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	}
	return 1;
}

DUK_INTERNAL duk_bool_t duk_bi_date_timeval_in_valid_range(duk_double_t x) {
	return (x >= -DUK_DATE_MSEC_100M_DAYS && x <= DUK_DATE_MSEC_100M_DAYS);
}

DUK_INTERNAL duk_bool_t duk_bi_date_timeval_in_leeway_range(duk_double_t x) {
	return (x >= -DUK_DATE_MSEC_100M_DAYS_LEEWAY && x <= DUK_DATE_MSEC_100M_DAYS_LEEWAY);
}

DUK_INTERNAL duk_bool_t duk_bi_date_year_in_valid_range(duk_double_t x) {
	return (x >= DUK_DATE_MIN_ECMA_YEAR && x <= DUK_DATE_MAX_ECMA_YEAR);
}

DUK_LOCAL duk_double_t duk__timeclip(duk_double_t x) {
	if (!DUK_ISFINITE(x)) {
		return DUK_DOUBLE_NAN;
	}

	if (!duk_bi_date_timeval_in_valid_range(x)) {
		return DUK_DOUBLE_NAN;
	}

	x = duk_js_tointeger_number(x);

	/* Here we'd have the option to normalize -0 to +0. */
	return x;
}

/* Integer division which floors also negative values correctly. */
DUK_LOCAL duk_int_t duk__div_floor(duk_int_t a, duk_int_t b) {
	DUK_ASSERT(b > 0);
	if (a >= 0) {
		return a / b;
	} else {
		/* e.g. a = -4, b = 5  -->  -4 - 5 + 1 / 5  -->  -8 / 5  -->  -1
		 *      a = -5, b = 5  -->  -5 - 5 + 1 / 5  -->  -9 / 5  -->  -1
		 *      a = -6, b = 5  -->  -6 - 5 + 1 / 5  -->  -10 / 5  -->  -2
		 */
		return (a - b + 1) / b;
	}
}

/* Compute day number of the first day of a given year. */
DUK_LOCAL duk_int_t duk__day_from_year(duk_int_t year) {
	/* Note: in integer arithmetic, (x / 4) is same as floor(x / 4) for non-negative
	 * values, but is incorrect for negative ones.
	 */
	return 365 * (year - 1970) + duk__div_floor(year - 1969, 4) - duk__div_floor(year - 1901, 100) +
	       duk__div_floor(year - 1601, 400);
}

/* Given a day number, determine year and day-within-year. */
DUK_LOCAL duk_int_t duk__year_from_day(duk_int_t day, duk_small_int_t *out_day_within_year) {
	duk_int_t year;
	duk_int_t diff_days;

	/* estimate year upwards (towards positive infinity), then back down;
	 * two iterations should be enough
	 */

	if (day >= 0) {
		year = 1970 + day / 365;
	} else {
		year = 1970 + day / 366;
	}

	for (;;) {
		diff_days = duk__day_from_year(year) - day;
		DUK_DDD(DUK_DDDPRINT("year=%ld day=%ld, diff_days=%ld", (long) year, (long) day, (long) diff_days));
		if (diff_days <= 0) {
			DUK_ASSERT(-diff_days < 366); /* fits into duk_small_int_t */
			*out_day_within_year = -diff_days;
			DUK_DDD(DUK_DDDPRINT("--> year=%ld, day-within-year=%ld", (long) year, (long) *out_day_within_year));
			DUK_ASSERT(*out_day_within_year >= 0);
			DUK_ASSERT(*out_day_within_year < (duk_bi_date_is_leap_year(year) ? 366 : 365));
			return year;
		}

		/* Note: this is very tricky; we must never 'overshoot' the
		 * correction downwards.
		 */
		year -= 1 + (diff_days - 1) / 366; /* conservative */
	}
}

/* Given a (year, month, day-within-month) triple, compute day number.
 * The input triple is un-normalized and may contain non-finite values.
 */
DUK_LOCAL duk_double_t duk__make_day(duk_double_t year, duk_double_t month, duk_double_t day) {
	duk_int_t day_num;
	duk_bool_t is_leap;
	duk_small_int_t i, n;

	/* Assume that year, month, day are all coerced to whole numbers.
	 * They may also be NaN or infinity, in which case this function
	 * must return NaN or infinity to ensure time value becomes NaN.
	 * If 'day' is NaN, the final return will end up returning a NaN,
	 * so it doesn't need to be checked here.
	 */

	if (!DUK_ISFINITE(year) || !DUK_ISFINITE(month)) {
		return DUK_DOUBLE_NAN;
	}

	year += DUK_FLOOR(month / 12.0);

	month = DUK_FMOD(month, 12.0);
	if (month < 0.0) {
		/* handle negative values */
		month += 12.0;
	}

	/* The algorithm in E5.1 Section 15.9.1.12 normalizes month, but
	 * does not normalize the day-of-month (nor check whether or not
	 * it is finite) because it's not necessary for finding the day
	 * number which matches the (year,month) pair.
	 *
	 * We assume that duk__day_from_year() is exact here.