"ashutosh dimri" <[EMAIL PROTECTED]> writes: > how to convert a hexadecimal into base 10 integer using haskell . I have > written a code but its not working for large values , please help
from http://pleac.sourceforge.net/pleac_haskell/numbers.html#AEN118 : -- "read" handles both octal and hexadecimal when prefixed with 0x or 0o -- here are versions adding the prefix and calling "read" hex s = read ("0x" ++ s) :: Integer oct s = read ("0o" ++ s) :: Integer -- hex "45" == 69 -- oct "45" == 37 -- hex "45foo" => Exception: Prelude.read: no parse -- calling explicitly readHex or readOct: hex = fst . head . Numeric.readHex oct = fst . head . Numeric.readOct _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
