#include <quad.h>
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);
}
