在不使用 VariantTimeToSystemTime 的情况下将 OLE 自动化日期 (OADate) double 转换为 struct tm
Convert OLE Automation Date (OADate) double to struct tm without using VariantTimeToSystemTime
我主要用标准 C++ (VS2010) 编写一个 Windows DLL，它不使用 MFC/ATL。
一个 parent 模块确实使用 MFC 并将 COleDateTime.m_dt 传递到我的 DLL，它作为 double 到达。我相信这是一个 OLE 自动化日期，也称为 OADate。
我想将其转换为任何类型的标准结构 (tm...)，无需将 MFC、OLE 等拉入我的 DLL。
之前有人问过这个问题 (Convert Date/Time (as Double) to struct* tm in C++) 但是，答案总是使用 VariantTimeToSystemTime()，这忽略了问题的重点 - 不使用 MFC / OLE 等
VariantTimeToSystemTime 的要求是：
Header - OleAuto.h
图书馆 - OleAut32.lib
DLL-OleAut32.dll
目前我的 DLL 基本上没有任何依赖关系，所以我不想将 OleAut32.dll 拉入此转换。
到目前为止，我发现的最好的东西是这个 C# mono code，我可以将其转换为 C++。
我有 2 个解决方案，第一个是使用实现 gmtime_r 的函数，这样这个解决方案就不会使用任何标准函数。第二种解决方案是使用标准函数 gmtime_r.
1.第一个解决方案：自己实现 gmtime_r（01-Jan-1601 到 31-Dec-9999）：
它适用于 01-Jan-1601 和 31-Dec-9999 之间的日期。我已经实现了一个 fromOADate 函数，该函数使用此 answer on SO 中的 SecondsSinceEpochToDateTime 函数，将 01-Jan-1970 之前或之后的秒数转换为 tm 结构，但仅适用于01-Jan-1601 上。
我通过添加一个 ULL 后缀更改了该答案的功能，使其也适用于 32 位。这要求 long long 类型为 64 位宽，否则此解决方案将不起作用。
如果您需要 1601 年之前的日期，您可以更改 SecondsSinceEpochToDateTime，因为它有据可查。
要测试不同的值，这个 online conversion 非常好，它还支持 unix 时间戳和 OADate 类型。
完整的工作代码和example on ideone:
#include <iostream>
#include <ctime>
#include <cstring>
struct tm* SecondsSinceEpochToDateTime(struct tm* pTm, uint64_t SecondsSinceEpoch)
{
uint64_t sec;
unsigned int quadricentennials, centennials, quadrennials, annuals/*1-ennial?*/;
unsigned int year, leap;
unsigned int yday, hour, min;
unsigned int month, mday, wday;
static const unsigned int daysSinceJan1st[2][13]=
{
{0,31,59,90,120,151,181,212,243,273,304,334,365}, // 365 days, non-leap
{0,31,60,91,121,152,182,213,244,274,305,335,366} // 366 days, leap
};
/*
400 years:
1st hundred, starting immediately after a leap year that's a multiple of 400:
n n n l \
n n n l } 24 times
... /
n n n l /
n n n n
2nd hundred:
n n n l \
n n n l } 24 times
... /
n n n l /
n n n n
3rd hundred:
n n n l \
n n n l } 24 times
... /
n n n l /
n n n n
4th hundred:
n n n l \
n n n l } 24 times
... /
n n n l /
n n n L <- 97'th leap year every 400 years
*/
// Re-bias from 1970 to 1601:
// 1970 - 1601 = 369 = 3*100 + 17*4 + 1 years (incl. 89 leap days) =
// (3*100*(365+24/100) + 17*4*(365+1/4) + 1*365)*24*3600 seconds
sec = SecondsSinceEpoch + 11644473600ULL;
wday = (uint)((sec / 86400 + 1) % 7); // day of week
// Remove multiples of 400 years (incl. 97 leap days)
quadricentennials = (uint)(sec / 12622780800ULL); // 400*365.2425*24*3600
sec %= 12622780800ULL;
// Remove multiples of 100 years (incl. 24 leap days), can't be more than 3
// (because multiples of 4*100=400 years (incl. leap days) have been removed)
centennials = (uint)(sec / 3155673600ULL); // 100*(365+24/100)*24*3600
if (centennials > 3)
{
centennials = 3;
}
sec -= centennials * 3155673600ULL;
// Remove multiples of 4 years (incl. 1 leap day), can't be more than 24
// (because multiples of 25*4=100 years (incl. leap days) have been removed)
quadrennials = (uint)(sec / 126230400); // 4*(365+1/4)*24*3600
if (quadrennials > 24)
{
quadrennials = 24;
}
sec -= quadrennials * 126230400ULL;
// Remove multiples of years (incl. 0 leap days), can't be more than 3
// (because multiples of 4 years (incl. leap days) have been removed)
annuals = (uint)(sec / 31536000); // 365*24*3600
if (annuals > 3)
{
annuals = 3;
}
sec -= annuals * 31536000ULL;
// Calculate the year and find out if it's leap
year = 1601 + quadricentennials * 400 + centennials * 100 + quadrennials * 4 + annuals;
