[Haskell-cafe] Re: Why is type 'b' forced to be type 'm a' and not possibly 'm a -> m a'
Vivian McPhail wrote: > > > class Forkable a where > > > fork :: String -> a -> a -> a > What I would like to be able to do is > differentiate between Forkable (m a -> > b) and Forkable ( -> b). Have you tried this combination of instances? instance Forkable (IO a) where ... -- and similarly for all the concrete -- monad types you will use fork with instance (Forkable a, Forkable b) => Forkable (a -> b) where ... Alternatively, since the fork function seems to be all about propagating a value (the String), would Control.Monad.Reader serve your purpose? Regards, Tom ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: Why is type 'b' forced to be type 'm a' and not possibly 'm a -> m a'
> Vivian McPhail wrote: > > > > class Forkable a where > > > > fork :: String -> a -> a -> a > > > What I would like to be able to do is > > differentiate between Forkable (m a -> > > b) and Forkable ( -> b). > > Have you tried this combination of instances? > > instance Forkable (IO a) where ... > -- and similarly for all the concrete > -- monad types you will use fork with > > instance (Forkable a, Forkable b) => > Forkable (a -> b) where ... > > Alternatively, since the fork function seems to be all about > propagating a value (the String), would Control.Monad.Reader > serve your purpose? The value that gets 'forked' is not actually the string, it is the result of a monadic computation. > > Regards, > Tom > I have tried: > instance Forkable (USM NRef) where... Which is my Monad The problem lies with > instance (Forkable a, Forkable b) => Forkable (a -> b) where > fork n a1 a2 a = fork n (a1 a) (a2 a) because I need the arg a to be evaluated before it gets passed to a1 and a2. This definition does the right thing when type 'a' is a function type, because it is not a value, but with something like 'm a -> (m a -> m a) -> m a' with Forkable (a -> b) the first arg gets evaluated twice, to be more concrete: With (and golden white) eggs I want the 'eggs' that is passed to 'golden' to be the same as the 'eggs' that is passed to 'white', i.e. -> and1 (golden2 eggs3) (white4 eggs3) and not -> and1 (golden2 eggs3) (white4 eggs5) So to do this I need to be able to recognise the case where the 'a' of (a -> b) is of type 'm a' so that I can evaluate it -- doesn't typecheck instance (Monad m, Forkable (m a), Forkable b) => Forkable (m a -> b) where fork n a1 a2 a = do a' <- a fork n (a1 $ return a') (a2 $ return a') Tom suggested that I might be able to use the Reader monad, but I'm not clear as to how I could do this. Cheers, Vivian -- No virus found in this outgoing message. Checked by AVG Free Edition. Version: 7.1.405 / Virus Database: 268.12.5/451 - Release Date: 19/09/2006 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: Why is type 'b' forced to be type 'm a' and not possibly 'm a -> m a'
Vivian McPhail wrote: ... > I need the arg a to be evaluated before it gets > passed to a1 and a2. This definition does the right thing > when type 'a' is a function type, because it is not a > value, but with something like 'm a -> (m a -> m a) -> m > a' with Forkable (a -> b) the first arg gets evaluated > twice, to be more concrete: > > With > > (and golden white) eggs > > I want the 'eggs' that is passed to 'golden' to be the > same as the 'eggs' that is passed to 'white', i.e. ... Could you reduce the need for Forkable instances, by rewriting '(and golden white) eggs' as 'and golden white =<< eggs'? Or would the same piece of code also have to handle combinations such as monadic 'and golden white' and non-monadic eggs? [BTW, thanks for giving me a pretext to use the phrase non-monadic eggs!] > Tom suggested that I might be able to use the Reader monad > , but I'm not clear as to how I could do this. Please ignore that. I only mentioned it in case the sole purpose of fork was to propagate a String, which you've now explained is not so. Regards, Tom ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: Why is type 'b' forced to be type 'm a' and not possibly 'm a -> m a' (Anatoly Zaretsky)
> Message: 6
> Date: Fri, 15 Sep 2006 18:36:35 +0300
> From: "Anatoly Zaretsky" <[EMAIL PROTECTED]>
> Subject: Re: [Haskell-cafe] Why is type 'b' forced to be type 'm a'
> and not possibly 'm a -> m a'
> To: "Vivian McPhail" <[EMAIL PROTECTED]>
> Cc: Haskell Cafe
> Message-ID:
> <[EMAIL PROTECTED]>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> On 9/15/06, Vivian McPhail <[EMAIL PROTECTED]> wrote:
> >
> > class Forkable a where
> > fork :: String -> a -> a -> a
> >
> > ...
> > {-
> > instance (Monad m, Forkable (m a), Forkable b) => Forkable (m a -> b)
where
> > fork n a1 a2 a = do
> > a' <- a
> > fork n (a1 $ return a') (a2 $ return a')
> > -}
> >
>
> Let's do manual type checking.
> First, fork :: Forkable a => String -> a -> a -> a
> So for Forkable (m a -> b)
> fork :: String -> (m a -> b) -> (m a -> b) -> m a -> b
> Then
> fork n a1 a2 a :: b
> But you define it as
> fork n a1 a2 a = do {...}
> So it should be of type Monad t => t a, not just any `b'.
>
> Instead, you can define
> instance (Monad m, Forkable (m b)) => Forkable (m a -> m b) where
> ...
>
Well, I can partially instantiate what I am trying to achieve by enumerating
cases. Note that when the first type is a monadic type the computation gets
evaluated and then forked, but when the first type is a function it merely
gets passed. My problem is that there are a very large number of possible
cases. So in the case Forkable (m a -> b), a number of instances of which I
can implement (e.g. Forkable (m a -> m a -> m a), Forkable ((m a -> m a) ->
m a), and Forkable (m a -> (m a -> m a) -> m a)), I don't see why 'b' should
necessarily typecheck to 't t1'.
What I would like to be able to do is differentiate between Forkable (m a ->
b) and Forkable ( -> b).
By the way, the following code typechecks and runs correctly, my problem is
that enumerating all possible types requires five factorial (120) different
instances, and to a lazy functional programmer who can 'see' the pattern it
seems that there must be a nicer way of achieving my end.
\begin{code}
instance (Monad m, Forkable (m a)) => Forkable (m a -> m a) where
fork n a1 a2 a = do
a' <- a
fork n (a1 $ return a') (a2 $ return a')
instance (Monad m, Forkable (m a)) => Forkable (m a -> m a -> m a) where
fork n a1 a2 a b = do
a' <- a
fork n (a1 $ return a') (a2 $ return a') b
instance (Monad m, Forkable (m a)) => Forkable ((m a -> m a) -> m a) where
fork n a1 a2 a = do
fork n (a1 a) (a2 a)
instance (Monad m, Forkable (m a)) => Forkable (m a -> (m a -> m a) -> m a)
where
fork n a1 a2 a b = do
a' <- a
fork n (a1 $ return a') (a2 $ return a') b
\end{code}
> Note that to compile it you also need -fallow-undecidable-instances
> and -fallow-overlapping-instances.
>
> --
> Tolik
>
Thanks for your help so far!
>
>
Cheers,
Vivian
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