将本地时间转换为时间戳

Question

将由GetLocalTime返回的SYSTEMTIME结构转换为时间戳（自1970年以来的秒数ULONG）会怎样？

Answer 1

一种方法是将参考时间和目标时间都转换为FILETIME结构，然后您可以根据它们的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;