16 Apr
2013
16 Apr
'13
8:57 a.m.
* Christopher Howard
So, I'm doing something like this
foldl (>>=) someA list :: Monad m => m a
where list :: Monad m => [a -> m a], someA :: Monad m => m a
Is there a more concise way to write this? I don't think foldM is what I want -- or is it?
I don't think it can get any more concise. (No, foldM isn't what you want.) But you most probably should prefer the right fold in this case. (Unless you don't care about efficiency.) Something like someA >>= foldr (>=>) return list should do. Roman