如何测量C代码的运行时间比较？

Question

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?

Answer 1

If you're running under a system that includes it you can just execute it under time :

$ time ./benchmark1

and

$ time ./benchmark2

Answer 2

#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;
}