C how to measure time correctly?

user1319384 picture user1319384 · Apr 14, 2012 · Viewed 10.8k times · Source

This is the "algorithm", but when I want to measure the execution time it gives me zero. Why?

#define ARRAY_SIZE 10000
...

clock_t start, end;

start = clock();

for( i = 0; i < ARRAY_SIZE; i++) 
{
non_parallel[i] = vec[i] * vec[i];
}
end = clock();
printf( "Number of seconds: %f\n", (end-start)/(double)CLOCKS_PER_SEC );

So What should i do to measure the time?

Answer

Mysticial picture Mysticial · Apr 14, 2012

Two things:

  1. 10000 is not a lot on a modern computer. Therefore that loop will run in probably less than a millisecond - less than the precision of clock(). Therefore it will return zero.

  2. If you aren't using the result of non_parallel its possible that the entire loop will be optimized out by the compiler.

Most likely, you just need a more expensive loop. Try increasing ARRAY_SIZE to something much larger.


Here's a test on my machine with a larger array size:

#define ARRAY_SIZE 100000000

int main(){

    clock_t start, end;

    double *non_parallel = (double*)malloc(ARRAY_SIZE * sizeof(double));
    double *vec          = (double*)malloc(ARRAY_SIZE * sizeof(double));

    start = clock();

    for(int i = 0; i < ARRAY_SIZE; i++) 
    {
        non_parallel[i] = vec[i] * vec[i];
    }

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


    free(non_parallel);
    free(vec);
    return 0;
}

Output:

Number of seconds: 0.446000