Re: [Haskell-beginners] foldr on infinite list to decide prime number

Thank you for kind response. See you in another thread. :-) 2016-02-02 22:33 GMT+09:00 Gesh :

On February 2, 2016 3:36:03 AM GMT+02:00, Chul-Woong Yang wrote:

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I feel sorry for posting mis-formatted code. I re-post the question. --

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?

Any helps will be deeply appreciated.

Thank you.

2016-02-02 10:32 GMT+09:00 Chul-Woong Yang :

...

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?

Any helps will be deeply appreciated.

Thank you.

Chul-Woong

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Note that || is lazy, specifically:

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True || _ = True False || x = x Hence, once foldr reaches a prime larger than sqrt n, the first branch of || will be True, short-circuiting the entire infinite computation.

HTH, Gesh _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners