Re: [Haskell-cafe] expanding do notation / StateMonad example from tutorial all about monads

31 Dec 2005


      On 12/31/05, Marc Weber  wrote:
...
Hi.. I'm still struggling with monads ;-)
In the tutorial "All about monads" there is the function getOne used:
getOne:: (Random a) => (a,a) -> State StdGen a
getOne bound = do g <-get
                  (x,g')= return $ randomR bounds g
                  put g'
                  return x
In the haskell "Yet another Haskell tutorial" by Hal Daume it is said
that
    do  {fp <- e; es}
is translated to:
  let ok p = do  {es} ;
      ok _ = fail "..."
  in e >>= ok
Using this rule I was able to expand the first "<-" of getOne:
  getOne2 :: (Random a) => (a,a) -> State StdGen a
  getOne2 bounds = let ok g = do  (x,g') <- return $ randomR bounds g
                                  put g'
                                  return x
                       ok _ = fail "asd"
                   in get >>= ok
but I couldn't manage to also expand the second.
My attempt is:
  getOne3 :: (Random a) => (a,a) -> State StdGen a
  getOne3 bounds = let ok g = do  let ok2 (x,g') = do { put g'; return x } --
                                      ok2 _ = fail "2"
                                  in (return $ randomR bounds g >>= ok2
                       ok _ = fail "1"
                   in get >>= ok
But I get this compile error in at the line marked by a trailing --:
 "The last statement in a 'do' construct must be an expression"
Can you help me?
Omit the 'do' in 'do let ok2'.

--
Friendly,
  Lemmih