M'luds As counsel for the defence, I would advise my learned friend to consult the source code. On 6 Oct 2010, at 21:02, Conor McBride wrote:
Exhibit A, from http://haskell.org/ghc/docs/6.12.2/html/libraries/random-1.0.0.2/System-Rand...
"In addition, read may be used to map an arbitrary string (not necessarily one produced by show) onto a value of type StdGen. In general, the read instance of StdGen has the following properties:
* It guarantees to succeed on any string."
instance Read StdGen where readsPrec _p = \ r -> case try_read r of r'@[_] -> r' _ -> [stdFromString r] -- because it shouldn't ever fail. where try_read r = do (s1, r1) <- readDec (dropWhile isSpace r) (s2, r2) <- readDec (dropWhile isSpace r1) return (StdGen s1 s2, r2) {- If we cannot unravel the StdGen from a string, create one based on the string given. -} stdFromString :: String -> (StdGen, String) stdFromString s = (mkStdGen num, rest) where (cs, rest) = splitAt 6 s num = foldl (\a x -> x + 3 * a) 1 (map ord cs) This is clearly guaranteed to succeed on any String, provided you don't use big complicated words. Prelude System.Random Text.Read Numeric> read "jam" :: StdGen 1382 1 Prelude System.Random Text.Read Numeric> read "marmalade" :: StdGen *** Exception: Prelude.read: no parse At least this explains how to work around the problem. take 6 Conor