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tm_to_time.c
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C/C++ Source or Header
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1995-06-12
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1KB
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55 lines
#include <sys/types.h>
#include <sys/time.h>
/* Return 1 if `y' is a leap year, 0 otherwise.
*/
static int leap (y) int y; {
y += 1900;
if (y % 400 == 0)
return (1);
if (y % 100 == 0)
return (0);
return (y % 4 == 0);
}
/* Return the number of days between Jan 1, 1970 and the given
* broken-down time.
*/
static int ndays (p) struct tm *p; {
register n = p->tm_mday;
register m, y;
register char *md = "\37\34\37\36\37\36\37\37\36\37\36\37";
for (y = 70; y < p->tm_year; ++y) {
n += 365;
if (leap (y)) ++n;
}
for (m = 0; m < p->tm_mon; ++m)
n += md[m] + (m == 1 && leap (y));
return (n);
}
/* Convert a broken-down time (such as returned by localtime())
* back into a `time_t'.
*/
time_t tm_to_time (tp) struct tm *tp; {
register int m1, m2;
time_t t;
struct tm otm;
t = (ndays (tp) - 1) * 86400L + tp->tm_hour * 3600L
+ tp->tm_min * 60 + tp->tm_sec;
/*
* Now the hard part -- correct for the time zone:
*/
otm = *tp;
tp = localtime (&t);
m1 = tp->tm_hour * 60 + tp->tm_min;
m2 = otm.tm_hour * 60 + otm.tm_min;
t -= ((m1 - m2 + 720 + 1440) % 1440 - 720) * 60L;
return (t);
}