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

[Daniel Peebles]

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 <nrolle at web.de> wrote:
> super nice.
> best solution for me so far.
> big thanks.
> regards
>
> 2009/5/6 Victor Nazarov <asviraspossible at gmail.com>:
>> On Tue, May 5, 2009 at 8:49 PM, 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)
>>
>> 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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