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

[Gesh]

On February 2, 2016 3:36:03 AM GMT+02:00, Chul-Woong Yang <cwyang at aranetworks.com> wrote:
>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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>
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Note that || is lazy, specifically:
> 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