/* From OpenBSD */

#include <sys/types.h>
/* #include <sys/limits.h> */

#define CHAR_BIT 8

union uu {
	quad_t	q;		/* as a (signed) quad */
	u_quad_t uq;		/* as an unsigned quad */
	long	sl[2];		/* as two signed longs */
	u_long	ul[2];		/* as two unsigned longs */
};



/*
 * Define high and low longwords.
 */
#define	H		_QUAD_HIGHWORD
#define	L		_QUAD_LOWWORD

/*
 * Total number of bits in a quad_t and in the pieces that make it up.
 * These are used for shifting, and also below for halfword extraction
 * and assembly.
 */
#define	QUAD_BITS	(sizeof(quad_t) * CHAR_BIT)
#define	LONG_BITS	(sizeof(long) * CHAR_BIT)
#define	HALF_BITS	(sizeof(long) * CHAR_BIT / 2)

/*
 * Extract high and low shortwords from longword, and move low shortword of
 * longword to upper half of long, i.e., produce the upper longword of
 * ((quad_t)(x) << (number_of_bits_in_long/2)).  (`x' must actually be u_long.)
 *
 * These are used in the multiply code, to split a longword into upper
 * and lower halves, and to reassemble a product as a quad_t, shifted left
 * (sizeof(long)*CHAR_BIT/2).
 */
#define	HHALF(x)	((u_long)(x) >> HALF_BITS)
#define	LHALF(x)	((u_long)(x) & (((long)1 << HALF_BITS) - 1))
#define	LHUP(x)		((u_long)(x) << HALF_BITS)

extern u_quad_t __qdivrem(u_quad_t u, u_quad_t v, u_quad_t *rem);

/*
 * XXX
 * Compensate for gcc 1 vs gcc 2.  Gcc 1 defines ?sh?di3's second argument
 * as u_quad_t, while gcc 2 correctly uses int.  Unfortunately, we still use
 * both compilers.
 */
#if __GNUC__ >= 2
typedef unsigned int	qshift_t;
#else
typedef u_quad_t	qshift_t;
#endif

#define	B	((long)1 << HALF_BITS)	/* digit base */

/* Combine two `digits' to make a single two-digit number. */
#define	COMBINE(a, b) (((u_long)(a) << HALF_BITS) | (b))

/* select a type for digits in base B: use unsigned short if they fit */
#if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff
typedef unsigned short digit;
#else
typedef u_long digit;
#endif

static void shl(digit *p, int len, int sh);


quad_t
__divdi3(a, b)
	quad_t a, b;
{
	u_quad_t ua, ub, uq;
	int neg;

	if (a < 0)
		ua = -(u_quad_t)a, neg = 1;
	else
		ua = a, neg = 0;
	if (b < 0)
		ub = -(u_quad_t)b, neg ^= 1;
	else
		ub = b;
	uq = __qdivrem(ua, ub, (u_quad_t *)0);
	return (neg ? -uq : uq);
}

quad_t
__moddi3(a, b)
	quad_t a, b;
{
	u_quad_t ua, ub, ur;
	int neg;

	if (a < 0)
		ua = -(u_quad_t)a, neg = 1;
	else
		ua = a, neg = 0;
	if (b < 0)
		ub = -(u_quad_t)b, neg ^= 1;
	else
		ub = b;
	(void)__qdivrem(ua, ub, &ur);
	return (neg ? -ur : ur);
}

u_quad_t
__udivdi3(a, b)
	u_quad_t a, b;
{

	return (__qdivrem(a, b, (u_quad_t *)0));
}

u_quad_t
__umoddi3(a, b)
	u_quad_t a, b;
{
	u_quad_t r;

	(void)__qdivrem(a, b, &r);
	return (r);
}



/*
 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
 *
 * We do this in base 2-sup-HALF_BITS, so that all intermediate products
 * fit within u_long.  As a consequence, the maximum length dividend and
 * divisor are 4 `digits' in this base (they are shorter if they have
 * leading zeros).
 */
u_quad_t
__qdivrem(uq, vq, arq)
	u_quad_t uq, vq, *arq;
{
	union uu tmp;
	digit *u, *v, *q;
	register digit v1, v2;
	u_long qhat, rhat, t;
	int m, n, d, j, i;
	digit uspace[5], vspace[5], qspace[5];

	/*
	 * Take care of special cases: divide by zero, and u < v.
	 */
	if (vq == 0) {
		/* divide by zero. */
		static volatile const unsigned int zero = 0;

		tmp.ul[H] = tmp.ul[L] = 1 / zero;
		if (arq)
			*arq = uq;
		return (tmp.q);
	}
	if (uq < vq) {
		if (arq)
			*arq = uq;
		return (0);
	}
	u = &uspace[0];
	v = &vspace[0];
	q = &qspace[0];

