c - Windows API : Get micro seconds between milli seconds

Question

GetSystemTimePreciseAsFileTime:以尽可能高的精度 (<1us) 检索当前系统日期和时间。

这给了我们一个 FILETIME

然后我们将它传递给:文件时间到系统时间这给了我们一个 SYSTEMTIME

但是 SYSTEMTIME 只向我们显示毫秒精度。

typedef struct _SYSTEMTIME {
  WORD wYear;
  WORD wMonth;
  WORD wDayOfWeek;
  WORD wDay;
  WORD wHour;
  WORD wMinute;
  WORD wSecond;
  WORD wMilliseconds;
} SYSTEMTIME, *PSYSTEMTIME, *LPSYSTEMTIME;

那么我如何获得 (<1us) 部分呢？我希望能够打印类似的东西

printf( "milli:[%u] micro:[%u] nano:[%u]\n", 
         st.wMilliseconds, st.wMicro, st.Nano );

谢谢

我试过以下方法:

int main()
{
    FILETIME ft = { 0 };
    SYSTEMTIME st = { 0 };

    UINT32 nsT = 0;
    UINT32 us = 0;
    UINT32 ns = 0;

    for (int i = 0; i < 32; i++)
    {

        GetSystemTimePreciseAsFileTime(&ft);

        FileTimeToSystemTime(&ft, &st);

        nsT = (ft.dwLowDateTime & 0x0FFFFFFFFLL) | (ft.dwHighDateTime << 32LL);

        us = (nsT % 10000LL) / 10LL;
        ns = (nsT % 10LL);

        wprintf(L"DT:[%u%02u%02u %02u:%02u:%02u:%03u:%03u:%u]\n",
            st.wYear, st.wMonth, st.wDay,
            st.wHour, st.wMinute, st.wSecond, st.wMilliseconds, us, ns);
    }
}

但是它会产生乱序的时间

DT:[20190921 12:48:51:152:735:4]
DT:[20190921 12:48:51:164:561:4]
DT:[20190921 12:48:51:169:126:3]
DT:[20190921 12:48:51:172:595:0]
DT:[20190921 12:48:51:173:428:7]
DT:[20190921 12:48:51:173:415:9]
DT:[20190921 12:48:51:177:446:2]
DT:[20190921 12:48:51:181:309:8]
DT:[20190921 12:48:51:182:386:7]
DT:[20190921 12:48:51:184:650:6]
DT:[20190921 12:48:51:184:740:6]
DT:[20190921 12:48:51:189:271:8]
DT:[20190921 12:48:51:190:471:4]
DT:[20190921 12:48:51:194:746:6]
DT:[20190921 12:48:51:197:049:5]
DT:[20190921 12:48:51:198:793:4]
DT:[20190921 12:48:51:200:823:4]
DT:[20190921 12:48:51:200:529:1]
DT:[20190921 12:48:51:202:606:3]
DT:[20190921 12:48:51:210:913:4]
DT:[20190921 12:48:51:214:764:6]
DT:[20190921 12:48:51:214:815:4]
DT:[20190921 12:48:51:219:079:4]
DT:[20190921 12:48:51:222:580:2]
DT:[20190921 12:48:51:223:959:5]
DT:[20190921 12:48:51:224:933:8]
DT:[20190921 12:48:51:225:844:7]
DT:[20190921 12:48:51:226:810:7]
DT:[20190921 12:48:51:228:811:1]
DT:[20190921 12:48:51:230:139:1]
DT:[20190921 12:48:51:253:539:1]
DT:[20190921 12:48:51:253:653:9]

请注意以下对于给定的毫 (173)，我们有递减的微单位(428 到 415)415 应该大于 428。

DT:[20190921 12:48:51:173:428:7]
DT:[20190921 12:48:51:173:415:9]

DT:[20190921 12:48:51:200:823:4]
DT:[20190921 12:48:51:200:529:1]

然后我将字符串冲刺到全局数组并在后续循环中打印它以确保没有标准输出竞争条件，但它仍然在更精细的范围内乱序:

DT:[20190921 13:05:01:228:534:1]
DT:[20190921 13:05:01:228:534:3]
DT:[20190921 13:05:01:228:534:3]
DT:[20190921 13:05:01:228:534:1]
DT:[20190921 13:05:01:228:534:3]
DT:[20190921 13:05:01:228:534:1]
DT:[20190921 13:05:01:228:534:3]

