[Haskell-cafe] Re: C-like Haskell

[John Lato]

On Thu, Jan 29, 2009 at 11:30 AM, Duncan Coutts
<duncan.coutts at worc.ox.ac.uk> wrote:
> On Thu, 2009-01-29 at 10:35 +0000, John Lato wrote:
>> > On Wed, 2009-01-28 at 20:11 -0500, sam lee wrote:
>> >> Did you print it? I'm using same code with ghc --make -O2  and it
>> >> takes forever to finish.
>> >
>> > Yes, you can see in the output that it prints the same answer in each
>> > case. I was using r = 10^9 as you suggested.
>>
>> I wouldn't call these answers the same.  Can this be ascribed to an
>> overflow error?
>
> Oh yes, you're quite right. I didn't notice because it only affects the
> middle digits :-)
>
>> >> > C version:
>> >> > $ time ./circ
>> >> > 3141592649589764829
>>                       ^^
>> >
>> >> > Haskell version:
>> >> > time ./circ2
>> >> > 3141592653589764829
>
> Compare:
>
> circ2 r = (1+4*r) + 4 * (circ2' (rs+1) r 1 0)
>
> vs
>
>    # the main loop
>  }
>  sum*=4;
>  sum++;
>
> Looks like the C version misses the +4*r at the end.
>
>  3141592649589764829 + 4*10^9
> = 3141592653589764829
>
> Assuming that's right, the mistake only costs constant-time and doesn't
> invalidate the timings.
>

Thank you.  That explains it nicely.  Giving credit where it's due,
Lennart A. also pointed out that the difference is 4e9 and suspected a
boundary case.

I think I need more coffee today :)

John Lato