[Haskell-cafe] IO is not a monad

[Yitzchak Gale]

Hi,

Lennart Augustsson wrote:
> Could you explain why would a class Seq not be sufficient?
> If there were a class Seq, I'd not want functions to be in
> that class.

Oh, I see. Well that is pretty much the same
as ignoring seq altogether. I am hoping to get
a better answer than that - where we can see
how strictness questions fit in with the theory
of monads.

Anyway - why do you not want seq for functions?
Are you making functions second-class citizens
in Haskell, so that function-valued expressions
can no longer be lazy? If not, then how do you
force strictness in a function-valued expression?

For example:

Prelude Data.List> let fs = [const 0, const 1]::[Int->Int]
Prelude Data.List> let nextf f g = fs !! ((f (0::Int) + g (0::Int))`mod`2)
Prelude Data.List> :t nextf
nextf :: (Int -> Int) -> (Int -> Int) -> Int -> Int
Prelude Data.List> foldl' nextf id (concat $ replicate 100000 fs) 5
0
Prelude Data.List> foldl nextf id (concat $ replicate 100000 fs) 5
*** Exception: stack overflow

Regards,
Yitz