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

Nico Rolle writes:

A function is recieving a lambda expression like this: (\ x y -> x > y) or like this (\ x y z a -> (x > y) && (z < a)

And the type of that function is..?

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

Would this work? data LambdaExpr a = L0 a | L1 (a -> a) | L2 (a -> a -> a) | ... apply :: LambdaExpr a -> [a] -> a apply (L0 x) _ = x apply (L1 f) (x:_) = f x apply (L2 f) (x1:x2:_) = f x1 s2 : -k -- If I haven't seen further, it is by standing in the footprints of giants