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Home/ Questions/Q 8372237
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T14:22:38+00:00 2026-06-09T14:22:38+00:00

I’m working my way through Real World Haskell (I’m in chapter 4) and to

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I’m working my way through Real World Haskell (I’m in chapter 4) and to practice a bit off-book I’ve created the following program to calculate the nth prime.

import System.Environment

isPrime primes test = loop primes test
    where
        loop (p:primes) test
            | test `mod` p == 0 = False
            | p * p > test = True
            | otherwise = loop primes test


primes = [2, 3] ++ loop [2, 3] 5
    where 
        loop primes test
            | isPrime primes test = test:(loop primes' test')
            | otherwise = test' `seq` (loop primes test')
            where
               test' = test + 2
               primes' = primes ++ [test]

main :: IO()
main = do
    args <- getArgs
    print(last  (take (read (head args) :: Int) primes))

Obviously since I’m saving a list of primes this is not a constant space solution. The problem is when I try to get a very large prime say ./primes 1000000 I receive the error:

    Stack space overflow: current size 8388608 bytes.
    Use `+RTS -Ksize -RTS' to increase 

I’m fairly sure that I got the tail recursion right; reading http://www.haskell.org/haskellwiki/Stack_overflow and the various responses here lead me to believe that it’s a byproduct of lazy evaluation, and thunks are building up until it overflows, but so far I’ve been unsuccessful in fixing it. I’ve tried using seq in various places to force evaluation, but it hasn’t had an effect. Am I on the right track? Is there something else I’m not getting?

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  1. Editorial Team
    Editorial Team
    2026-06-09T14:22:39+00:00Added an answer on June 9, 2026 at 2:22 pm

    As I said in my comment, you shouldn’t be building a list by appending a single element list to the end of a really long list (your line primes' = primes ++ [test]). It is better to just define the infinite list, primes, and let lazy evaluation do it’s thing. Something like the below code:

    primes = [2, 3] ++ loop 5
        where.
            loop test
                | isPrime primes test = test:(loop test')
                | otherwise = test' `seq` (loop test')
                where
                    test' = test + 2
    

    Obviously you don’t need to parameterize the isPrime function by primes either, but that’s just a nit. Also, when you know all the numbers are positive you should use rem instead of mod – this results in a 30% performance increase on my machine (when finding the millionth prime).

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