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)