leap = !(year % 4) && (year % 100 || !(year % 400));
// Calculate the day of the year and the time
yday = sec / 86400;
sec %= 86400;
hour = sec / 3600;
sec %= 3600;
min = sec / 60;
sec %= 60;
// Calculate the month
for (mday = month = 1; month < 13; month++)
{
if (yday < daysSinceJan1st[leap][month])
{
mday += yday - daysSinceJan1st[leap][month - 1];
break;
}
}
// Fill in C's "struct tm"
memset(pTm, 0, sizeof(*pTm));
pTm->tm_sec = sec; // [0,59]
pTm->tm_min = min; // [0,59]
pTm->tm_hour = hour; // [0,23]
pTm->tm_mday = mday; // [1,31] (day of month)
pTm->tm_mon = month - 1; // [0,11] (month)
pTm->tm_year = year - 1900; // 70+ (year since 1900)
pTm->tm_wday = wday; // [0,6] (day since Sunday AKA day of week)
pTm->tm_yday = yday; // [0,365] (day since January 1st AKA day of year)
pTm->tm_isdst = -1; // daylight saving time flag
return pTm;
}
struct tm* fromOADate(struct tm* p_Tm, double p_OADate)
{
static const int64_t OA_UnixTimestamp = -2209161600; /* 30-Dec-1899 */
if (!( -109205 <= p_OADate /* 01-Jan-1601 */
&& p_OADate <= 2958465)) /* 31-Dec-9999 */
{
throw std::string("OADate must be between 109205 and 2958465!");
}
int64_t OADatePassedDays = p_OADate;
double OADateDayTime = p_OADate - OADatePassedDays;
int64_t OADateSeconds = OA_UnixTimestamp
+ OADatePassedDays * 24LL * 3600LL
+ OADateDayTime * 24.0 * 3600.0;
return SecondsSinceEpochToDateTime(p_Tm, OADateSeconds);
}
int main()
{
struct tm timeVal;
std::cout << asctime(fromOADate(&timeVal, -109205)); /* 01-Jan-1601 00:00:00 */
std::cout << asctime(fromOADate(&timeVal, 0)); /* 30-Dec-1899 00:00:00 */
std::cout << asctime(fromOADate(&timeVal, 25569)); /* 01-Jan-1970 00:00:00 */
std::cout << asctime(fromOADate(&timeVal, 50424.134803241)); /* 19-Jan-2038 03:14:07 */
std::cout << asctime(fromOADate(&timeVal, 2958465)); /* 31-Dec-9999 00:00:00 */
return 0;
}
2。第二种解决方案：使用 gmtime_r（01-Jan-1970 到 19-Jan-2038/31-Dec-9999（32/64 位））：
如前所述，此解决方案的范围不如上面的变体，而是仅使用标准函数 (full working example at ideone)：
#include <iostream>
#include <ctime>
struct tm* fromOADate(struct tm* p_Tm, double p_OADate)
{
static const int64_t OA_UnixTimestamp = -2209161600; /* 30-Dec-1899 */
if (!( 25569 <= p_OADate /* 01-Jan-1970 00:00:00 */
&& p_OADate <= 2958465)) /* 31-Dec-9999 00:00:00 */
{
throw std::string("OADate must be between 25569 and 2958465!");
}
time_t OADatePassedDays = p_OADate;
double OADateDayTime = p_OADate - OADatePassedDays;
time_t OADateSeconds = OA_UnixTimestamp
+ OADatePassedDays * 24LL * 3600LL
+ OADateDayTime * 24.0 * 3600.0;
/* date was greater than 19-Jan-2038 and build is 32 bit */
if (0 > OADateSeconds)
{
throw std::string("OADate must be between 25569 and 50424.134803241!");
}
return gmtime_r(&OADateSeconds, p_Tm);
}
int main()
{
struct tm timeVal;
std::cout << asctime(fromOADate(&timeVal, 25569)); /* 01-Jan-1970 00:00:00 */
std::cout << asctime(fromOADate(&timeVal, 50424.134803241)); /* 19-Jan-2038 03:14:07 */
return 0;
}
我主要用标准 C++ (VS2010) 编写一个 Windows DLL，它不使用 MFC/ATL。
一个 parent 模块确实使用 MFC 并将 COleDateTime.m_dt 传递到我的 DLL，它作为 double 到达。我相信这是一个 OLE 自动化日期，也称为 OADate。
我想将其转换为任何类型的标准结构 (tm...)，无需将 MFC、OLE 等拉入我的 DLL。
之前有人问过这个问题 (Convert Date/Time (as Double) to struct* tm in C++) 但是，答案总是使用 VariantTimeToSystemTime()，这忽略了问题的重点 - 不使用 MFC / OLE 等
VariantTimeToSystemTime 的要求是：
Header - OleAuto.h
图书馆 - OleAut32.lib
DLL-OleAut32.dll
目前我的 DLL 基本上没有任何依赖关系，所以我不想将 OleAut32.dll 拉入此转换。
到目前为止，我发现的最好的东西是这个 C# mono code，我可以将其转换为 C++。
我有 2 个解决方案，第一个是使用实现 gmtime_r 的函数，这样这个解决方案就不会使用任何标准函数。第二种解决方案是使用标准函数 gmtime_r.
