这就是算法，但是当我要测量的执行时间，它给了我零。为什么呢？

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 );

所以，我应该怎么做来测量时间？

So What should i do to measure the time?

推荐答案

两件事情：

10000 不是很多现代的计算机上。因此，该循环将运行大概不到一毫秒 - 比时钟precision以下（）。因此，它会返回零。

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.

如果您不使用 non_parallel 的可能，整个循环将被优化掉的编译器。

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

最有可能的，你只需要一个更昂贵的循环。尝试增加 ARRAY_SIZE 的东西要多得多。

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

这是一个测试我的机器上有一个较大的数组大小：

#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;
}

输出：

Number of seconds: 0.446000