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Home/ Questions/Q 7195927
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T20:40:42+00:00 2026-05-28T20:40:42+00:00

Example code 1: const int N=100000; for(int j=0;j<N;j++){ arr1[j] += a1[j]; arr2[j] += a2[j];

  • 0

Example code 1:

const int N=100000;

for(int j=0;j<N;j++){
    arr1[j] += a1[j];
    arr2[j] += a2[j];
}

Example code 2:

for(int j=0;j<N;j++){
    arr1[j] += a1[j];
}
for(int j=0;j<N;j++){
    arr2[j] += a2[j];
}

I need to calculate the running time of these block of code. Is there any tool(benchmark) to calculate it?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-28T20:40:43+00:00Added an answer on May 28, 2026 at 8:40 pm
    #include <sys/time.h>
    #include <stdio.h>
    #include <unistd.h>
    
    const int N=100000;
    
    void time_first() {
      struct timeval start, mid, end;
      long mtime, seconds, useconds;    
    
      gettimeofday(&start, NULL);
      for(int j=0;j<N;j++){
        arr1[j] += a1[j];
        arr2[j] += a2[j];
      }  
      gettimeofday(&end, NULL);
    
      seconds  = end.tv_sec  - start.tv_sec;
      useconds = end.tv_usec - start.tv_usec;
    
      mtime = ((seconds) * 1000 + useconds/1000.0) + 0.5;
    
      printf("First elapsed time: %ld milliseconds\n", mtime);
    }
    
    void time_second() {
      struct timeval start, mid, end;
      long mtime, seconds, useconds;    
    
      gettimeofday(&start, NULL);
      for(int j=0;j<N;j++){
        arr1[j] += a1[j];
      }
      for(int j=0;j<N;j++){
        arr2[j] += a2[j];
      }
      gettimeofday(&end, NULL);
    
      seconds  = end.tv_sec  - start.tv_sec;
      useconds = end.tv_usec - start.tv_usec;
    
      mtime = ((seconds) * 1000 + useconds/1000.0) + 0.5;
    
      printf("Second elapsed time: %ld milliseconds\n", mtime);
    }
    
    int main() {
      initialize arr1, a1 and a2
    
      time_first();
      time_second();
      return 0;
    }
    
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