Haskell processing [IO String]

Question

I've got the following function:

lines' :: [IO String]
lines' = getLine : lines'

I was hoping I could just use all the mighty list functions on this list, like filter etc. But my knowledge about the IO monad in haskell is improvable.

The list-of-io_stuff-concept convinced me after using Rx for C#.

Is there any way to do what I want in haskell ? Something like:

ten_lines :: [IO String]
ten_lines = take 10 lines'

proc_lines :: [IO String]
proc_lines = [ (l, length l) | l <- lines' ]

Thanks!

Answer 1

There are a whole bunch of normal list functions modified to work with monads in Control.Monad. Of particular interest to your question:

sequence :: Monad m => [m a] -> m [a]
mapM     :: Monad m => (a -> m b) -> [a] -> m [b]
filterM  :: Monad m => (a -> m Bool) -> [a] -> m [a]
foldM    :: Monad m => (a -> b -> m a) -> a -> [b] -> m a

(sequence and mapM are actually exported by the prelude and available by default.)

For example, let's take a look at the type of your take 10 lines' example:

Prelude Control.Monad> :t take 10 lines'
take 10 lines' :: [IO String]

We want to turn this [IO String] into a single IO [String] action. This is exactly what sequence does! We can tell this by the type signature. So:

sequence $ take 10 lines'

will do what you want.

Most of these functions also have a version ending in _, like sequence_. This has exactly the same effect as the normal function except it throws away the result, returning () instead. That is, sequence_ :: [m a] -> m (). This is a good choice whenever you don't actually care about the result for two reasons: it's more explicit about your intentions and the performance can be better.

So if you wanted to print 10 lines rather than get them, you would write something like this:

printLines = putStrLn "foo" : printLines

main = sequence_ $ take 10 printLines