
20 Aug
2011
20 Aug
'11
3:21 p.m.
Hi.
On 20 August 2011 19:36, Manfred Lotz
On Sat, 20 Aug 2011 14:08:43 -0400 David Place
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