[Haskell-cafe] couple of questions on monads

Daryoush Mehrtash dmehrtash at gmail.com
Mon Oct 13 17:31:23 EDT 2008

Is there a write up on what makes an implementation lazy vs strict?

I  like to better understand the trade off between the two and use cases
where one is better than the other.

I noticed that some functions in the lazy implementation uses "~" .    For

evalStateT :: (Monad m) => StateT s m a -> s -> m a
evalStateT m s = do
    ~(a, _) <- runStateT m s
    return a

What does ~ do?



On Mon, Oct 13, 2008 at 1:34 PM, Jonathan Cast <jonathanccast at fastmail.fm>wrote:

> On Mon, 2008-10-13 at 13:37 -0700, Daryoush Mehrtash wrote:
> > Question 1: Why are there lazy and strict modules of some monads?
> > (e.g. Control.Monad.State)
> Because both are useful, for different purposes.  (For the same reason
> that it's helpful, in general, to have both eager and lazy evaluation in
> the same language --- sometimes one version is more efficient, sometimes
> the other one is).
> >
> > Question 2:  If I define a new monad (say XYZ), does it have to be as
> > Control.Monad.XYZ  module?
> No.  Haskell has neither a requirement nor a convention that monads go
> in Control.Monad.  Control.Monad.* is simply the home of the MTL
> library, which contains a number of exceptionally useful monads; new
> monads that aren't as exceptionally general-purpose as MTL probably
> shouldn't go there, to reduce clutter.
> jcc
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