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diff --git a/kernel/kernel/time/timeconv.c b/kernel/kernel/time/timeconv.c
new file mode 100644
index 000000000..86628e755
--- /dev/null
+++ b/kernel/kernel/time/timeconv.c
@@ -0,0 +1,127 @@
+/*
+ * Copyright (C) 1993, 1994, 1995, 1996, 1997 Free Software Foundation, Inc.
+ * This file is part of the GNU C Library.
+ * Contributed by Paul Eggert (eggert@twinsun.com).
+ *
+ * The GNU C Library is free software; you can redistribute it and/or
+ * modify it under the terms of the GNU Library General Public License as
+ * published by the Free Software Foundation; either version 2 of the
+ * License, or (at your option) any later version.
+ *
+ * The GNU C Library is distributed in the hope that it will be useful,
+ * but WITHOUT ANY WARRANTY; without even the implied warranty of
+ * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
+ * Library General Public License for more details.
+ *
+ * You should have received a copy of the GNU Library General Public
+ * License along with the GNU C Library; see the file COPYING.LIB.  If not,
+ * write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
+ * Boston, MA 02111-1307, USA.
+ */
+
+/*
+ * Converts the calendar time to broken-down time representation
+ * Based on code from glibc-2.6
+ *
+ * 2009-7-14:
+ *   Moved from glibc-2.6 to kernel by Zhaolei<zhaolei@cn.fujitsu.com>
+ */
+
+#include <linux/time.h>
+#include <linux/module.h>
+
+/*
+ * Nonzero if YEAR is a leap year (every 4 years,
+ * except every 100th isn't, and every 400th is).
+ */
+static int __isleap(long year)
+{
+	return (year) % 4 == 0 && ((year) % 100 != 0 || (year) % 400 == 0);
+}
+
+/* do a mathdiv for long type */
+static long math_div(long a, long b)
+{
+	return a / b - (a % b < 0);
+}
+
+/* How many leap years between y1 and y2, y1 must less or equal to y2 */
+static long leaps_between(long y1, long y2)
+{
+	long leaps1 = math_div(y1 - 1, 4) - math_div(y1 - 1, 100)
+		+ math_div(y1 - 1, 400);
+	long leaps2 = math_div(y2 - 1, 4) - math_div(y2 - 1, 100)
+		+ math_div(y2 - 1, 400);
+	return leaps2 - leaps1;
+}
+
+/* How many days come before each month (0-12). */
+static const unsigned short __mon_yday[2][13] = {
+	/* Normal years. */
+	{0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365},
+	/* Leap years. */
+	{0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366}
+};
+
+#define SECS_PER_HOUR	(60 * 60)
+#define SECS_PER_DAY	(SECS_PER_HOUR * 24)
+
+/**
+ * time_to_tm - converts the calendar time to local broken-down time
+ *
+ * @totalsecs	the number of seconds elapsed since 00:00:00 on January 1, 1970,
+ *		Coordinated Universal Time (UTC).
+ * @offset	offset seconds adding to totalsecs.
+ * @result	pointer to struct tm variable to receive broken-down time
+ */
+void time_to_tm(time_t totalsecs, int offset, struct tm *result)
+{
+	long days, rem, y;
+	const unsigned short *ip;
+
+	days = totalsecs / SECS_PER_DAY;
+	rem = totalsecs % SECS_PER_DAY;
+	rem += offset;
+	while (rem < 0) {
+		rem += SECS_PER_DAY;
+		--days;
+	}
+	while (rem >= SECS_PER_DAY) {
+		rem -= SECS_PER_DAY;
+		++days;
+	}
+
+	result->tm_hour = rem / SECS_PER_HOUR;
+	rem %= SECS_PER_HOUR;
+	result->tm_min = rem / 60;
+	result->tm_sec = rem % 60;
+
+	/* January 1, 1970 was a Thursday. */
+	result->tm_wday = (4 + days) % 7;
+	if (result->tm_wday < 0)
+		result->tm_wday += 7;
+
+	y = 1970;
+
+	while (days < 0 || days >= (__isleap(y) ? 366 : 365)) {
+		/* Guess a corrected year, assuming 365 days per year. */
+		long yg = y + math_div(days, 365);
+
+		/* Adjust DAYS and Y to match the guessed year. */
+		days -= (yg - y) * 365 + leaps_between(y, yg);
+		y = yg;
+	}
+
+	result->tm_year = y - 1900;
+
+	result->tm_yday = days;
+
+	ip = __mon_yday[__isleap(y)];
+	for (y = 11; days < ip[y]; y--)
+		continue;
+	days -= ip[y];
+
+	result->tm_mon = y;
+	result->tm_mday = days + 1;
+}
+EXPORT_SYMBOL(time_to_tm);