Re: [Haskell-cafe] foild function for expressions
I believe the exercise is about understanding folds. There are two references that are related to the exercise: A tutorial on the universality and expressiveness of fold, by Graham Hutton. Dealing with large bananas, by Ralf Lammel, etc. The last paper motivates well the need to gather all the function parameters to the fold (ie, the algebra) in a separate record structure. I don't think being told what to write will help you understand what is going on, which is simpler than it seems. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foild function for expressions
Ryan Ingram wrote: On 12/3/07, Kalman Noel [EMAIL PROTECTED] wrote: You're confusing sum and product types. I'm not so sure; it looks like they already have that type (Exp) and wants to use AlgExp to hold the folding functions used. Ah, I didn't catch that on the first read. I suppose Carlo should then tell us what Exp exactly looks like; and it would be nice, too, to explain to me what the function in question is supposed to achieve then. He doesn't seem to want to reduce the expression, after all. Kalman -- Free pop3 email with a spam filter. http://www.bluebottle.com/tag/5 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foild function for expressions
Brent Yorgey wrote: One comment: it looks like (add exp1 exp2), (and exp1 exp2) and so on above are not correct. The second argument of foldExp is a value of type Exp, so you are pattern-matching on the constructors of Exp, and constructors are always uppercase. Perhaps Exp has constructors named Add, And, and so on? Then you would want to do something like Yepp, it's my fault I wasn't too precise in my explanation. There's a declaration of the data type EXP, which is not difficult to gess with the things I've said: data Exp = LitI Int |LitB Bool |Add Exp Exp |And Exp Exp |If Exp Exp Exp I'm going to read carefully all of your answer now (thanks to all!!!), and then I'll tell you. ;) -- View this message in context: http://www.nabble.com/foild-function-for-expressions-tf4932877.html#a14155122 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foild function for expressions
On Dec 3, 2007 12:18 PM, Carlo Vivari [EMAIL PROTECTED] wrote:
Hi! I'm a begginer in haskell and I have a problem with an exercise, I
expect
someone could help me:
In one hand I have a declaration of an algebra data, like this:
data AlgExp a = AlgExp
{ litI :: Int - a,
litB :: Bool - a,
add :: a - a - a,
and :: a - a - a,
ifte :: a - a - a - a}
(being ifte an 'ifthenelse' expresion...)
What I want to do is to write a fold function for expressions, something
like this:
foldExp :: AlgExp a - Exp - a
foldExp alg (LitI i) = litI alg i
foldExp alg (LitB i) = litB alg i
foldExp alg (add exp1 exp2) = ¿¿¿???
foldExp alg (and exp1 exp2) = ¿¿¿???
foldExp alg (ifte exp1 exp2 exp3) = ¿¿¿???
You'll want something like this:
foldExp alg (Add e1 e2) = add alg (foldExp alg e1) (foldExp alg e2)
--
Dave Menendez [EMAIL PROTECTED]
http://www.eyrie.org/~zednenem/
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Re: [Haskell-cafe] foild function for expressions
Yes,as I said before to other of you the exp data type was also declared in the exercise (my fault not to say it), and look as this: data Exp = LitI Int |LitB Bool |Add Exp Exp |And Exp Exp |If Exp Exp Exp -- View this message in context: http://www.nabble.com/foild-function-for-expressions-tf4932877.html#a14158061 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foild function for expressions
Carlo Vivari wrote:
data AlgExp a = AlgExp
{ litI :: Int - a,
litB :: Bool - a,
add :: a - a - a,
and :: a - a - a,
ifte :: a - a - a - a}
You're confusing sum and product types. That is, you're using a product type,
but you probably need a sum type, like this:
data Exp1 = LitI Int
| LitB Bool
| Add Exp1 Exp1
| And Exp1 Exp1
| IfThenElse Exp1 Exp1 Exp1
But in this case, using GADTs (beware: not Haskell 98, but a very popular
extension) makes for a more elegant solution. Note the strong types, disallowing
e. g. the addition of a number to a boolean value:
data Exp2 a where
LitI:: Int - Exp2 Int
LitB:: Bool - Exp2 Bool
Add :: Exp2 Int - Exp2 Int - Exp2 Int
And :: Exp2 Bool - Exp2 Bool - Exp2 Bool
IfThenElse :: Exp2 Bool - Exp2 a - Exp2 a - Exp2 a
Kalman
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Re: [Haskell-cafe] foild function for expressions
foldExp :: AlgExp a - Exp - a foldExp alg (LitI i) = litI alg i foldExp alg (LitB i) = litB alg i foldExp alg (add exp1 exp2) = ¿¿¿??? foldExp alg (and exp1 exp2) = ¿¿¿??? foldExp alg (ifte exp1 exp2 exp3) = ¿¿¿??? One comment: it looks like (add exp1 exp2), (and exp1 exp2) and so on above are not correct. The second argument of foldExp is a value of type Exp, so you are pattern-matching on the constructors of Exp, and constructors are always uppercase. Perhaps Exp has constructors named Add, And, and so on? Then you would want to do something like foldExp alg (Add exp1 exp2) = ??? and so on. For the ??? part, you want to pull out an appropriate function from your alg record and apply it to exp1 and exp2. -Brent ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foild function for expressions
On 12/3/07, Kalman Noel [EMAIL PROTECTED] wrote: You're confusing sum and product types. That is, you're using a product type, but you probably need a sum type, like this: I'm not so sure; it looks like they already have that type (Exp) and wants to use AlgExp to hold the folding functions used. Carlo, I think you're on the right track. Think of it this way: you have some Exps and you want to get some things of type a to pass to the functions in alg. How could you get those things with what you have so far? -- ryan ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] foild function for expressions
On Mon, Dec 03, 2007 at 09:27:45PM -0600, Derek Elkins wrote:
On Mon, 2007-12-03 at 19:13 -0800, Stefan O'Rear wrote:
On Mon, Dec 03, 2007 at 09:18:18AM -0800, Carlo Vivari wrote:
Hi! I'm a begginer in haskell and I have a problem with an exercise, I
expect
someone could help me:
In one hand I have a declaration of an algebra data, like this:
data AlgExp a = AlgExp
{ litI :: Int - a,
litB :: Bool - a,
add :: a - a - a,
and :: a - a - a,
ifte :: a - a - a - a}
(being ifte an 'ifthenelse' expresion...)
