[Haskell-beginners] learning lazy evaluation

Tue Oct 21 15:43:07 UTC 2014

That is an interesting observation. Haskell is defined to have non-strict
evaluation, which basically means the evaluation must happen from the
outside in. It does not guarantee laziness, so even in interpreted settings
it doesn't *have* to create a thunk. What you're seeing is simply an
implementation detail of GHC that's consistent with the Haskell language
but inconsistent with the expectation that it should be maximally lazy.

On Mon, Oct 20, 2014 at 2:18 PM, Dimitri DeFigueiredo <
defigueiredo at ucdavis.edu> wrote:

>  Thanks Bob.
>
> The (lack of) exponential explosion makes sense now. And, yes, I was using
> both ghci and ghc 7.6.3 without any flags.
>
> When memoization takes place is still a little murky to me, though.
> Figure 2.2 in that chapter of Simon Marlow's book has the kind of
> "aliasing" with the number '1' (two pointers pointing to the same location)
> that I thought would also happen with 'minimum_bug xs' , but I guess 1::Int
> is a special case that is different from other terms without optimization
> or the depiction is not accurate.
>
> I have a further question on that is troubling me. It seems that
> constructors are treated differently from other expressions when it comes
> to evaluation. The examples in the book show that:
>
> Prelude> let x = 1 + 2 :: Int
> Prelude> :sprint x
> x = _
>
>
> In other words, x is just an unevaluated thunk.
> But at the same time:
>
> Prelude> let x = 1 + 2 :: Int
> Prelude> let z = (x,x)
>
> Prelude> :sprint z
> z = (_,_)
>
>
> So, 'z' has been evaluated to WHNF.
> I was expecting to get:
>
> Prelude> :sprint z
> z = _
>
>
> Meaning, an unevaluated expression. Just like we did with 'x', but it
> seems z has already been partially evaluated. Am I getting this right? Are
> expressions with constructors evaluated differently? Why is :sprint z
> different from 'z = _' in this case?
>
>
> Thanks again,
>
>
> Dimitri
>
>
> On 20/10/14 13:21, Bob Ippolito wrote:
>
>
>
> On Mon, Oct 20, 2014 at 11:57 AM, Dimitri DeFigueiredo <
> defigueiredo at ucdavis.edu> wrote:
>
>> I found an interesting situation while making recursive calls that I am
>> trying to understand. I re-wrote the 'minimum' function from the prelude as
>> such:
>>
>> minimum_bug :: (Ord a) => [a] -> a
>> minimum_bug  [x] = x
>> minimum_bug  (x:xs) | x > minimum_bug xs = minimum_bug xs
>>                     | otherwise          = x
>>
>> (I understand I should be using something like "minbug xs = foldr1 min
>> xs" for this, but bare with me)
>>
>> The interesting thing is that the function has exponential time behavior.
>> In a way, that makes sense as I am making two recursive calls. But I was
>> expecting GHC to be smart enough to transform it into something like:
>>
>> minimum_bug' :: (Ord a) => [a] -> a
>> minimum_bug'  [x] = x
>> minimum_bug'  (x:xs) | x > y      = y
>>                      | otherwise  = x
>>                         where y = minimum_bug' xs
>>
>> (This one always works as expected)
>>
>> I understand that lazy evaluation implies memoization in some cases.
>> When does GHC only use memoization to avoid this kind of behavior?
>>
>
>  I assume you're trying this with ghci or compiling without
> optimizations. In these scenarios GHC isn't doing anything clever at all,
> so there will be two separate calls to minimum_bug in each recursion. Using
> a variable ensures that this value is explicitly shared between the guard
> and the result in the list.
>
>  This section of Parallel and Concurrent Programming in Haskell helped me
> understand Haskell's non-strict evaluation model:
> http://chimera.labs.oreilly.com/books/1230000000929/ch02.html#sec_par-eval-whnf
>
>  Another interesting fact is that in my simple tests the exponential
>> behavior does NOT occur when I pass the function a list in sorted order.
>>
>> main = do putStrLn "started - in order list"
>>           putStrLn $ show $ minimum_bug [1..30000]  -- no problem
>>           putStrLn "started - out of order list"
>>           putStrLn $ show $ minimum_bug [27,26..1]  -- don't do this with
>> large numbers!
>>           putStrLn "done!"
>>
>> It's not clear to me how the sorted list is able to escape the
>> exponential slow down.
>>
>
>  Because displaying the value doesn't have any recursion to it, since
> it's `x` and not `minumum_bug x`.
>
>  You can walk through the execution.
>
>  -- Original call
> minimum_bug [1, 2]
> -- Expand the first matching pattern
> minimum_bug (1:[2]) | 1 > minimum_bug [2]
> -- minimum_bug [2] is forced due to the guard
> -- Evaluate minimum_bug [2]
> minimum_bug [2]
> -> 2
> -- Step back to the guard we were trying to evaluate
> minimum_bug (1:[2]) | 1 > 2
> 1 > 2
> -> False
> -- Guard fails, go to the otherwise case
>   | otherwise = 1
> -> 1
>
>  The `1` is already in normal form, so `putStrLn . show` doesn't have to
> do any extra reductions.
>
>  If you walk through with `minimum_bug [2, 1]` you'll end up with
> `minimum_bug [1]` as the result, which is not in normal form and thus
> requires additional evaluation when `putStrLn . show` attempts to display
> it. If you use a larger list and walk through, you can see that the amount
> of redundant evaluation explodes when the list is in descending order.
>
>  -bob
>
>
>
>
> _______________________________________________
> Beginners mailing listBeginners at haskell.orghttp://www.haskell.org/mailman/listinfo/beginners
>
>
>
> _______________________________________________
> Beginners mailing list
> Beginners at haskell.org
> http://www.haskell.org/mailman/listinfo/beginners
>
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://www.haskell.org/pipermail/beginners/attachments/20141021/d2c5ba23/attachment.html>