So I'm making a list of prime numbers to help me learn haskell using simple trial division (no fancy stuff until I get better with the language). I'm trying to use the following code:
primes = 2 : [ x | x <- [3..], all (\p -> (mod x p) /= 0) primes]
This is loaded without an error. However:
>take 2 primes
[2ERROR - C stack overflow
I tried the same thing with nested list comprehensions. It doesn't work. I would guess that I'm making too many recursive calls, but this shouldn't be the case if i'm only computing one prime. In my mind the lazy evaluation should make it so that take 2 primes does something along the lines of:
primes = 2 : [ 3 | all (\p -> (mod 3 p) /= 0) [2] ]
Which doesn't require all that much computation - mod 3 2 == True, so all (\p -> (mod 3 p) /= 0) == True, which means take 2 primes == [2, 3], right? I don't understand why this isn't working. Hopefully someone much more versed in the black magic of functional programming can help me...
This is on HUGS, if that makes any difference.
EDIT- I was able to come up with this solution (not pretty):
primes = 2 : [ x | x <- [3..], all (\p -> (mod x p) /= 0) (takeWhile (<= (ceiling (sqrt (fromIntegral x)))) primes)]
EDIT2- The program works fine when interpreted through HUGS or GHCi, but when I try to compile it with GHC, it outputs test: <<loop>>. Anybody know what the problem is?
Hugs shouldn't do this, but the code is broken anyway so it doesn't matter. Consider:
primes = 2 : [ x | x <- [3..], all (\p -> (mod x p) /= 0) primes]
How do you determine if 3 is prime? well, does mod 3 2 == 0? No. Does mod 3 ??? == 0? OOPS! What is the next element of primes after two? we don't know, we are trying to compute it. You need to add an ordering constraint that adds 3 (or any other x) once all p elem primes less than sqrt x have been tested.