Re: [Haskell-beginners] understanding curried function calls

20 Aug 2014

      This is exactly what I did, but it doesn't give me a systematic way to 
evaluate the expressions.
In other words, I don't really know why it works or what else to try in 
different circumstances.
The problem arises when I have curried multi-argument functions.

People complain about pointer notation in type declarations in C. Well, 
this is pointer notation++.
And it is worse if I can't find resources to learn it.

Eventually, I would like to be able to grok famous beasts like this one.

foldl  ::  (a->  b->  a)  ->  a->  [b]  ->  a
foldl  f a bs=
    foldr  (\b g x->  g(f x b))  id  bs a

But I need a consistent approach.

Cheers,

Dimitri

Em 20/08/14 04:10, John Wiegley escreveu:
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Dimitri DeFigueiredo  writes:
Here's a simple exercise from Stephanie Weirich's class [1] that I am having
a hard time with.
consider
doTwice :: (a -> a) -> a -> a
doTwice f x = f (f x)
what does this do?
ex1 :: (a -> a) -> a -> a
ex1 = doTwice doTwice
Another way to write doTwice is:
doTwice :: (a -> a) -> (a -> a)
     doTwice f = \x -> f (f x)
This is equivalent.  If you stare it for a while, it should answer the rest of
your questions.
John
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