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Is there some reason that localtime takes a pointer to time_t instead of just time_t? I can't see that the time_t ever gets modified and the time function returns a time_t not a pointer, so you can't make a call like localtime(time(NULL),&tmStruct).
Just curious...
Function prototipe:
struct tm *localtime(const time_t *timep);
struct tm *localtime_r(const time_t *timep, struct tm *result);
Man page:
The localtime() function converts the calendar time timep to broken-
time representation, expressed relative to the user’s specified time‐
zone. The function acts as if it called tzset(3) and sets the external
variables tzname with information about the current timezone, timezone
with the difference between Coordinated Universal Time (UTC) and local
standard time in seconds, and daylight to a non-zero value if daylight
savings time rules apply during some part of the year. The return
value points to a statically allocated struct which might be overwrit‐
ten by subsequent calls to any of the date and time functions. The
localtime_r() function does the same, but stores the data in a user-
supplied struct. It need not set tzname, timezone, and daylight.
Appreciate the reply, but I had no confusion about the function or how to use it. I was curious if anyone knew the rationale behind passing a pointer to time_t instead of just passing time_t.
in C structs are mostly passed by const pointer to save stack space (as in C++ they are passed by const reference). This Also prevents unneeded copying which can be time-consuming for big-structs
Agreed, possibly time_t is a struct in some implementations (none that I've seen, but...)
nobody wants to change API when internal implementation changes... so APIs for types that could be changed to structs in the future are made as if the types are already structsQuote:
Originally Posted by rjhanby