[Haskell-beginners] foldr on infinite list to decide prime number
Francesco Ariis
fa-ml at ariis.it
Tue Feb 2 02:01:23 UTC 2016
On Tue, Feb 02, 2016 at 10:32:10AM +0900, Chul-Woong Yang wrote:
> Hi, all.
>
> While I know that foldr can be used on infinite list to generate infinite
> list,
> I'm having difficulty in understaind following code:
>
> isPrime n = n > 1 && -- from haskell wiki foldr (\p r -> p*p > n || ((n
> `rem` p) /= 0 && r)) True primes primes = 2 : filter isPrime [3,5..]
>
> primes is a infinite list of prime numbers, and isPrime does foldr to get a
> boolean value.
> What causes foldr to terminate folding?
foldr _immediately_ calls the passed function, hence /it can short
circuit/, that isn't the case for foldl.
I wrote an article to explain it [1]. It was drafted in a time when
foldr and friends were monomorphic (i.e. they only worked with lists),
but it should illustrate the point nicely.
Current polymorphic implementation of foldr is:
foldr :: (a -> b -> b) -> b -> t a -> b
foldr f z t = appEndo (foldMap (Endo #. f) t) z
and I must admit I have problems explaining why it terminates
early (as it does).
[1] http://ariis.it/static/articles/haskell-laziness/page.html (more
complex cases section)
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