CODEWITHSUNDEEP
haskelltail-recursion
Hynek -Pichi- Vychodil
Hynek -Pichi- Vychodil

Reputation: 26131

How does Haskell tail recursion work?

I wrote this snippet of code and I assume len is tail-recursive, but a stack overflow still occurs. What is wrong?

myLength :: [a] -> Integer

myLength xs = len xs 0
    where len [] l = l
          len (x:xs) l = len xs (l+1)

main = print $ myLength [1..10000000]

Upvotes: 48

Views: 9736

Answers (6)

jmg
jmg

Reputation: 7414

The simplest solution to your problem is turning on optimization.

I have your source in a file called tail.hs.

jmg$ ghc --make tail.hs -fforce-recomp
[1 of 1] Compiling Main             ( tail.hs, tail.o )
Linking tail ...
jmg$ ./tail 
Stack space overflow: current size 8388608 bytes.
Use `+RTS -Ksize -RTS' to increase it.
girard:haskell jmg$ ghc -O --make tail.hs -fforce-recomp
[1 of 1] Compiling Main             ( tail.hs, tail.o )
Linking tail ...
jmg$ ./tail 
10000000
jmg$

@Hynek -Pichi- Vychodil The tests above were done on Mac OS X Snow Leopard 64bit with a GHC 7 and GHC 6.12.1, each in a 32 bit version. After you're downvote, I repeated this experiment on Ubuntu Linux with the following result:

jmg@girard:/tmp$ cat length.hs
myLength :: [a] -> Integer

myLength xs = len xs 0
    where len [] l = l
          len (x:xs) l = len xs (l+1)

main = print $ myLength [1..10000000]

jmg@girard:/tmp$ ghc --version
The Glorious Glasgow Haskell Compilation System, version 6.12.1
jmg@girard:/tmp$ uname -a
Linux girard 2.6.35-24-generic #42-Ubuntu SMP Thu Dec 2 02:41:37 UTC 2010 x86_64 GNU/Linux
jmg@girard:/tmp$ ghc --make length.hs -fforce-recomp
[1 of 1] Compiling Main             ( length.hs, length.o )
Linking length ...
jmg@girard:/tmp$ time ./length 
Stack space overflow: current size 8388608 bytes.
Use `+RTS -Ksize -RTS' to increase it.

real    0m1.359s
user    0m1.140s
sys 0m0.210s
jmg@girard:/tmp$ ghc -O --make length.hs -fforce-recomp
[1 of 1] Compiling Main             ( length.hs, length.o )
Linking length ...
jmg@girard:/tmp$ time ./length 
10000000

real    0m0.268s
user    0m0.260s
sys 0m0.000s
jmg@girard:/tmp$

So, if you're interested we can continue to find out what is the reason, that this fails for you. I guess, GHC HQ, would accept it as a bug, if such a straight forward recursion over lists is not optimized into an efficient loop in this case.

Upvotes: 2

Michael Steele
Michael Steele

Reputation: 15772

eelco.lempsink.nl answers the question you asked. Here's a demonstration of Yann's answer:

module Main
    where

import Data.List
import System.Environment (getArgs)

main = do
  n <- getArgs >>= readIO.head
  putStrLn $ "Length of an array from 1 to " ++ show n
               ++ ": " ++ show (myLength [1..n])

myLength :: [a] -> Int
myLength = foldl' (const . succ) 0

foldl' goes through the list from left to right each time adding 1 to an accumulator which starts at 0.

Here's an example of running the program:

C:\haskell>ghc --make Test.hs -O2 -fforce-recomp
[1 of 1] Compiling Main             ( Test.hs, Test.o )
Linking Test.exe ...

C:\haskell>Test.exe 10000000 +RTS -sstderr
Test.exe 10000000 +RTS -sstderr

Length of an array from 1 to 10000000: 10000000
     401,572,536 bytes allocated in the heap
          18,048 bytes copied during GC
           2,352 bytes maximum residency (1 sample(s))
          13,764 bytes maximum slop
               1 MB total memory in use (0 MB lost due to fragmentation)

  Generation 0:   765 collections,     0 parallel,  0.00s,  0.00s elapsed
  Generation 1:     1 collections,     0 parallel,  0.00s,  0.00s elapsed

  INIT  time    0.00s  (  0.00s elapsed)
  MUT   time    0.27s  (  0.28s elapsed)
  GC    time    0.00s  (  0.00s elapsed)
  EXIT  time    0.00s  (  0.00s elapsed)
  Total time    0.27s  (  0.28s elapsed)

  %GC time       0.0%  (0.7% elapsed)

  Alloc rate    1,514,219,539 bytes per MUT second

  Productivity 100.0% of total user, 93.7% of total elapsed


C:\haskell>

Upvotes: 1

Yann Vernier
Yann Vernier

Reputation:

The foldl carries the same problem; it builds a thunk. You can use foldl' from Data.List to avoid that problem:

import Data.List
myLength = foldl' (const.succ) 0

The only difference between foldl and foldl' is the strict accumulation, so foldl' solves the problem in the same way as the seq and $! examples above. (const.succ) here works the same as (\a b -> a+1), though succ has a less restrictive type.

Upvotes: 4

Alex Fort
Alex Fort

Reputation: 18819

Just so you know, there's a much easier way to write this function:

myLength xs = foldl step 0 xs where step acc x = acc + 1

Alex

Upvotes: 1

eelco
eelco

Reputation: 1607

Remember that Haskell is lazy. Your computation (l+1) will not occur until it's absolutely necessary.

The 'easy' fix is to use '$!' to force evaluation:

myLength :: [a] -> Integer
myLength xs = len xs 0
where len [] l = l
      len (x:xs) l = len xs $! (l+1)

      main = print $ myLength [1..10000000]

Upvotes: 40

mattiast
mattiast

Reputation: 1944

Seems like laziness causes len to build thunk:

len [1..100000] 0
-> len [2..100000] (0+1)
-> len [3..100000] (0+1+1)

and so on. You must force len to reduce l every time:

len (x:xs) l = l `seq` len xs (l+1)

For more information, look http://haskell.org/haskellwiki/Stack_overflow.

Upvotes: 14

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