[EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> I've been trying to decipher some formulae I found online which
> convert hex to decimal numbers, with a view to re-writing it in
> lingo, but my head is hurting already.
Hi Chris,
You'll find below the fastest 32-bit conversion handlers that I am know of.
The hex() handler is much slower with negative numbers than with positive,
and it returns "" for 0, so there is still room for improvement there.
Cheers,
James
----------------------------------------------------------------------------
on hexToDecimal(hex) ------------------------------------------------
-- Returns the integer value corresponding to any 8-digit
-- hexadecimal string. Functions correctly with negative numbers
-- (> 7FFFFFFF). Expects the <hex> string to contain 8 characters
-- or less, in the range "0123456789ABCDEF"... Or you can uncomment
-- the error checking lines below, in which case a self-explanatory
-- error symbol will be returned instead of an (invalid) integer.
-- Inspired by code by Andrew White.
-------------------------------------------------------------------
-- -- Error checking
-- hex = cleanString(hex)
-- if not stringP(hex) then
-- return hex -- #integerOverflow | #invalidHexString
-- end if
-- -- End of error checking
counter = hex.char.count
decimal = 0
repeat with i = 1 to counter
decimal = (decimal*16) + (offset(hex.char[i], "123456789ABCDEF"))
end repeat
return decimal
end hexToDecimal
on cleanString(hex)
-- Strips the string <hex> of all spaces, initial "#" or "0"
-- characters, and checks that the figures are all in the range
-- "0123456789ABCDEF". Returns the cleaned <hex> string, or a
-- self-explanatory error symbol.
if not stringP(hex) then
return #invalidString
end if
repeat while TRUE
set spacePlace = offset (" ", hex)
if spacePlace then
delete char spacePlace of hex
else
exit repeat
end if
end repeat
repeat while TRUE
case hex.char[1] of
"0", "#":
delete hex.char[1]
otherwise
exit repeat
end case
end repeat
i = hex.char.count
repeat while i
if not offset(hex.char[i], "0123456789ABCDEF") then
return #invalidHexString
end if
i = i - 1
end repeat
if i > 8 then
return #integerOverflow
else
return hex
end if
end cleanString
on hex(aDecimal)
-- Returns the hexadecimal string corresponding to any 4-byte
-- integer, including negative numbers. If you uncomment the
-- first three lines of code and <aDecimal> is not an
-- integer a self-explanatory error symbol is returned.
-- NOTE: returns "" for 0
-------------------------------------------------------------------
-- -- Error checking
-- if not integerP(aDecimal) then
-- return #invalidInteger
-- end if
-- -- End of error checking
if aDecimal > 0 then
-- Use recursion if the number is positive
hexDigit = aDecimal mod 16
if hexDigit < 10 then
return hex(aDecimal / 16) & hexDigit
else
return hex(aDecimal / 16) & ("ABCDEF").char[hexDigit - 9]
end if
else if aDecimal then -- The number is negative
-- Convert the number's positive counterpart, ...
inverse = hex(aDecimal + the maxInteger + 1)
-- ... pad with zeros if necessary, ...
i = 8 - inverse.char.count
if i then
i = i - 1
repeat while i
put "0" before inverse
i = i - 1
end repeat
-- ... then convert back to a negative number
put 8 before char 1 of inverse
else
char1 = ("9ABCDEF").char[value(inverse.char[1])]
put char1 into char 1 of inverse
end if
return inverse
else -- aDecimal is 0
return ""
end if
end hex
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