Re: question about memcpy in libc

Hi,

I am not sure if this is the right list for this mail or not, but another good one would be: linux-c-programming@vger.kernel.org

On 8 November 2012 07:09, 牟刚 mugang@actions-semi.com wrote:

       I’m confused when I use memcpy() in pandaboard android. The test

code is very simple, as below:

#define PER_COPY_SIZE (1024*1024*4)

int main(int argc, char *argv[ ]){

     char *src;

     char *dst;

     char *p;



     src=(char *)malloc(PER_COPY_SIZE);

     dst=(char*)malloc(PER_COPY_SIZE);

     p=src;

#if 1

     for(int i=0; i<PER_COPY_SIZE; i++){

         *p++=i%255;

     }

#endif

     for(int j=0; j<1000; j++){

               memcpy(dst, src, PER_COPY_SIZE);

     }

     free(src);

     free(dst);

     return 0;

}

   It copies 4M byte data from source to dest repeatedly and takes 12

seconds. But when I get rid of the part of code in red color, it takes only 4 seconds. Why does the writing to source firstly has influence the memcpy( ) time so greatly?

I tested it on Vexpress TC2 with 2 A15's and 3 A7's.

Interesting. I added additional if, endif on the second loop to and checked their assembly in three conditions: - Both loops are enabled: full: Time: 3.7 sec - Only first loop is enabled: first: .189 sec - Only second loop is enabled: second: 1.6 sec

Find assembly of these attached.

I don't see any trick in assembly that can do it. That that is out of the way.

What is left is Cache. But yes, i must admit, i am still not able to solve this puzzle.

Lets see if Arnd can help here :)

-- viresh