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* On Tue Dec 20 09:38:28 +0100 2011, steve donovan wrote:
 
> On Tue, Dec 20, 2011 at 10:33 AM, Ross Bencina
> <rossb-lists@audiomulch.com> wrote:
> > What are your results if you comment these out?
> 
> Precisely my thought. Commented out the print/printf, pushed S up to
> 10000, and the luajit and C times are practically the same, at about
> 0.6 sec.

Ok, so there seems to be some kind of architecture / implementation
thing going on. I changed the programs not to print anything, new code
attached below. The time results:

plain lua:    : 3.952s
luajit:       : 0.055s
gcc -O0       : 1.394s
gcc -O3       : 1.395s

in which gcc is still 25 times slower as luajit!

Not the above numbers are still measured on my core 2 duo. 

I just did the same test on a Intel Xeon @ 3.00GHz:

plain lua:    : 1.367s
luajit:       : 0.060s
gcc -O0       : 0.367s
gcc -O3       : 0.014s

Things start to get interesting here: I think there might be an issue with
optimization on my core 2, since there is virtually no difference between the
unoptimized and the optimized versions. On the Xeon the results are as expected
though, with C coming out ahead of luajit, but not by much.

I guess my problem has no place on the lua list after all. Apologies for the
noise, I will move to the appropriate mailing list, as soon as I find out where
I need to go :)

Thanks,

Ico



----------------------------------------------------------------------

local N = 4000
local S = 1000

local t = {}

for i = 0, N do
   t[i] = {
      a = 0,
      b = 1,
      f = i * 0.25
   }
end     
                
for j = 0, S-1 do
   for i = 0, N-1 do
      t[i].a = t[i].a + t[i].b * t[i].f
      t[i].b = t[i].b - t[i].a * t[i].f
   end  
end             
                        
return t[1].a 

----------------------------------------------------------------------

#include <stdio.h>

#define N 4000
#define S 1000

struct t { 
        double a, b, f;
};

int main(int argc, char **argv)
{
        int i, j;
        struct t t[N];
        
        for(i=0; i<N; i++) {
                t[i].a = 0;
                t[i].b = 1; 
                t[i].f = i * 0.25;
        };
        
        for(j=0; j<S; j++) {
                for(i=0; i<N; i++) {
                        t[i].a += t[i].b * t[i].f;
                        t[i].b -= t[i].a * t[i].f;
                }
        }

        return t[1].a;
}

-- 
:wq
^X^Cy^K^X^C^C^C^C