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

[Chul-Woong Yang]

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 <cwyang at aranetworks.com>:

> 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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