Question

我正在做一个涉及比较编程语言的项目。我在计算Ackermann函数。我测试了Java，Python和Ruby，并获得了10到30毫秒的响应。但C ++似乎需要125毫秒。这是正常的，还是gettimeofday()的问题？ Gettimeofday()在time.h。

我正在测试（虚拟）Ubuntu Natty Narwhal 32位。我不是短处理能力（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;
}

Answer 1

假设您正在讨论^（a）的数据的解决方案，gettimeofday州的the POSIX specification：

未指定系统时钟的分辨率。

这是因为系统可能具有广泛变化的跟踪小时段的容量。即使是ISO标准clock()功能也包括这样的警告。

如果您正在讨论如何 long 将其命名为^（a），则该标准不保证这些行的性能。一个实现完全可以等待125 分钟然后给你时间，虽然我怀疑这样的实现会有很大的市场成功： - ）

作为有限分辨率的示例，我输入以下代码进行检查：

#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))
        {
            gettimeofday (&tv[i], NULL);
            count[i]++;
        }
    }

    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;
}

代码基本上记录了基础时间的变化，并计算了实际更改时间gettimeofday()所需的调用次数。这是在我的新款i7 grunter笔记本电脑上，所以它的处理能力并不缺乏（计数表示它能够为每个时间段调用gettimeofday()的频率，达到5,000分）。

输出结果为：

 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

显示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施和/或环境。

^（a）如果你在谈论我没有想过的 else ，你应该更详细地充实你的问题。我们在SO上相当不错，但我们并不是无所不知。

gettimeofday（）C ++不一致

1 个答案: