[Haskell-beginners] Problems inferring instances

[Dave Bayer]

Here's a very indirect answer, more a hunch:

I'm not sure how, but what you're trying to do reminds me of  
Control.Applicative. Go take a look at the documentation and/or source  
code for that library, then follow the link to

	Applicative Programming with Effects
	Conor McBride and Ross Paterson
	http://www.soi.city.ac.uk/~ross/papers/Applicative.html

which is one of the most beautiful papers ever written on Haskell.  
Even if I'm sending you on a wild goose chase, you'll enjoy the paper.

I've had similar monumental struggles trying to push the type system  
past my understanding of how it works. I find that invariably, if I  
roll back one click on my ambitions, type "darcs revert", step outside  
for 30 seconds, then what I want to do works without incident on the  
next try. A good example of this is the "wrapper" class Sum in  
Data.Monoid. You'd think that one could just tell the type system that  
a Num is a Monoid, but the type system _really_ likes something to  
chew on, hence the wrapper. I spent way too long contemplating GHC  
error messages proposing the option -XLetGravityFailButDontBlameUs,  
before accepting that if there was a better way, it would be in the  
library code.

So the key to maintaining momentum as a Haskell beginner is to see the  
simplification, one-click compromise that makes your obstacle trivial.  
Here, if I were you I'd first write your code for practice with the  
left- (or right-?) most === a different operator. By analogy with  
Applicative, or with the . . . . $ pattern one sees everywhere when  
composing. Then maybe it will be clear how to write it the way you want.

On Jan 5, 2009, at 6:40 PM, dcmorse+haskell at gmail.com wrote:

> Learning from the example of "read" and also Real World Haskell, I
> come across the idea to overload my function's return types. Trying to
> think of an application for this, I've always wanted to write ==
> applications like in Icon, that is
>
> a === b === c means a == b && b == c.