1.第一个解决方案：自己实现 gmtime_r（01-Jan-1601 到 31-Dec-9999）：
它适用于 01-Jan-1601 和 31-Dec-9999 之间的日期。我已经实现了一个 fromOADate 函数，该函数使用此 answer on SO 中的 SecondsSinceEpochToDateTime 函数，将 01-Jan-1970 之前或之后的秒数转换为 tm 结构，但仅适用于01-Jan-1601 上。
我通过添加一个 ULL 后缀更改了该答案的功能，使其也适用于 32 位。这要求 long long 类型为 64 位宽，否则此解决方案将不起作用。
如果您需要 1601 年之前的日期，您可以更改 SecondsSinceEpochToDateTime，因为它有据可查。
要测试不同的值，这个 online conversion 非常好，它还支持 unix 时间戳和 OADate 类型。
完整的工作代码和example on ideone:
#include <iostream>
#include <ctime>
#include <cstring>
struct tm* SecondsSinceEpochToDateTime(struct tm* pTm, uint64_t SecondsSinceEpoch)
{
uint64_t sec;
unsigned int quadricentennials, centennials, quadrennials, annuals/*1-ennial?*/;
unsigned int year, leap;
unsigned int yday, hour, min;
unsigned int month, mday, wday;
static const unsigned int daysSinceJan1st[2][13]=
{
{0,31,59,90,120,151,181,212,243,273,304,334,365}, // 365 days, non-leap
{0,31,60,91,121,152,182,213,244,274,305,335,366} // 366 days, leap
};
/*
400 years:
1st hundred, starting immediately after a leap year that's a multiple of 400:
n n n l \
n n n l } 24 times
... /
n n n l /
n n n n
2nd hundred:
n n n l \
n n n l } 24 times
... /
n n n l /
n n n n
3rd hundred:
n n n l \
n n n l } 24 times
... /
n n n l /
n n n n
4th hundred:
n n n l \
n n n l } 24 times
... /
n n n l /
n n n L <- 97'th leap year every 400 years
*/
// Re-bias from 1970 to 1601:
// 1970 - 1601 = 369 = 3*100 + 17*4 + 1 years (incl. 89 leap days) =
// (3*100*(365+24/100) + 17*4*(365+1/4) + 1*365)*24*3600 seconds
sec = SecondsSinceEpoch + 11644473600ULL;
wday = (uint)((sec / 86400 + 1) % 7); // day of week
// Remove multiples of 400 years (incl. 97 leap days)
quadricentennials = (uint)(sec / 12622780800ULL); // 400*365.2425*24*3600
sec %= 12622780800ULL;
// Remove multiples of 100 years (incl. 24 leap days), can't be more than 3
// (because multiples of 4*100=400 years (incl. leap days) have been removed)
centennials = (uint)(sec / 3155673600ULL); // 100*(365+24/100)*24*3600
if (centennials > 3)
{
centennials = 3;
}
sec -= centennials * 3155673600ULL;
// Remove multiples of 4 years (incl. 1 leap day), can't be more than 24
// (because multiples of 25*4=100 years (incl. leap days) have been removed)
quadrennials = (uint)(sec / 126230400); // 4*(365+1/4)*24*3600
if (quadrennials > 24)
{
quadrennials = 24;
}
sec -= quadrennials * 126230400ULL;
// Remove multiples of years (incl. 0 leap days), can't be more than 3
// (because multiples of 4 years (incl. leap days) have been removed)
annuals = (uint)(sec / 31536000); // 365*24*3600
if (annuals > 3)
{
annuals = 3;
}
sec -= annuals * 31536000ULL;
// Calculate the year and find out if it's leap
year = 1601 + quadricentennials * 400 + centennials * 100 + quadrennials * 4 + annuals;
leap = !(year % 4) && (year % 100 || !(year % 400));
// Calculate the day of the year and the time