如果我添加一些代码来比较计算的毫秒与 ST.wMilliseconds，我们会发现不一致:

ms:[140][0x8C] != st.wMilliseconds:[177][0xB1]
ms:[284][0x11C] != st.wMilliseconds:[184][0xB8]
ms:[337][0x151] != st.wMilliseconds:[236][0xEC]

感谢 Adrian，以下代码似乎可以完美运行:

int main()
{
    WCHAR b[MAX][256] = { 0 };

    FILETIME ft = { 0 };
    SYSTEMTIME st = { 0 };

    UINT64 ftLo = 0;
    UINT64 ftHi = 0;
    UINT64 myTime = 0;
    UINT64 ms = 0;
    UINT64 us = 0;
    UINT64 ns = 0;

    for (int i = 0; i < MAX; i++)
    {
        GetSystemTimePreciseAsFileTime(&ft);

        FileTimeToSystemTime(&ft, &st);

        ftLo = ft.dwLowDateTime;
        ftHi = ft.dwHighDateTime;
        myTime = ftLo | (ftHi << 32uLL);

        ms = (myTime % 10000000uLL) / 10000uLL;
        us = (myTime % 10000uLL) / 10uLL;
        ns = (myTime % 10uLL);

        if (ms != st.wMilliseconds)
            wprintf(L"ms:[%llu][0x%llX] != st.wMilliseconds:[%u][0x%X]\n",
                    ms, ms, st.wMilliseconds, st.wMilliseconds);

        wprintf(L"%u%02u%02u %02u:%02u:%02u:%03llu:%03llu:%llu\n",
            st.wYear, st.wMonth, st.wDay,
            st.wHour, st.wMinute, st.wSecond,
            ms, us, ns);


    }

    return 0;
}

结果:

20190922 10:51:20:625:917:4
20190922 10:51:20:626:065:5
20190922 10:51:20:628:774:2
20190922 10:51:20:631:084:5
20190922 10:51:20:631:628:4
20190922 10:51:20:632:223:0
20190922 10:51:20:635:252:7
20190922 10:51:20:637:841:8
20190922 10:51:20:641:058:9
20190922 10:51:20:649:869:0
20190922 10:51:20:651:283:0
20190922 10:51:20:652:677:9
20190922 10:51:20:652:749:7
20190922 10:51:20:652:808:6
20190922 10:51:20:654:621:0
20190922 10:51:20:656:662:3
20190922 10:51:20:659:048:9
20190922 10:51:20:660:926:0
20190922 10:51:20:664:202:1
20190922 10:51:20:666:506:9
20190922 10:51:20:666:592:6
20190922 10:51:20:666:712:8
20190922 10:51:20:670:800:7
20190922 10:51:20:680:442:7
20190922 10:51:20:680:522:0
20190922 10:51:20:681:207:1
20190922 10:51:20:682:988:3
20190922 10:51:20:684:476:8
20190922 10:51:20:685:727:1
20190922 10:51:20:685:777:0
20190922 10:51:20:686:526:3
20190922 10:51:20:686:919:9

Answer 1

正如您所注意到的，您不能通过使用 SYSTEMTIME 结构来执行此操作。您需要从返回的 FILETIME 结构中提取数据并将其作为 64 位整数处理:

FILETIME fTime;
GetSystemTimePreciseAsFileTime(&fTime);
uint64_t ftLo = uint64_t(fTime.dwLowDateTime);
uint64_t ftHi = uint64_t(fTime.dwHighDateTime);
uint64_t myTime = ftLo | (ftHi << 32uLL);

myTime 变量将保存自 1601 年 1 月 1 日以来 100 纳秒间隔的数量。

要从这个“epoch”值获取 SYSTEMTIME 中“缺失”的 us 和 ns 字段，您可以简单地使用下面的代码。您还应该获得自己的毫秒值，以防转换为 SYSTEMTIME 时“舍入”了 dwMilliseconds 值(不知道是否会发生这种情况):

uint64_t millis = (myTime % 10000000uLL) / 10000uLL;
uint64_t micros = (myTime % 10000uLL) / 10uLL;
uint64_t nanos = (myTime % 10uLL) * 100uLL; // Will have a resolution of only 100ns

(假设自 1601 年 1 月 1 日以来的所有天数都是整数毫秒，我认为这是一个合理的假设!)

注意:另一种选择是将 FILETIME 结构复制(或强制转换)为 ULARGE_INTEGER，然后分配该 union 的 QuadPart 元素到 我的时间。这将也有效，但是，在标准 C++ 中，写入 union 的一部分然后从另一部分读取在形式上是未定义的行为。