+/* We don't use `isdigit' here since the locale dependent
+ interpretation is not what we want here. We only need to accept
+ the arabic digits in the ASCII range. One day there is perhaps a
+ more reliable way to accept other sets of digits. */
+#define ISDIGIT(Ch) ((unsigned int) (Ch) - '0' <= 9)
+
+static char *memcpy_lowcase __P ((char *dest, const char *src, size_t len));
+
+static char *
+memcpy_lowcase (dest, src, len)
+ char *dest;
+ const char *src;
+ size_t len;
+{
+ while (len-- > 0)
+ dest[len] = TOLOWER (src[len]);
+ return dest;
+}
+
+static char *memcpy_uppcase __P ((char *dest, const char *src, size_t len));
+
+static char *
+memcpy_uppcase (dest, src, len)
+ char *dest;
+ const char *src;
+ size_t len;
+{
+ while (len-- > 0)
+ dest[len] = TOUPPER (src[len]);
+ return dest;
+}
+
+#if ! HAVE_TM_GMTOFF
+/* Yield the difference between *A and *B,
+ measured in seconds, ignoring leap seconds. */
+static int tm_diff __P ((const struct tm *, const struct tm *));
+static int
+tm_diff (a, b)
+ const struct tm *a;
+ const struct tm *b;
+{
+ /* Compute intervening leap days correctly even if year is negative.
+ Take care to avoid int overflow in leap day calculations,
+ but it's OK to assume that A and B are close to each other. */
+ int a4 = (a->tm_year >> 2) + (TM_YEAR_BASE >> 2) - ! (a->tm_year & 3);
+ int b4 = (b->tm_year >> 2) + (TM_YEAR_BASE >> 2) - ! (b->tm_year & 3);
+ int a100 = a4 / 25 - (a4 % 25 < 0);
+ int b100 = b4 / 25 - (b4 % 25 < 0);
+ int a400 = a100 >> 2;
+ int b400 = b100 >> 2;
+ int intervening_leap_days = (a4 - b4) - (a100 - b100) + (a400 - b400);
+ int years = a->tm_year - b->tm_year;
+ int days = (365 * years + intervening_leap_days
+ + (a->tm_yday - b->tm_yday));
+ return (60 * (60 * (24 * days + (a->tm_hour - b->tm_hour))
+ + (a->tm_min - b->tm_min))
+ + (a->tm_sec - b->tm_sec));
+}
+#endif /* ! HAVE_TM_GMTOFF */