Re: [Haskell-cafe] Re: Re: monad subexpressions

david48

3 Aug 2007 3 Aug '07

3:48 p.m.

Sorry for the double post, I posted with the wrong email address and haskell-cafe rejected it. On 8/3/07, Neil Mitchell wrote:

...

...
Right. In effect, as a matter of fact, the notation

x <- a

would become equivalent to

let x = (<- a)

Hmm, interesting. Consider:

let x = 12 let x = (<- x)

Wouldn't that be forbidden ? I'd expect the x in ( <- x ) have to be of type m a. If you meant : x <- return 12 let x = ( <- x ) Then I imagine it would turn into x <- return 12 x >>= \tx -> let x = tx in .... Isn't that correct ?

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Chris Smith

3 Aug 3 Aug

3:55 p.m.

New subject: monad subexpressions

david48 wrote:

...

On 8/3/07, Neil Mitchell wrote:

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Hmm, interesting. Consider:

let x = 12 let x = (<- x)

Wouldn't that be forbidden ?

I'd expect the x in ( <- x ) have to be of type m a.

Yes, unless of course you did: instance (Monad m, Num n) => Num (m n) or some such nonsense. :)

...

If you meant :

x <- return 12 let x = ( <- x )

This would be equally wrong. Perhaps you meant: do let x = return 12 let x = (<- x) ... Then this would become: do let x = return 12 t1 <- x let x = t1 ... Which is, in turn: let x = return 12 in x >>= (\t1 -> let x = t1 in ...) -- Chris Smith

Antoine Latter

9:13 p.m.

New subject: monad subexpressions

On 8/3/07, Chris Smith wrote:

...

Yes, unless of course you did:

instance (Monad m, Num n) => Num (m n)

or some such nonsense. :)

I decided to take this as a dare - at first I thought it would be easy to declare (Monad m, Num n) => m n to be an instance of Num (just lift or return the operators as necessary), but I ran into trouble once I realized I needed two things I wasn't going to get: An instance of Eq (m n), and an instance of Show (m n) for all monads m. Eq would need a function of the form: (==) :: Monad m => m a -> m a -> Bool and Show would need a function of type m a -> String There's no way I'm getting a function of those types using return and join to operate on the monad. So, there went that idea. -Antoine

Twan van Laarhoven

9:41 p.m.

New subject: monad subexpressions

Antoine Latter wrote:

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On 8/3/07, Chris Smith wrote:

...
Yes, unless of course you did:

instance (Monad m, Num n) => Num (m n)

or some such nonsense. :)

I decided to take this as a dare - at first I thought it would be easy to declare (Monad m, Num n) => m n to be an instance of Num (just lift or return the operators as necessary), but I ran into trouble once I realized I needed two things I wasn't going to get:

An instance of Eq (m n), and an instance of Show (m n) for all monads m. Eq would need a function of the form:

(==) :: Monad m => m a -> m a -> Bool

and Show would need a function of type m a -> String

What about Eq1 and Show1 classes? In the same vein as Typeable1:

...

class Eq1 f where eq1 :: Eq a => f a -> f a -> Bool neq1 :: Eq a => f a -> f a -> Bool

...

class Show1 f where show1 :: Show a => f a -> String showsPrec1 :: Show a => Int -> f a -> ShowS

Now you can declare the Num instance:

...

instance (Monad m, Eq1 m, Show1 m, Num n) => Num (m n) where (+) = liftM2 (+) (-) = liftM2 (-) (*) = liftM2 (*) abs = liftM abs signum = liftM signum negate = ligtM negate fromInteger = return . fromInteger

And just to show that such instances can exist:

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instance Eq1 [] where eq1 = (==) neq1 = (/=)

...

instance Show1 [] where show1 = show showsPrec1 = showsPrec

Note: All of this is untested code. Twan

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Antoine Latter
Chris Smith
david48
Twan van Laarhoven