Do statement under a where clause

Question

I'm trying to convert IO [String] to [String] with <- binding; however, I need to use a do block to do that under a where statement, but Haskell complains about the indentation all the time. Here is the code:

decompEventBlocks :: IO [String] -> IO [[String]]
decompEventBlocks words
 | words' /= [] = block : (decompEventBlocks . drop $ (length block) words')
 | otherwise = []
  where 
   do
    words' <- words
    let block = (takeWhile (/="END") words')

What is the reason for that ? And how can we use do block in a where statement ? Moreover, is there any chance that we can have some statements before the guards ?

Answer 1

Do notation is used to write expressions of the general form

ex :: Monad m => m t
let ex = do 
          {  x <- foo         -- foo        :: Monad m => m a,   x :: a
          ;  y <- bar x       -- bar  x     :: Monad m => m b,   y :: b
          ;  z <- baz x y     -- baz  x y   :: Monad m => m c,   z :: c
          ;  quux x y z       -- quux x y z :: Monad m => m t
          }

Notice all the ms are the same, and a, b, c, ... can be different, though the t in the last do sub-expression's type and the overall do expression's type is the same.

The do notation variables are said to be "bound" by the <- construct. They come into scope at their introduction (to the left of <-) and remain in scope for all the subsequent do sub-expressions.

One built-in monadic expression available for any monad is return :: Monad m => a -> m a. Thus x <- return v binds x to v, so that x will be available in the subsequent sub-expressions, and will have the value of v.

All the do variables are confined to that do block, and can not be used outside it. Each variable's scope is all the code in the same do block, below / after the variable's binding.

This also means that <-'s is a non-recursive binding, since the variable can't go on its right hand side as well as the left: it will be two different variables with the same name, in that case, and the variable on the right will have to have been established somewhere above that point.

There are a few general patterns here:

do { _ <- p ; _ <- q ; r }    ===   do { p ; q ; r }
do { x <- p ; return x }      ===   do { p }          ===   p
do { x <- return v ; foo x }  ===   do { foo v }      ===   foo v
do { p ; q ; r }              ===   do { p ; do { q ; r } }
                              ===   do { do { p ; q } ; r }
do { x <- p ;                 ===   do { x <- p ;
     y <- q x ;                          z <- do { y <- q x ;
     return (foo x y) }                            return (foo x y) } ;
                                         return z }

All the Monad m => m a expressions are just that, expressions, and so can in particular be an if - then - else expressions with both the consequent and the alternative branches being of the same monadic type (which is often confusing for beginners):

    do { x <- p ;
         y <- if (pred x) then (foo x) else (bar x) ;
         return (baz x y) }

update: One of the main points of the monad is its total separation of effects from the pure calculations. Once in a monad, you can't "get out". Monadic computations can use pure calculations, but not vice versa.

Answer 2

You cannot convert for IO String to a String.

What you can do, however, is bind the contents of IO String to a 'variable', but that will still result in the whole computation being embedded inside IO.

foo = do
   x <- baz -- here baz is the IO String
   let x' = doStuff x
   return x' -- embeds the String inside IO, as otherwise the computation would result in IO ()

To answer your question

foo x = baz x -- x here is your 'IO String'
  where
    baz x = do
      x' <- x
      return $ doStuff x'

Answer 3

Remember: do-blocks are syntactic sugar for monadic notation. This means the following applies:

do {a; b} = a >> b
dp {a <- b; c} = b >>= \a -> c

In other words, when using do-notation, you are actually producing values. This is why you can't just have a do-block in the top level of your where statement.

The way to solve this is to put the function into a do-block:

decompEventBlocks :: IO [String] -> IO [[String]]
decompEventBlocks words = do
    -- We unwrap the IO [String], but we keep it in the do-block,
    -- because it must be kept in a monadic context!
    words' <- words 
    let block = (takeWhile (/="END") words')
    -- This is equivalent to the guards you had in your function.
    -- NB return :: Monad m => a -> m a, to keep it in a monadic context!
    if not $ null words'
        then do 
          -- Since the recursion is monadic, we must bind it too:
          rest <- decompEventBlocks $ return $ drop (length block) words'
          return $ block : rest
        else return []

To learn about monads, do-notation, >>=, and >>, I highly reccommend reading the LYAH chapters to gain a good understanding before attempting more monadic code.

Answer 4

As a slightly different angle on AJFarmar's answer: The only things you can have in a where are declarations. do blocks aren't declarations, they are expressions. I.e. this is the same as if you tried to write where 2+5. If you want to declare block in a where, it must be

where
  // can have other declarations, even mutually recursive
  block = ...