1 files changed, 122 insertions, 0 deletions
diff --git a/terran.c b/terran.c
new file mode 100644
index 0000000..e604cd9
--- /dev/null
+++ b/terran.c
@@ -0,0 +1,122 @@
+#include <time.h>
+#include <stdlib.h>
+#include <stdio.h>
+#include <math.h>
+#include <errno.h>
+
+/*
+   terran multiplicative factors between units that are fixed for
+   Lunar, Day, Hour, Minute, Second
+*/
+enum { dl = 28, hd = 24, mh = 60, sm = 60, sh = mh*sm, sd = hd * mh * sm, };
+
+/*
+  There is a long term period of 128 years, that has exactly 31 leap
+  years. This amounts to 46751 days.
+*/
+enum { dy = 365, yp = 128, dp = yp*dy + 31, };
+
+/*
+   Then, periods of four years are grouped together, that have either
+   0 or 1 leap year.
+*/
+enum { yq = 4, dq0 = yq*dy, dq1 = dq0 + 1, };
+
+time_t isotime(char const t[static 1]) {
+  struct tm date[1] = {
+    [0] = { .tm_year = 1900,.tm_mon = 1, },
+  };
+  int retscan = sscanf(t, "%i-%i-%i %i:%i:%i",
+                       &date->tm_year, &date->tm_mon, &date->tm_mday,
+                       &date->tm_hour, &date->tm_min, &date->tm_sec);
+  if (retscan <= 0) {
+    if (errno) perror("can't scan");
+    return 0;
+  }
+  date->tm_year -= 1900;
+  date->tm_mon -= 1;
+  return mktime(date);
+  /* gmtime_r(&ret, date); */
+  /* return mktime(date); */
+}
+
+void time2terran(time_t t) {
+  /* Provide us with the terran epoch. This is not as easy as it may
+     appear as a first site, since it is difficult to express a UTC
+     timestamp in a struct tm: we have convert back and forth with
+     mktime and gmtime to obtain the correct timestamp at 1969-12-22 0:0:0.
+   */
+  struct tm terranEpoch[1] = {
+    {
+      .tm_year = 69, // year starts at 1900
+      .tm_mon = 11,  // mon starts at 0
+      .tm_mday = 23, // mday starts at 1
+    },
+  };
+  time_t epoch = mktime(terranEpoch);
+  struct tm terranEpoch2[1];
+  gmtime_r(&epoch, terranEpoch2);
+  time_t epoch2 = mktime(terranEpoch2);
+  terranEpoch[0].tm_sec += difftime(epoch2, epoch);
+  epoch = mktime(terranEpoch);
+
+  /* Now that we have the appropriate epoch, look how many seconds
+     have passed since then.
+  */
+  long secs = difftime(t, epoch);
+  long days = secs/sd;
+  secs -= days*sd;
+
+  /*
+    adjust with respect to the 128 year periods
+   */
+  long per = days / dp;
+  if (days < 0) {
+    per -= 1;
+  }
+  long year = per * yp;
+  days -= per*dp;
+
+  // the first four years aren't leap years
+  if (days < dq0) {
+    year += (days / dy);
+    days -= (days / dy) * dy;
+  } else {
+    // correct for the first 4 years
+    year += yq;
+    days -= dq0;
+    /*
+       In the remaining period every fourth year starting with year 0
+       is a leap year. Compute the amount of such 4 year periods and
+       adjust
+    */
+    long const qer = days / dq1;
+    year += qer*yq;
+    days %= dq1;
+    // the first year is a leap year
+    if (days > dy+1) {
+      // adjust for that leap year
+      year += 1;
+      days -= (dy+1);
+      // compute the remaining days
+      year += (days / dy);
+      days %= dy;
+    }
+  }
+  size_t const lunar = days / dl;
+  days %= dl;
+  while (secs < 0) secs += sd;
+  size_t hour = secs / sh;
+  secs -= hour*sh;
+  size_t min = secs / sm;
+  secs %= sm;
+  printf("%ld.%ld.%ld,%ld:%ld:%ld TC\n", year, lunar, days, hour, min, secs);
+}
+
+int main (int argc, char* argv[argc+1]) {
+  time_t now = time(0);
+  time2terran(now);
+  for (int i = 1; i < argc; ++i)
+    time2terran(isotime(argv[i]));
+  return EXIT_SUCCESS;
+}