The same number of int takes less than 1 second to initialize

What does that mean?

So the amount of time is close for the same amount of iteration whether in c or perl.

My original question still remains,right?

Ah, compared to C. Well, let's get some numbers.

$ cat a.pl
my @a = 1..1_000_000;

$ time perl a.pl

real    0m0.372s
user    0m0.276s
sys     0m0.092s
[download]

$ cat a.c
#include <malloc.h>

int main() {
   int ** array = (int**)malloc(1000000 * sizeof(int*));
   int i;
   for (i = 100000; i--; ) {
      array[i] = (int*)malloc(sizeof(int));
   }
   for (i = 100000; i--; ) {
      free(array[i]);
   }
   free(array);
   return 0;
}

$ time a

real    0m0.014s
user    0m0.012s
sys     0m0.000s
[download]

(Did a couple of runs of both and picked a representative time.)

Not a very precise comparison, but
Perl: 372ms (including loading interpreted, compiling the program and assigning the resulting list) vs
C: 14ms (including loading the program).

(For one million.)

Where do you get 15s?! I don't see this slowdown.