calloc speed difference

[Christian Franke]

Lee wrote:
> Why is the cygwin gcc calloc so much slower than the
> i686-w64-mingw32-gcc calloc?
>    1:12 vs 0:11
>
> $cat calloc-test.c
> #include <stdio.h>
> #include <stdlib.h>
> #define ALLOCATION_SIZE (100 * 1024 * 1024)
> int main (int argc, char *argv[]) {
>      for (int i = 0; i < 10000; i++) {
>          void *temp = calloc(ALLOCATION_SIZE, 1);
>          if ( temp == NULL ) {
>             printf("drat! calloc returned NULL\n");
>             return 1;
>          }
>          free(temp);
>      }
>      return 0;
> }
>

Could reproduce the difference on an older i7-2600K machine:

Cygwin: ~20s
MinGW: ~4s

Timing [cm]alloc() calls without actually using the allocated memory 
might produce misleading results due to lazy page allocation and/or 
zero-filling.

MinGW binaries use calloc() from msvcrt.dll. This calloc() does not call 
malloc() and then memset(). It directly calls:

   mem = HeapAlloc(_crtheap, HEAP_ZERO_MEMORY, size);

which possibly only reserves allocate-and-zero-fill-on-demand pages for 
later.

Cygwin's calloc() is different.

This variant of the above code adds one write access to each 4KiB page 
(guarded by "volatile" to prevent dead assignment optimization):

#include <stdio.h>
#include <stdlib.h>
#define ALLOCATION_SIZE (100 * 1024 * 1024)
int main (int argc, char *argv[]) {
     for (int i = 0; i < 1000; i++) {
         void *temp = calloc(ALLOCATION_SIZE, 1);
         if ( temp == NULL ) {
            printf("drat! calloc returned NULL\n");
            return 1;
         }
         for (int j = 0; j < ALLOCATION_SIZE; j += 4096)
           ((volatile char *)temp)[j] = (char)i;
         free(temp);
     }
     return 0;
}

Results:

Cygwin: ~310s
MinGW: ~210s

Christian


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