[Haskell-cafe] Where to start about writing web/HTTP apps ?

[gary ng]

Ah, lambda, function as first order object and curry.
A million thanks, that clears my mind. It is very hard
for me to shake away the things that I have learnt for
over 20 years. I still think from time to time that
the "accumulator" in foldr/foldl as a constant being
folded and forget that it can be anything. In this
case, because 'n' is not binded during the fold, it(g
(x*n)) becomes a function needing one more parameter
and keeps getting folded. The '1' at the end becomes
this final parameter that makes all these stacked
function in function all being evaluated back to
value.

Your explanation really taught me about these concepts
as I can understand simple tutorials(I used them in my
python programs) about them but only this kind deep
recursive I can see the advantage of curry.

I have already gone through the links you suggested
before I tried to see the various factorial
implementation, but I am sure I need to go back to
them again and again as haskell(and the idea behind it
in programming) is not something I can understand in
one go.

IMO, every CS course should teach haskell, as one can
easily pick up C/Pascal etc. without any help but
haskell is all too different.

--- Thomas Spriggs <thomasspriggs at hotmail.com> wrote:

> Well, here goes. The way the lambda function/foldr
> thing evaluates, the 
> resulting foldl like function needs a new identity
> parameter hence the 
> additional "1". To demonstrate something like how
> this is evaluated for a 
> low number eg 3: (Please would someone correct me if
> I have made a mistake 
> in this)
> fac 3 = (foldr (\x g n -> g (x*n)) id [1..3]) 1
> fac 3 = (foldr (\x g n -> g (x*n)) id [1,2,3]) 1
> fac 3 = (foldr (\x g n -> g (x*n)) (\n -> id (3*n))
> [1,2])) 1
> fac 3 = (foldr (\x g n -> g (x*n)) (\n -> (\n -> id
> (3*n)) (2*n)) [1]) 1
> fac 3 = (foldr (\x g n -> g (x*n)) (\n -> (\n -> (\n
> -> id (3*n)) (2*n)) 
> (1*n)) []) 1
> fac 3 = (\n -> (\n -> (\n -> id (3*n)) (2*n)) (1*n))
> 1
> fac 3 = (\n -> (\n -> id (3*n)) (2*n)) (1*1)
> fac 3 = (\n -> (\n -> id (3*n)) (2*n)) 1
> fac 3 = (\n -> id (3*n)) (2*1)
> fac 3 = (\n -> id (3*n)) 2
> fac 3 = id (3*2)
> fac 3 = id 6
> fac 3 = 6
> I would suggest that you use something other than
> the "evolution of a 
> haskell programmer" to learn haskell as the versions
> of factorial get 
> complicated very quickly and its largely use less as
> you should probably 
> just use:
> fac n = product [1..n]
> anyway. A better introduction would be something
> like 
> http://www.cse.unsw.edu.au/~cs1011/05s2/ and use 
>
http://zvon.org/other/haskell/Outputglobal/index.html
> and 
> http://www.haskell.org/tutorial/ if you want to
> learn something in specific 
> or are strugling. All links from
> http://www.haskell.org/learning.html of 
> course.
> >
> >thanks for help in advance.
> You're welcome.
> >
> >regards,
> >
> >gary
> >
> >
> >
> >
>
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> Well good luck furthering your knowledge of haskell,
> 
> Thomas
> 
>
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