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Home/ Questions/Q 7788377
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T20:58:47+00:00 2026-06-01T20:58:47+00:00

This is the algorithm, but when I want to measure the execution time it

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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?

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  1. Editorial Team
    Editorial Team
    2026-06-01T20:58:49+00:00Added an answer on June 1, 2026 at 8:58 pm

    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
    
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