Hi.

On 20 August 2011 19:36, Manfred Lotz <manfred.lotz@arcor.de> wrote:
On Sat, 20 Aug 2011 14:08:43 -0400
David Place <d@vidplace.com> wrote:

> sequence_ :: Monad m => [m a] -> m ()

Thanks. I found sequence but I have to learn to add an underscore to a
funtion in order to see if this could be it.

Actually in this case sequence and sequence_ are identical. You have IO for m, and () for a, so:

sequence  :: [IO ()] -> IO ()
sequence_ :: [IO ()] -> IO ()

Ozgur