将 UNIX 纪元时间转换为 SYSTEMTIME

Question

在我的 C++ Windows 应用程序中，我将 SYSTEMTIME 转换为格式化字符串，如下所示：

    SYSTEMTIME systemTime;
    GetSystemTime(&systemTime);
    char pSystemTime[50];
    sprintf_s(pSystemTime, 50, "%04d%02d%02d%02d%02d%02d%03u\0", systemTime.wYear, systemTime.wMonth,
                                                                              systemTime.wDay, systemTime.wHour,
                                                                              systemTime.wMinute, systemTime.wSecond,
                                                                              systemTime.wMilliseconds);

现在我有一个自 UNIX 纪元以来以毫秒为单位的时间。

long time = 1442524186;  //for example

如何将这么长的纪元时间转换为 SYSTEMTIME 以便我也可以将其格式化为字符串？

Answer 1

static UINT64 FileTimeToMillis(const FILETIME &ft)
{
    ULARGE_INTEGER uli;
    uli.LowPart = ft.dwLowDateTime; // could use memcpy here!
    uli.HighPart = ft.dwHighDateTime;

    return static_cast<UINT64>(uli.QuadPart/10000);
}

static void MillisToSystemTime(UINT64 millis, SYSTEMTIME *st)
{
    UINT64 multiplier = 10000;
    UINT64 t = multiplier * millis;

    ULARGE_INTEGER li;
    li.QuadPart = t;
    // NOTE, DON'T have to do this any longer because we're putting
    // in the 64bit UINT directly
    //li.LowPart = static_cast<DWORD>(t & 0xFFFFFFFF);
    //li.HighPart = static_cast<DWORD>(t >> 32);

    FILETIME ft;
    ft.dwLowDateTime = li.LowPart;
    ft.dwHighDateTime = li.HighPart;

    ::FileTimeToSystemTime(&ft, st);
}

来源：https ://stackoverflow.com/a/11123106/2385309

编辑：另请参阅：https ://stackoverflow.com/a/26486180/2385309 您可能需要为 Epoch 时间毫秒添加/减去 11644473600000