On Tue, May 5, 2009 at 10:05 AM, Nico Rolle wrote:

I dont't understand why u ask me that but the lambda expression will only get values not functions as a parameter. a = 1 b = 2 c = 3

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What if I call a (+)?

this won't happen in my use case. regards nico

Here's an example for a single fixed argument and return types: class LambdaApply lam where lamApply :: lam -> [Integer] -> Bool instance LambdaApply Bool where lamApply b [] = b lamApply _ _ = error "Too many arguments in argument list" instance LambdaApply r => LambdaApply (Integer -> r) where lamApply _ [] = error "Not enough arguments in argument list" lamApply f (x:xs) = lamApply (f x) xs The problem is generalizing this to any type, because of the base case: instance LambdaApply r where lamApply r [] = r lamApply _ _ = error "Too many arguments in argument list" overlaps with function types such as (Integer -> a). Since you say this is only for value types, if you're willing to encode each value type you care about, you can make this work in a somewhat more general way. But unless you are just using it for syntactic sugar (which seems unlikely for this use case), it's usually a signal that you are doing something wrong. -- ryan

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Von: Ryan Ingram Gesendet: 05.05.09 18:58:32 An: Nico Rolle CC: haskell-cafe@haskell.org Betreff: Re: [Haskell-cafe] calling a variable length parameter lambda  expression

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This is a Hard Problem in Haskell.

Let me ask you, how many parameters does this function take? a = (\x -> x)

How many parameters does this function take? b = (\f x -> f x)

How many parameters does this function take? c = (\f x y -> f x y)

What if I call a (+)?

  -- ryan

On Tue, May 5, 2009 at 9:49 AM, Nico Rolle wrote:

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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)

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