calling a variable length parameter lambda expression

Nico Rolle

5 May 2009 5 May '09

12:49 p.m.

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

Show replies by date

Miguel Mitrofanov

5 May 5 May

12:56 p.m.

New subject: calling a variable length parameter lambda expression

Short answer: that's impossible. Well, with some oleging it should be possible, but the very fact that you're trying to do something like this indicates that you're doing something wrong. Where did this list of parameters came from? May be, you can apply your function to them one at a time, as they arrive? On 5 May 2009, at 20:49, Nico Rolle 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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

Ryan Ingram

12:58 p.m.

New subject: calling a variable length parameter lambda expression

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:

...

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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

Ketil Malde

2:53 p.m.

New subject: 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

Richard O'Keefe

10:20 p.m.

New subject: calling a variable length parameter lambda expression

On 6 May 2009, at 4:49 am, Nico Rolle 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)

Your first function has type Ord a => a -> a -> Bool so your list of parameters must have type Ord a => [a] and length 2. Your second function -- and why do you have the excess parentheses? they make it harder to read -- has type Ord a => a -> a -> a -> Bool so your list of parameters must have type Ord a => [a] and length 3.

...

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)

but in both cases lambda_ex wants a single Ord, so unfold_parameters parameters would have to return just that Ord. You _could_ fake something up for this case using type classes, but the big question is WHY do you want to do this? It makes sense for Lisp or Scheme, but not very much sense for a typed language. Why are you building a list [x,y] or [x,y,z] rather than a function (\f -> f x y) or (\f -> f x y z) so that you can do parameters lambda_ex?

Victor Nazarov

6 May 6 May

5:55 a.m.

New subject: calling a variable length parameter lambda expression

On Tue, May 5, 2009 at 8:49 PM, Nico Rolle 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

Daniel Peebles

11:40 a.m.

New subject: calling a variable length parameter lambda expression

isn't doLam just id with an Ord restriction there? On Wed, May 6, 2009 at 5:55 AM, Victor Nazarov wrote:

...

On Tue, May 5, 2009 at 8:49 PM, Nico Rolle 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

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

Nico Rolle

11:41 a.m.

New subject: calling a variable length parameter lambda expression

super nice. best solution for me so far. big thanks. regards 2009/5/6 Victor Nazarov :

...

On Tue, May 5, 2009 at 8:49 PM, Nico Rolle 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

Daniel Peebles

11:44 a.m.

New subject: 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 :

...
On Tue, May 5, 2009 at 8:49 PM, Nico Rolle 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

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

Victor Nazarov

7 May 7 May

4:40 a.m.

New subject: calling a variable length parameter lambda expression

On Wed, May 6, 2009 at 7:44 PM, Daniel Peebles wrote:

...

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).

Yes, program will crash with error "Pattern match failure" or smth like that. And yes, doLam is just an id with restricted type. My solution is close to what seems the most common way of passing arguments in Scheme, and the payoff is Scheme's dynamic typing... This solution is not the Haskell way and should be avoid. -- Victor Nazarov

...

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 :

...
On Tue, May 5, 2009 at 8:49 PM, Nico Rolle 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

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

5857

Age (days ago)

5859

Last active (days ago)

List overview

Download

9 comments

7 participants

participants (7)

Daniel Peebles
Ketil Malde
Miguel Mitrofanov
Nico Rolle
Richard O'Keefe
Ryan Ingram
Victor Nazarov