix
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pop hl
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ret
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.sDollar:
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.db '$', 0
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