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

Ryan Ingram ryani.spam at gmail.com
Tue May 5 12:58:10 EDT 2009


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 <nrolle at 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)
>
> best regards
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