如何将GetLocalTime返回的SYSTEMTIME结构转换为时间戳(自1970年以来的秒数,ULONG)?
答案 0 :(得分:2)
一种方法是将参考时间和目标时间都转换为FILETIME
结构,然后让您对其值执行简单的算术运算。这些以100ns为单位进行测量,因此您需要将最终结果除以10000000以获得秒数。
// reference time, convert to FILETIME
SYSTEMTIME stRef{};
stRef.wYear = 1970;
stRef.wMonth = 1;
stRef.wDay = 1;
FILETIME ftRef;
SystemTimeToFileTime(&stRef, &ftRef);
// target time, convert to FILETIME
SYSTEMTIME stTarget;
GetLocalTime(&stTarget);
FILETIME ftTarget;
SystemTimeToFileTime(&stTarget, &ftTarget);
// convert both to ULARGE_INTEGER
ULARGE_INTEGER ulRef, ulTarget;
ulRef.HighPart = ftRef.dwHighDateTime;
ulRef.LowPart = ftRef.dwLowDateTime;
ulTarget.HighPart = ftTarget.dwHighDateTime;
ulTarget.LowPart = ftTarget.dwLowDateTime;
// subtract reference time from target time
ulTarget.QuadPart -= ulRef.QuadPart;
// convert to seconds (divide by 10000000)
DWORD dwSecondsSinceRefTime = ulTarget.QuadPart / 10000000i64;