So...there's just no good way to avoid the duplication? On Tue, Jan 20, 2009 at 11:10 PM, wren ng thornton wrote:

Andrew Wagner wrote:

...

Strange little bit of code: http://moonpatio.com:8080/fastcgi/hpaste.fcgi/view?id=829#a829

If I do any of the following, all of which seem natural to me, it fails to typecheck:

1. move f out of the 'where' clause (with or without a type signature) 2. put the same type signature on f as is on (/\) 3. replace f with (/\) completely

What's going on here?

...

:t (nub .) . (++) (nub .) . (++) :: (Eq a) => [a] -> [a] -> [a]

...

:t foldr (map . (nub .) . (++)) foldr (map . (nub .) . (++)) :: (Eq a) => [[a]] -> [[a]] -> [[a]]

The type you give to (/\) is more restrictive than the type of the expression, and f uses the generality of the expression.

-- Live well, ~wren _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

Or if the specificity of (/\) is important to you, you could define f at the global scope using that type, not export it, then declare (/\) as equal to f but with a more restrictive type. On Wed, Jan 21, 2009 at 5:45 AM, Alexander Dunlap wrote:

Instead of declaring (/\) :: Eq a => Sentence a -> Sentence a -> Sentence a, you could say (/\) :: Eq a -> [a] -> [a] -> [a]. Then it would work in both places. ([a] -> [a] -> [a] is a more general type than [[Term a]] -> [[Term a]] -> [[Term a]], so functions with the former type can be used in place of functions of the latter type but not vice versa.)

Alex

2009/1/20 Andrew Wagner :

...

So...there's just no good way to avoid the duplication?

On Tue, Jan 20, 2009 at 11:10 PM, wren ng thornton wrote:

...

Andrew Wagner wrote:

...

Strange little bit of code: http://moonpatio.com:8080/fastcgi/hpaste.fcgi/view?id=829#a829

If I do any of the following, all of which seem natural to me, it fails to typecheck:

1. move f out of the 'where' clause (with or without a type signature) 2. put the same type signature on f as is on (/\) 3. replace f with (/\) completely

What's going on here?

...

:t (nub .) . (++) (nub .) . (++) :: (Eq a) => [a] -> [a] -> [a]

...

:t foldr (map . (nub .) . (++)) foldr (map . (nub .) . (++)) :: (Eq a) => [[a]] -> [[a]] -> [[a]]

The type you give to (/\) is more restrictive than the type of the expression, and f uses the generality of the expression.

-- Live well, ~wren _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe