c operation takes 0 seconds

Question

I am trying to measure how much time takes serial program to reverse an array. My code:

#include <time.h>

int main() {
   int n = 100000;
   int c, d, a[n], b[n];
   clock_t start, end;

   for (c = 0; c < n ; c++)
      a[c] = c;

   start = clock();

   for (c = n - 1, d = 0; c >= 0; c--, d++)
      b[d] = a[c];

   end = clock();
   printf("Number of seconds: %f\n", (end-start)/(double)CLOCKS_PER_SEC );

   return 0;
}

However, it is always takes 0.000000 seconds to run. Why? If I increase n, I get segmentation errors.

You should be aware that this does not reverse your array as you reverse twice! Think about (for example) the first iteration which reverse c=n-1, d=0 and the last iteration which reverse c=0, d=n-1... — Jean-Baptiste Yunès, Jan 03 '15 at 17:50

score 3 · Answer 1 · answered Jan 03 '15 at 15:01

3

The loop:

   for (c = n - 1, d = 0; c >= 0; c--, d++)
      b[d] = a[c];

is optimized out by your compiler. Your compiler figured out b and a are not further used in your program.

To avoid the compiler optimizing out the loop you can pass b and a as arguments to a dummy external function. You could also use volatile qualifiers but this is likely to bias your result.

answered Jan 03 '15 at 15:01

ouah

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So, how can I fix it? – Bob Jan 03 '15 at 15:02
1

You could print a random element of the arrays near end of program. – Basile Starynkevitch Jan 03 '15 at 15:07
@ouha wouldn't `-O0` on GCC remedy that? – Qix - MONICA WAS MISTREATED Jan 03 '15 at 15:09
1

@Qix could or couldn't be as even with `-O0` `gcc` performs some optimizations. But the point is if you want to measure performances, you don't want to compile a program with `-O0`. – ouah Jan 03 '15 at 15:14
I added printing a random element near the end of a program - did not help. – Bob Jan 03 '15 at 16:06
@Bob as maxpovver mentioned in his answer, your loop with only 100000 iterations is also too fast for `clock` to mesure it, you need to repeat the loop. – ouah Jan 03 '15 at 16:11

score 3 · Accepted Answer · edited May 23 '17 at 10:26

Try many iterations:

#include <time.h>

int main() {
int n = 100000;
int c, d, a[n], b[n];
clock_t start, end;
start = clock();
for(int i =0; i< 100000000;i++) {
  for (c = 0; c < n ; c++)
   a[c] = c;
for (c = n - 1, d = 0; c >= 0; c--, d++)
  b[d] = a[c];
}
end = clock();
printf("Number of seconds: %f\n", (end-start)/(double)CLOCKS_PER_SEC/100000000 );
return 0;
}

but better read this: C++ clock stays zero

c operation takes 0 seconds

2 Answers2