[Haskell-beginners] Why Integral makes a function much slower than Int?

8 Sep 2018

      Hi all,

I implemented a simple function to calculate the length of a list. 
However, I noticed that the speed of the function changes significantly 
if I make I minor change to the type signature but keep the 
implementation the same otherwise. I was wondering why this happens.

Contents of mylength.hs:

module MyLength where

-- Here is the first implementation:

myLength :: Integral n => [a] -> n
myLength xs = run xs 0
   where
     run [] n = n
     run (_:ys) n = run ys (n+1)

-- Then, the other implementation just fixes the return type without
-- modifying the actual implementation:

myLength' :: [a] -> Int
myLength' = myLength

When I compile these implementations with:

ghc -dynamic -O2 mylength.hs

Then, launch ghci:

ghci -fobject-code

And in ghci:

:load mylength.hs
:set +s

(myLength [1..10000000]) :: Int
10000000
(2.30 secs, 1,612,737,944 bytes)

(myLength' [1..10000000]) :: Int
10000000
(0.38 secs, 720,077,536 bytes)

So, the first implementation is much worse, although I've fixed the 
return type to Int too.

What's going on? Does this mean one shouldn't use Integral but Int 
instead most of the time?

Thanks for help!

Regards,
Jaakko

[Haskell-beginners] Why Integral makes a function much slower than Int?

Jaakko Luttinen