Re: [Haskell-cafe] calling a variable length parameter lambda expression

Keep in mind that using lists for your parameters means you lose static guarantees that you've passed the correct number of arguments to a function (so you could crash at runtime if you pass too few or too many parameters to a function). On Wed, May 6, 2009 at 11:41 AM, Nico Rolle wrote:

super nice. best solution for me so far. big thanks. regards

2009/5/6 Victor Nazarov :

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On Tue, May 5, 2009 at 8:49 PM, 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)

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

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