memcpy question

memcpy question

[Michael.Schmidt]

Hello,

I apologize in advance if this is the wrong message board.  I've noticed
that memcpy seems to take a while (sometimes) for large chunks of memory
(about 1 MB).  

Here's a small sample application:

#include<iostream>
#include<time.h>
#include<cmath>
#include<iomanip>
#include <sys/timeb.h>

using namespace std;

double getTime()
{
  struct timeb tb ;

  ftime ( &tb );
  return (double) tb.time + (double) tb.millitm / 1000.0 ;
}

int main()
{
  char src[1048576] = {0};
  char dst[1048576];
  
  for(unsigned int i = 0; i < 25; ++i)
  {
    double start_time = getTime();
    memcpy(dst, src, 1048576);
    double end_time = getTime();
    cout << "Time: " << 1000.0*(end_time - start_time) << endl;
  }

  return 0;
}

Here's the output:

Time: 0.999928
Time: 0
Time: 0
Time: 0
Time: 0
Time: 0.999928
Time: 0
Time: 0
Time: 0
Time: 0
Time: 0
Time: 0
Time: 0
Time: 0.999928
Time: 0
Time: 0
Time: 0
Time: 0
Time: 0
Time: 0
Time: 1.00017
Time: 0
Time: 0
Time: 0
Time: 0

I would be eternally grateful if anyone could explain what's happening.
Run on a Core 2 Duo E6550, 4 GB RAM.

Thanks,
Mike

Re: memcpy question

[Michael.Schmidt]

Hello again,

My time function was imprecise.

double getTime()
{
  timeval tv ;

  gettimeofday ( & tv, NULL ) ;

  return (double) tv.tv_sec + (double) tv.tv_usec / 1000000.0 ;
}

Using the function above gives times of around 0.13 ms, which is much
better than what I've seen in my project.  I saw times that were more
than 2 ms.  What could slow memcpy down?

Thanks,
Mike


On Tue, 2008-10-07 at 12:50 -0500, Michael.Schmidt@L-3com.com wrote:
> Hello,
> 
> I apologize in advance if this is the wrong message board.  I've noticed
> that memcpy seems to take a while (sometimes) for large chunks of memory
> (about 1 MB).  
> 
> Here's a small sample application:
> 
> #include<iostream>
> #include<time.h>
> #include<cmath>
> #include<iomanip>
> #include <sys/timeb.h>
> 
> using namespace std;
> 
> double getTime()
> {
>   struct timeb tb ;
> 
>   ftime ( &tb );
>   return (double) tb.time + (double) tb.millitm / 1000.0 ;
> }
> 
> int main()
> {
>   char src[1048576] = {0};
>   char dst[1048576];
>   
>   for(unsigned int i = 0; i < 25; ++i)
>   {
>     double start_time = getTime();
>     memcpy(dst, src, 1048576);
>     double end_time = getTime();
>     cout << "Time: " << 1000.0*(end_time - start_time) << endl;
>   }
> 
>   return 0;
> }
> 
> Here's the output:
> 
> Time: 0.999928
> Time: 0
> Time: 0
> Time: 0
> Time: 0
> Time: 0.999928
> Time: 0
> Time: 0
> Time: 0
> Time: 0
> Time: 0
> Time: 0
> Time: 0
> Time: 0.999928
> Time: 0
> Time: 0
> Time: 0
> Time: 0
> Time: 0
> Time: 0
> Time: 1.00017
> Time: 0
> Time: 0
> Time: 0
> Time: 0
> 
> I would be eternally grateful if anyone could explain what's happening.
> Run on a Core 2 Duo E6550, 4 GB RAM.
> 
> Thanks,
> Mike
> 

Re: memcpy question

[John Love-Jensen]

Hi Michael,

It appears that on your system, your routine is often running faster than
the granularity of your time source.

One way to compensate is by wrapping the memcpy in a for(int j = 0; j < 100;
++j) so that the routine does not run too fast for the granularity.

--Eljay