What I want to do is to write a fold function for expressions, something
like this:
foldExp :: AlgExp a - Exp - a
foldExp alg (LitI i) = litI alg i
foldExp alg (LitB i) = litB alg i
foldExp alg (add exp1 exp2) = ¿¿¿???
foldExp alg (and exp1 exp2) = ¿¿¿???
foldExp alg (ifte exp1 exp2 exp3) = ¿¿¿???
..ETC
the fact is that I have no idea of what to do with the other expresions
(add, and, and ifte)... I really don' t understand how to do this... It's
clear that a fold function should colapse in one valour, but how can I
espress it in the terms of the exercise?
For further information about the problem after this, it's suposed that I
have to rewrite some functions for expresions but in terms of foldexp (the
one I should write before)
The problem is that AlgExp defines an arbitrary algebra, but in order to
fold you need a universal algebra. So it makes the most sense to add
foldExp to the reccord.
This is an unusually poor post for you. Presuming you mean initial or
free or term for universal then it does (presumably) have it with
Exp. Assuming Exp is the expected thing (an AST) it is (combined with
the constructors) the initial algebra. foldExp should quite definitely
be the type it is and not be part of the record and the its
implementation is so far correct (modulo syntactical errors that are
potentially indicative of a deeper confusion).
Oh right, I misread the foldExp sketch. I thought he was trying to go
*from* a member of his algebra *to* Exp. Sorry.
Stefan
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Re: [Haskell-cafe] foild function for expressions
On Mon, Dec 03, 2007 at 09:18:18AM -0800, Carlo Vivari wrote:
Hi! I'm a begginer in haskell and I have a problem with an exercise, I expect
someone could help me:
In one hand I have a declaration of an algebra data, like this:
data AlgExp a = AlgExp
{ litI :: Int - a,
litB :: Bool - a,
add :: a - a - a,
and :: a - a - a,
ifte :: a - a - a - a}
(being ifte an 'ifthenelse' expresion...)
What I want to do is to write a fold function for expressions, something
like this:
foldExp :: AlgExp a - Exp - a
foldExp alg (LitI i) = litI alg i
foldExp alg (LitB i) = litB alg i
foldExp alg (add exp1 exp2) = ¿¿¿???
foldExp alg (and exp1 exp2) = ¿¿¿???
foldExp alg (ifte exp1 exp2 exp3) = ¿¿¿???
..ETC
the fact is that I have no idea of what to do with the other expresions
(add, and, and ifte)... I really don' t understand how to do this... It's
clear that a fold function should colapse in one valour, but how can I
espress it in the terms of the exercise?
For further information about the problem after this, it's suposed that I
have to rewrite some functions for expresions but in terms of foldexp (the
one I should write before)
The problem is that AlgExp defines an arbitrary algebra, but in order to
fold you need a universal algebra. So it makes the most sense to add
foldExp to the reccord.
Stefan
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Re: [Haskell-cafe] foild function for expressions
On Mon, 2007-12-03 at 19:13 -0800, Stefan O'Rear wrote:
On Mon, Dec 03, 2007 at 09:18:18AM -0800, Carlo Vivari wrote:
Hi! I'm a begginer in haskell and I have a problem with an exercise, I
expect
someone could help me:
In one hand I have a declaration of an algebra data, like this:
data AlgExp a = AlgExp
{ litI :: Int - a,
litB :: Bool - a,
add :: a - a - a,
and :: a - a - a,
ifte :: a - a - a - a}
(being ifte an 'ifthenelse' expresion...)
What I want to do is to write a fold function for expressions, something
like this:
foldExp :: AlgExp a - Exp - a
foldExp alg (LitI i) = litI alg i
foldExp alg (LitB i) = litB alg i
foldExp alg (add exp1 exp2) = ¿¿¿???
foldExp alg (and exp1 exp2) = ¿¿¿???
foldExp alg (ifte exp1 exp2 exp3) = ¿¿¿???
..ETC
the fact is that I have no idea of what to do with the other expresions
(add, and, and ifte)... I really don' t understand how to do this... It's
clear that a fold function should colapse in one valour, but how can I
espress it in the terms of the exercise?
For further information about the problem after this, it's suposed that I
have to rewrite some functions for expresions but in terms of foldexp (the
one I should write before)
The problem is that AlgExp defines an arbitrary algebra, but in order to
fold you need a universal algebra. So it makes the most sense to add
foldExp to the reccord.
This is an unusually poor post for you. Presuming you mean initial or
free or term for universal then it does (presumably) have it with
Exp. Assuming Exp is the expected thing (an AST) it is (combined with
the constructors) the initial algebra. foldExp should quite definitely
be the type it is and not be part of the record and the its
implementation is so far correct (modulo syntactical errors that are
potentially indicative of a deeper confusion).
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