提交 c5faf1bf 编写于 作者: R Rich Felker

implement week numbers and half of the week-based-year logic for strftime

output for plain week numbers (%U and %W) has been sanity-checked, and
output for the week-based-year week numbers (%V) has been checked
extensively against known-good data for the full non-negative range of
32-bit time_t.

year numbers for week-based years (%g and %G) are not yet implemented.
上级 1e2281b8
......@@ -8,6 +8,14 @@
const char *__langinfo(nl_item);
static int is_leap(int y)
{
/* Avoid overflow */
if (y>INT_MAX-1900) y -= 2000;
y += 1900;
return !(y%4) && ((y%100) || !(y%400));
}
size_t strftime(char *restrict s, size_t n, const char *restrict f, const struct tm *restrict tm)
{
nl_item item;
......@@ -116,10 +124,37 @@ do_fmt:
fmt = "%d";
goto number;
case 'U':
case 'V':
val = (tm->tm_yday + 7 - tm->tm_wday) / 7;
fmt = "%02d";
goto number;
case 'W':
// FIXME: week number mess..
continue;
val = (tm->tm_yday + 7 - (tm->tm_wday+6)%7) / 7;
fmt = "%02d";
goto number;
case 'V':
val = (tm->tm_yday + 7 - (tm->tm_wday+6)%7) / 7;
/* If 1 Jan is just 1-3 days past Monday,
* the previous week is also in this year. */
if ((tm->tm_wday - tm->tm_yday - 2 + 371) % 7 <= 2)
val++;
if (!val) {
val = 52;
/* If 31 December of prev year a Thursday,
* or Friday of a leap year, then the
* prev year has 53 weeks. */
int dec31 = (tm->tm_wday - tm->tm_yday - 1 + 7) % 7;
if (dec31 == 4 || (dec31 == 5 && is_leap(tm->tm_year%400-1)))
val++;
} else if (val == 53) {
/* If 1 January is not a Thursday, and not
* a Wednesday of a leap year, then this
* year has only 52 weeks. */
int jan1 = (tm->tm_wday - tm->tm_yday + 371) % 7;
if (jan1 != 4 && (jan1 != 3 || !is_leap(tm->tm_year)))
val = 1;
}
fmt = "%02d";
goto number;
case 'w':
val = tm->tm_wday;
fmt = "%d";
......
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