yday = sec / 86400;
sec %= 86400;
hour = sec / 3600;
sec %= 3600;
min = sec / 60;
sec %= 60;
// Calculate the month
for (mday = month = 1; month < 13; month++)
{
if (yday < daysSinceJan1st[leap][month])
{
mday += yday - daysSinceJan1st[leap][month - 1];
break;
}
}
// Fill in C's "struct tm"
memset(pTm, 0, sizeof(*pTm));
pTm->tm_sec = sec; // [0,59]
pTm->tm_min = min; // [0,59]
pTm->tm_hour = hour; // [0,23]
pTm->tm_mday = mday; // [1,31] (day of month)
pTm->tm_mon = month - 1; // [0,11] (month)
pTm->tm_year = year - 1900; // 70+ (year since 1900)
pTm->tm_wday = wday; // [0,6] (day since Sunday AKA day of week)
pTm->tm_yday = yday; // [0,365] (day since January 1st AKA day of year)
pTm->tm_isdst = -1; // daylight saving time flag
return pTm;
}
struct tm* fromOADate(struct tm* p_Tm, double p_OADate)
{
static const int64_t OA_UnixTimestamp = -2209161600; /* 30-Dec-1899 */
if (!( -109205 <= p_OADate /* 01-Jan-1601 */
&& p_OADate <= 2958465)) /* 31-Dec-9999 */
{
throw std::string("OADate must be between 109205 and 2958465!");
}
int64_t OADatePassedDays = p_OADate;
double OADateDayTime = p_OADate - OADatePassedDays;
int64_t OADateSeconds = OA_UnixTimestamp
+ OADatePassedDays * 24LL * 3600LL
+ OADateDayTime * 24.0 * 3600.0;
return SecondsSinceEpochToDateTime(p_Tm, OADateSeconds);
}
int main()
{
struct tm timeVal;
std::cout << asctime(fromOADate(&timeVal, -109205)); /* 01-Jan-1601 00:00:00 */
std::cout << asctime(fromOADate(&timeVal, 0)); /* 30-Dec-1899 00:00:00 */
std::cout << asctime(fromOADate(&timeVal, 25569)); /* 01-Jan-1970 00:00:00 */
std::cout << asctime(fromOADate(&timeVal, 50424.134803241)); /* 19-Jan-2038 03:14:07 */
std::cout << asctime(fromOADate(&timeVal, 2958465)); /* 31-Dec-9999 00:00:00 */
return 0;
}
2。第二种解决方案：使用 gmtime_r（01-Jan-1970 到 19-Jan-2038/31-Dec-9999（32/64 位））：
如前所述，此解决方案的范围不如上面的变体，而是仅使用标准函数 (full working example at ideone)：
#include <iostream>
#include <ctime>
struct tm* fromOADate(struct tm* p_Tm, double p_OADate)
{
static const int64_t OA_UnixTimestamp = -2209161600; /* 30-Dec-1899 */
if (!( 25569 <= p_OADate /* 01-Jan-1970 00:00:00 */
&& p_OADate <= 2958465)) /* 31-Dec-9999 00:00:00 */
{
throw std::string("OADate must be between 25569 and 2958465!");
}
time_t OADatePassedDays = p_OADate;
double OADateDayTime = p_OADate - OADatePassedDays;
time_t OADateSeconds = OA_UnixTimestamp
+ OADatePassedDays * 24LL * 3600LL
+ OADateDayTime * 24.0 * 3600.0;
/* date was greater than 19-Jan-2038 and build is 32 bit */
if (0 > OADateSeconds)
{
throw std::string("OADate must be between 25569 and 50424.134803241!");
}
return gmtime_r(&OADateSeconds, p_Tm);
}
int main()
{
struct tm timeVal;
std::cout << asctime(fromOADate(&timeVal, 25569)); /* 01-Jan-1970 00:00:00 */
std::cout << asctime(fromOADate(&timeVal, 50424.134803241)); /* 19-Jan-2038 03:14:07 */
return 0;
}