[Haskell-beginners] Sequential IO processing

[Sergey V. Mikhanov]

I tried this earlier as well:

sequenceIO [] = return []
sequenceIO (x : xs) = do result <- x
                         resultTail <- sequenceIO xs
                         return result : resultTail

This results in:

Couldn't match expected type `IO [a]' against inferred type `[m a]'
In the expression: return result : resultTail
In the expression:
    do result <- x
      resultTail <- sequenceIO xs
         return result : resultTail
In the definition of `sequenceIO':
    sequenceIO (x : xs)
                 = do result <- x
                      resultTail <- sequenceIO xs
                        return result : resultTail

>> sequenceIO [] = return []
>> sequenceIO (x : xs) = do result <- x
>>                         return result : sequenceIO xs
>
> The problem is indeed here. The type of 'sequenceIO xs' is IO [a], but the
> type of result is 'a'. You can't cons an 'a' onto an 'IO [a]'. Thus, what
> you need is something like this:
> sequenceIO [] = return []
> sequenceIO (x : xs) = do result <- x
>                                      xs' <- sequenceIO xs -- to take the
> list out of the IO Monad
>                                      return result : xs'