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我正在做一个涉及比较编程语言的项目。我正在计算阿克曼函数。我测试了 Java、Python 和 Ruby，得到了 10 到 30 毫秒的响应。但是 C++ 似乎需要 125 毫秒。这是正常的，还是有问题gettimeofday()？ Gettimeofday()是在时间.h。

我正在（虚拟）Ubuntu Natty Narwhal 32 位上进行测试。我不缺处理能力（四核 2.13 GHz Intel Xeon）。

我的代码在这里：

#include <iostream>
#include <sys/time.h>
using namespace std;
int a(int m,int n) {
    if (m == 0) {
    return n + 1;
    } else if (m > 0 and n == 0) {
    return a(m-1,1);
    } else if (m > 0 and n > 0) {
    return a(m-1,a(m,n-1));
    }
}

int main() {
    timeval tim;
    gettimeofday(&tim,NULL);
    double t1 = tim.tv_usec;
    int v = a(3,4);           
    gettimeofday(&tim,NULL);
    double t2 = tim.tv_usec;
    cout << v << endl << t2-t1;
    return 0;
}
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1 回答 1

11

假设您正在谈论返回数据的分辨率，则状态的 POSIX规范gettimeofday：

系统时钟的分辨率未指定。

这是因为系统在跟踪小时间段方面可能具有广泛变化的能力。甚至 ISO 标准clock()函数也包含这样的警告。

如果您谈论调用它需要多长时间，该标准不保证这些方面的性能。一个实现完全可以免费等待 125分钟，然后再给你时间，尽管我怀疑这样的实现会取得很大的市场成功 :-)

作为有限分辨率的示例，我输入了以下代码以在我的系统上进行检查：

#include <stdio.h>
#include <sys/time.h>

#define NUMBER 30

int main (void) {
    struct timeval tv[NUMBER];
    int count[NUMBER], i, diff;

    gettimeofday (&tv[0], NULL);

    for (i = 1; i < NUMBER; i++) {
        gettimeofday (&tv[i], NULL);
        count[i] = 1;
        while ((tv[i].tv_sec == tv[i-1].tv_sec) &&
            (tv[i].tv_usec == tv[i-1].tv_usec))
        {
            count[i]++;
            gettimeofday (&tv[i], NULL);
        }
    }

    printf ("%2d: secs = %d, usecs = %6d\n", 0, tv[0].tv_sec, tv[0].tv_usec);
    for (i = 1; i < NUMBER; i++) {
        diff = (tv[i].tv_sec - tv[i-1].tv_sec) * 1000000;
        diff += tv[i].tv_usec - tv[i-1].tv_usec;

        printf ("%2d: secs = %d, usecs = %6d, count = %5d, diff = %d\n",
            i, tv[i].tv_sec, tv[i].tv_usec, count[i], diff);
    }

    return 0;
}

代码基本上记录了底层时间的变化，记录了gettimeofday()实际变化所花费的调用次数。这是在一台相当强大的机器上，因此它的处理能力并不短缺（计数表明它能够调用gettimeofday()每个时间量子的频率，大约 5,800 标记，忽略第一个，因为我们不知道我们何时开始在那个量子测量值）。

输出是：

 0: secs = 1318554836, usecs = 990820
 1: secs = 1318554836, usecs = 991820, count =  5129, diff = 1000
 2: secs = 1318554836, usecs = 992820, count =  5807, diff = 1000
 3: secs = 1318554836, usecs = 993820, count =  5901, diff = 1000
 4: secs = 1318554836, usecs = 994820, count =  5916, diff = 1000
 5: secs = 1318554836, usecs = 995820, count =  5925, diff = 1000
 6: secs = 1318554836, usecs = 996820, count =  5814, diff = 1000
 7: secs = 1318554836, usecs = 997820, count =  5814, diff = 1000
 8: secs = 1318554836, usecs = 998820, count =  5819, diff = 1000
 9: secs = 1318554836, usecs = 999820, count =  5901, diff = 1000
10: secs = 1318554837, usecs =    820, count =  5815, diff = 1000
11: secs = 1318554837, usecs =   1820, count =  5866, diff = 1000
12: secs = 1318554837, usecs =   2820, count =  5849, diff = 1000
13: secs = 1318554837, usecs =   3820, count =  5857, diff = 1000
14: secs = 1318554837, usecs =   4820, count =  5867, diff = 1000
15: secs = 1318554837, usecs =   5820, count =  5852, diff = 1000
16: secs = 1318554837, usecs =   6820, count =  5865, diff = 1000
17: secs = 1318554837, usecs =   7820, count =  5867, diff = 1000
18: secs = 1318554837, usecs =   8820, count =  5885, diff = 1000
19: secs = 1318554837, usecs =   9820, count =  5864, diff = 1000
20: secs = 1318554837, usecs =  10820, count =  5918, diff = 1000
21: secs = 1318554837, usecs =  11820, count =  5869, diff = 1000
22: secs = 1318554837, usecs =  12820, count =  5866, diff = 1000
23: secs = 1318554837, usecs =  13820, count =  5875, diff = 1000
24: secs = 1318554837, usecs =  14820, count =  5925, diff = 1000
25: secs = 1318554837, usecs =  15820, count =  5870, diff = 1000
26: secs = 1318554837, usecs =  16820, count =  5877, diff = 1000
27: secs = 1318554837, usecs =  17820, count =  5868, diff = 1000
28: secs = 1318554837, usecs =  18820, count =  5874, diff = 1000
29: secs = 1318554837, usecs =  19820, count =  5862, diff = 1000

表明分辨率似乎限制在不超过一千微秒。当然，您的系统可能与此不同，最重要的是它取决于您的实现和/或环境。

绕过这种限制的一种方法是不要做一次，而是做N几次，然后将经过的时间除以N.

例如，假设您调用您的函数并且计时器说它花费了 125 毫秒，您怀疑这似乎有点高。我建议然后在一个循环中调用它一千次，测量整个一千次所花费的时间。

如果结果是 125 秒，那么，是的，它可能很慢。但是，如果它只需要 27 秒，那将表明您的计时器分辨率是导致看似大时间的原因，因为这相当于每次迭代 27 毫秒，与您从其他结果中看到的相同。

修改您的代码以考虑到这一点将遵循以下原则：

int main() {
    const int count = 1000;
    timeval tim;

    gettimeofday(&tim, NULL);
    double t1 = 1.0e6 * tim.tv_sec + tim.tv_usec;

    int v;
    for (int i = 0; i < count; ++i)
        v = a(3, 4);           

    gettimeofday(&tim, NULL);
    double t2 = 1.0e6 * tim.tv_sec + tim.tv_usec;

    cout << v << '\n' << ((t2 - t1) / count) << '\n';

    return 0;
}
于 2011-10-13T22:06:58.480 回答