rem is faster because it has slightly different behaviour to mod, and
there happens to be an intel instruction that maps more directly to
rem than to mod, thus making it much faster on intel processors.
Why do you expose perfect and divisors? Maybe if you just expose main,
perfect and divisors will be inlined (although this will only save
10,000 function entries, so will probably have negligible effect).
Rodrigo
On 29/10/2007, Peter Hercek
Derek Elkins wrote:
Try with rem instead of mod.
(What the heck is with bottom?)
The bottom was there by error and I was lazy to redo the tests so I rather posted exactly what I was doing. I do not know the compiler that good to be absolutely sure it cannot have impact on result ... so I rather did not doctor what I did :-)
Ok, rem did help quite a bit. Any comments why it is so?
Here are summary of results for those interested. I run all the tests once again. Haskell was about 13% slower than C++.
MS cl.exe options: /Ox /G7 /MD ghc options: -O2
C++ version: 1.000; 0.984; 0.984 Haskell version specialized to Int: 1.125; 1.125; 1.109 Haskell version specialized to Integer: 8.781; 8.813; 8.813 Haskell version specialized to Int64: 9.781; 9.766; 9.831
The code:
% cat b.hs module Main (divisors, perfect, main) where import Data.Int
divisors :: Int -> [Int] divisors i = [j | j<-[1..i-1], i `rem` j == 0]
perfect :: [Int] perfect = [i | i<-[1..10000], i == sum (divisors i)]
main = print perfect
% cat b.cpp #include <iostream> using namespace std;
int main() { for (int i = 1; i <= 10000; i++) { int sum = 0; for (int j = 1; j < i; j++) if (i % j == 0) sum += j; if (sum == i) cout << i << " "; } return 0; }
%
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