What is the best way to convert String to ByteString

Question

What is the best way to convert a String to a ByteString in Haskell?

My gut reaction to the problem is

import qualified Data.ByteString as B import Data.Char (ord)  packStr = B.pack . map (fromIntegral . ord) 

But this doesn't seem satisfactory.

Answer 1

Here is my cheat sheet for Haskell String/Text/ByteString strict/lazy conversion assuming the desired encoding is UTF-8. The Data.Text.Encoding library has other encodings available.

Please make sure to not write (using OverloadedStrings):

lazyByteString :: BL.ByteString lazyByteString = "lazyByteString ä ß" -- BAD! 

This will get encoded in an unexpected way. Try

lazyByteString = BLU.fromString "lazyByteString ä ß" -- good 

instead.

String literals of type 'Text' work fine with regard to encoding.

Cheat sheet:

import Data.ByteString.Lazy as BL import Data.ByteString as BS import Data.Text as TS import Data.Text.Lazy as TL import Data.ByteString.Lazy.UTF8 as BLU -- from utf8-string import Data.ByteString.UTF8 as BSU      -- from utf8-string import Data.Text.Encoding as TSE import Data.Text.Lazy.Encoding as TLE  -- String <-> ByteString  BLU.toString   :: BL.ByteString -> String BLU.fromString :: String -> BL.ByteString BSU.toString   :: BS.ByteString -> String BSU.fromString :: String -> BS.ByteString  -- String <-> Text  TL.unpack :: TL.Text -> String TL.pack   :: String -> TL.Text TS.unpack :: TS.Text -> String TS.pack   :: String -> TS.Text  -- ByteString <-> Text  TLE.encodeUtf8 :: TL.Text -> BL.ByteString TLE.decodeUtf8 :: BL.ByteString -> TL.Text TSE.encodeUtf8 :: TS.Text -> BS.ByteString TSE.decodeUtf8 :: BS.ByteString -> TS.Text  -- Lazy <-> Strict  BL.fromStrict :: BS.ByteString -> BL.ByteString BL.toStrict   :: BL.ByteString -> BS.ByteString TL.fromStrict :: TS.Text -> TL.Text TL.toStrict   :: TL.Text -> TS.Text 

Please +1 Peaker's answer, because he correctly deals with encoding.