c操作耗时0秒

Question

我正在尝试测量串行程序反转数组所需的时间。我的代码:

#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;
}

但是，它总是需要 0.000000 秒才能运行。为什么？如果我增加 n，就会出现段错误。

Answer 1

尝试多次迭代:

#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;
}

但最好读一下:C++ clock stays zero