	/*
	 * Break dividend and divisor into digits in base B, then
	 * count leading zeros to determine m and n.  When done, we
	 * will have:
	 *	u = (u[1]u[2]...u[m+n]) sub B
	 *	v = (v[1]v[2]...v[n]) sub B
	 *	v[1] != 0
	 *	1 < n <= 4 (if n = 1, we use a different division algorithm)
	 *	m >= 0 (otherwise u < v, which we already checked)
	 *	m + n = 4
	 * and thus
	 *	m = 4 - n <= 2
	 */
	tmp.uq = uq;
	u[0] = 0;
	u[1] = HHALF(tmp.ul[H]);
	u[2] = LHALF(tmp.ul[H]);
	u[3] = HHALF(tmp.ul[L]);
	u[4] = LHALF(tmp.ul[L]);
	tmp.uq = vq;
	v[1] = HHALF(tmp.ul[H]);
	v[2] = LHALF(tmp.ul[H]);
	v[3] = HHALF(tmp.ul[L]);
	v[4] = LHALF(tmp.ul[L]);
	for (n = 4; v[1] == 0; v++) {
		if (--n == 1) {
			u_long rbj;	/* r*B+u[j] (not root boy jim) */
			digit q1, q2, q3, q4;

			/*
			 * Change of plan, per exercise 16.
			 *	r = 0;
			 *	for j = 1..4:
			 *		q[j] = floor((r*B + u[j]) / v),
			 *		r = (r*B + u[j]) % v;
			 * We unroll this completely here.
			 */
			t = v[2];	/* nonzero, by definition */
			q1 = u[1] / t;
			rbj = COMBINE(u[1] % t, u[2]);
			q2 = rbj / t;
			rbj = COMBINE(rbj % t, u[3]);
			q3 = rbj / t;
			rbj = COMBINE(rbj % t, u[4]);
			q4 = rbj / t;
			if (arq)
				*arq = rbj % t;
			tmp.ul[H] = COMBINE(q1, q2);
			tmp.ul[L] = COMBINE(q3, q4);
			return (tmp.q);
		}
	}

	/*
	 * By adjusting q once we determine m, we can guarantee that
	 * there is a complete four-digit quotient at &qspace[1] when
	 * we finally stop.
	 */
	for (m = 4 - n; u[1] == 0; u++)
		m--;
	for (i = 4 - m; --i >= 0;)
		q[i] = 0;
	q += 4 - m;

	/*
	 * Here we run Program D, translated from MIX to C and acquiring
	 * a few minor changes.
	 *
	 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
	 */
	d = 0;
	for (t = v[1]; t < B / 2; t <<= 1)
		d++;
	if (d > 0) {
		shl(&u[0], m + n, d);		/* u <<= d */
		shl(&v[1], n - 1, d);		/* v <<= d */
	}
	/*
	 * D2: j = 0.
	 */
	j = 0;
	v1 = v[1];	/* for D3 -- note that v[1..n] are constant */
	v2 = v[2];	/* for D3 */
	do {
		register digit uj0, uj1, uj2;
		
		/*
		 * D3: Calculate qhat (\^q, in TeX notation).
		 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
		 * let rhat = (u[j]*B + u[j+1]) mod v[1].
		 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
		 * decrement qhat and increase rhat correspondingly.
		 * Note that if rhat >= B, v[2]*qhat < rhat*B.
		 */
		uj0 = u[j + 0];	/* for D3 only -- note that u[j+...] change */
		uj1 = u[j + 1];	/* for D3 only */
		uj2 = u[j + 2];	/* for D3 only */
		if (uj0 == v1) {
			qhat = B;
			rhat = uj1;
			goto qhat_too_big;
		} else {
			u_long n = COMBINE(uj0, uj1);
			qhat = n / v1;
			rhat = n % v1;
		}
		while (v2 * qhat > COMBINE(rhat, uj2)) {
	qhat_too_big:
			qhat--;
			if ((rhat += v1) >= B)
				break;
		}
		/*
		 * D4: Multiply and subtract.
		 * The variable `t' holds any borrows across the loop.
		 * We split this up so that we do not require v[0] = 0,
		 * and to eliminate a final special case.
		 */
		for (t = 0, i = n; i > 0; i--) {
			t = u[i + j] - v[i] * qhat - t;
			u[i + j] = LHALF(t);
			t = (B - HHALF(t)) & (B - 1);
		}
		t = u[j] - t;
		u[j] = LHALF(t);
		/*
		 * D5: test remainder.
		 * There is a borrow if and only if HHALF(t) is nonzero;
		 * in that (rare) case, qhat was too large (by exactly 1).
		 * Fix it by adding v[1..n] to u[j..j+n].
		 */
		if (HHALF(t)) {
			qhat--;
			for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
				t += u[i + j] + v[i];
				u[i + j] = LHALF(t);
				t = HHALF(t);
			}
			u[j] = LHALF(u[j] + t);
		}
		q[j] = qhat;
	} while (++j <= m);		/* D7: loop on j. */

	/*
	 * If caller wants the remainder, we have to calculate it as
	 * u[m..m+n] >> d (this is at most n digits and thus fits in
	 * u[m+1..m+n], but we may need more source digits).
	 */
	if (arq) {
		if (d) {
			for (i = m + n; i > m; --i)
				u[i] = (u[i] >> d) |
				    LHALF(u[i - 1] << (HALF_BITS - d));
			u[i] = 0;
		}
		tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
		tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
		*arq = tmp.q;
	}

	tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
	tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
	return (tmp.q);
}

/*
 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
 * `fall out' the left (there never will be any such anyway).
 * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
 */
static void
shl(p, len, sh)
	register digit *p;
	register int len;
	register int sh;
{
	register int i;

	for (i = 0; i < len; i++)
		p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
	p[i] = LHALF(p[i] << sh);
}