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

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