isn't doLam just id with an Ord restriction there?
On Wed, May 6, 2009 at 5:55 AM, Victor Nazarov <asviraspossi...@gmail.com> wrote: > On Tue, May 5, 2009 at 8:49 PM, Nico Rolle <nro...@web.de> wrote: >> >> Hi everyone. >> >> I have a problem. >> A function is recieving a lambda expression like this: >> (\ x y -> x > y) >> or like this >> (\ x y z a -> (x > y) && (z < a) >> >> my problem is now i know i have a list filled with the parameters for >> the lambda expression. >> but how can i call that expression? >> [parameters] is my list of parameters for the lambda expression. >> lambda_ex is my lambda expression >> >> is there a function wich can do smth like that? >> >> lambda _ex (unfold_parameters parameters) > > Why not: > > lam1 = \[x, y] -> x > y > lam2 = \[x, y, z, a] -> (x > y) && (z < a) > > doLam :: Ord a => ([a] -> Bool) -> [a] -> Bool > doLam lam params = lam params > > So, this will work fine: > > doLam lam1 [1, 2] > doLam lam2 [1,2,3,4] > > -- > Victor Nazarov > > _______________________________________________ > 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
