Why we have a stack overflow in a recursive function in haskell?

Question

where do i have an endless loop?

 f2 :: Int->Int->Int
 f2 n d 
    | d==2
       = 0 
    | n `mod` d ==0  && n`mod` d^3==0
       = 1 + (f2 n d-1)
    | otherwise
       = 0 + (f2 n d-1)

And what actually happened? Your title says stack overflow, but your actual question mentions an endless loop (presumably a non-responses program). These are different things and have different causes. — Robin Zigmond
– Robin Zigmond, Commented Mar 16, 2019 at 10:31
it shows C-stack overflow , so i guess the problem is in recursion — AngD
– AngD, Commented Mar 16, 2019 at 10:32
Incidentally, you don't need the n `mod` d == 0 check. If something's divisible by d^3 then it's automatically divisible by d — Robin Zigmond
– Robin Zigmond, Commented Mar 16, 2019 at 10:33
Stack overflows in Haskell are not caused by recursion itself, as they are in imperative languages. Haskell stack overflows are due to too much laziness. — Robin Zigmond
– Robin Zigmond, Commented Mar 16, 2019 at 10:35

sepp2k · Accepted Answer · 2019-03-16 10:40:45Z

9

This is a precedence issue. (f2 n d-1) is parsed as (f2 n d) - 1, which leads to infinite recursion because f2 is calling itself with the same arguments. You want f2 n (d - 1) instead.

answered Mar 16, 2019 at 10:40

sepp2k

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