You know, I was wondering... if Monads are a subset of Functors, and Applicative is a subset of Functors, and Monads are a subset of Applicative... shouldn't it be possible to tack on the definitions that automatically derive Functor and Applicative? Isn't it the case that there is really only one way to define Applicative for a Monad anyway? And isn't there only one way to define fmap for a Monad that makes sense? On Tue, Apr 14, 2009 at 2:39 PM, wrote:

Send Beginners mailing list submissions to beginners@haskell.org

To subscribe or unsubscribe via the World Wide Web, visit http://www.haskell.org/mailman/listinfo/beginners or, via email, send a message with subject or body 'help' to beginners-request@haskell.org

You can reach the person managing the list at beginners-owner@haskell.org

When replying, please edit your Subject line so it is more specific than "Re: Contents of Beginners digest..."

Today's Topics:

1. Re: Re: Thinking about monads (Brent Yorgey) 2. Re: How would you run a monad within another monad? (Arthur Chan) 3. Re: How would you run a monad within another monad? (Arthur Chan) 4. Re: How would you run a monad within another monad? (Arthur Chan) 5. Re: How would you run a monad within another monad? (Jason Dusek)

----------------------------------------------------------------------

Message: 1 Date: Tue, 14 Apr 2009 16:42:08 -0400 From: Brent Yorgey Subject: Re: [Haskell-beginners] Re: Thinking about monads To: beginners@haskell.org Message-ID: <20090414204207.GA16671@seas.upenn.edu> Content-Type: text/plain; charset=us-ascii

On Mon, Apr 13, 2009 at 03:11:42PM -0700, Michael Mossey wrote:

...

I know Maybe is both a functor and a monad, and I was thinking: what's

the

...

difference? They are both wrappers on types. Then I realized, the difference is: they have different class definitions.

In fact, every monad should be a functor, but not every functor is a monad. Being a monad is a much stronger condition than being a functor.

...

class Functor f where fmap :: (a->b) -> f a -> f b

(Note how fussy this definition would be in C++. It would be a kind of template, but would probably look a lot more complex and would require lengthy declarations.)

class Monad m where a >>= b :: m a -> (a -> m b) -> m b

Don't forget return :: a -> m a ! That's the other key method in the Monad class. (There are also >> and 'fail' but those are unimportant---the first is just a specialization of >>=, and fail is a hack).

-Brent

------------------------------

Message: 2 Date: Tue, 14 Apr 2009 14:05:22 -0700 From: Arthur Chan Subject: Re: [Haskell-beginners] How would you run a monad within another monad? To: Jason Dusek Cc: beginners@haskell.org Message-ID: <74cabd9e0904141405i1fbadb85u8b87ffb05d61c493@mail.gmail.com> Content-Type: text/plain; charset="utf-8"

Here's my contrived example that threw the error.

If you go into ghci, and do a `:t (foo' "blah" myDoohickey)`, you will get the type signature "IO ()". Doing the same for myOtherDoohickey returns "IO True"

So you would think that you'd be able to uncomment the code that makes IO an instance of Toplevel. foo' is a function that allows IO to run monadic values of type Doohickey. But it doesn't work.

---

import IO import Control.Monad.Reader

class (Monad n) => Doohickey n where putRecord :: String -> n ()

class (Monad m) => Toplevel m where foo :: (Doohickey n) => FilePath -> n a -> m a

newtype IOToplevelT a = IOToplevelT { runIOToplevelT :: ReaderT Handle IO a } deriving (Monad, MonadReader Handle, MonadIO)

instance Doohickey IOToplevelT where putRecord = liftIO . putStrLn

foo' s k = do f <- liftIO $ openFile s AppendMode runReaderT (runIOToplevelT k) f

--instance Toplevel IO where -- foo = foo'

myDoohickey = do putRecord "foo" putRecord "bar"

myOtherDoohickey = do putRecord "hello" putRecord "world" return True

On Mon, Apr 13, 2009 at 7:55 PM, Jason Dusek wrote:

...

Copypasting and loading your code doesn't throw an error. Please, pastebin an example that demonstrates the error.

-- Jason Dusek