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Handle (64-bit) / (32-bit) = (64-bit), i.e. one step beyond that 

provided by the divl instruction.
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Michael Brown 2007-01-28 19:34:17 +00:00
parent bd873525ff
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src/arch/i386/core/udivmod64.c 100644
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/*
* Copyright (C) 2007 Michael Brown <mbrown@fensystems.co.uk>.
*
* This program is free software; you can redistribute it and/or
* modify it under the terms of the GNU General Public License as
* published by the Free Software Foundation; either version 2 of the
* License, or any later version.
*
* This program 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
* General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with this program; if not, write to the Free Software
* Foundation, Inc., 675 Mass Ave, Cambridge, MA 02139, USA.
*/
/** @file
*
* 64-bit division
*
* The x86 CPU (386 upwards) has a divl instruction which will perform
* unsigned division of a 64-bit dividend by a 32-bit divisor. If the
* resulting quotient does not fit in 32 bits, then a CPU exception
* will occur.
*
* Unsigned integer division is expressed as solving
*
* x = d.q + r 0 <= q, 0 <= r < d
*
* given the dividend (x) and divisor (d), to find the quotient (q)
* and remainder (r).
*
* The x86 divl instruction will solve
*
* x = d.q + r 0 <= q, 0 <= r < d
*
* given x in the range 0 <= x < 2^64 and 1 <= d < 2^32, and causing a
* hardware exception if the resulting q >= 2^32.
*
* We can therefore use divl only if we can prove that the conditions
*
* 0 <= x < 2^64
* 1 <= d < 2^32
* q < 2^32
*
* are satisfied.
*
*
* Case 1 : 1 <= d < 2^32
* ======================
*
* We express x as
*
* x = xh.2^32 + xl 0 <= xh < 2^32, 0 <= xl < 2^32 (1)
*
* i.e. split x into low and high dwords. We then solve
*
* xh = d.qh + r' 0 <= qh, 0 <= r' < d (2)
*
* which we can do using a divl instruction since
*
* 0 <= xh < 2^64 since 0 <= xh < 2^32 from (1) (3)
*
* and
*
* 1 <= d < 2^32 by definition of this Case (4)
*
* and
*
* d.qh = xh - r' from (2)
* d.qh <= xh since r' >= 0 from (2)
* qh <= xh since d >= 1 from (2)
* qh < 2^32 since xh < 2^32 from (1) (5)
*
* Having obtained qh and r', we then solve
*
* ( r'.2^32 + xl ) = d.ql + r 0 <= ql, 0 <= r < d (6)
*
* which we can do using another divl instruction since
*
* xl <= 2^32 - 1 from (1), so
* r'.2^32 + xl <= ( r' + 1 ).2^32 - 1
* r'.2^32 + xl <= d.2^32 - 1 since r' < d from (2)
* r'.2^32 + xl < d.2^32 (7)
* r'.2^32 + xl < 2^64 since d < 2^32 from (4) (8)
*
* and
*
* 1 <= d < 2^32 by definition of this Case (9)
*
* and
*
* d.ql = ( r'.2^32 + xl ) - r from (6)
* d.ql <= r'.2^32 + xl since r >= 0 from (6)
* d.ql < d.2^32 from (7)
* ql < 2^32 since d >= 1 from (2) (10)
*
* This then gives us
*
* x = xh.2^32 + xl from (1)
* x = ( d.qh + r' ).2^32 + xl from (2)
* x = d.qh.2^32 + ( r'.2^32 + xl )
* x = d.qh.2^32 + d.ql + r from (3)
* x = d.( qh.2^32 + ql ) + r (11)
*
* Letting
*
* q = qh.2^32 + ql (12)
*
* gives
*
* x = d.q + r from (11) and (12)
*
* which is the solution.
*
*
* This therefore gives us a two-step algorithm:
*
* xh = d.qh + r' 0 <= qh, 0 <= r' < d (2)
* ( r'.2^32 + xl ) = d.ql + r 0 <= ql, 0 <= r < d (6)
*
* which translates to
*
* %edx:%eax = 0:xh
* divl d
* qh = %eax
* r' = %edx
*
* %edx:%eax = r':xl
* divl d
* ql = %eax
* r = %edx
*
* Note that if
*
* xh < d
*
* (which is a fast dword comparison) then the first divl instruction
* can be omitted, since the answer will be
*
* qh = 0
* r = xh
*
*
* Case 2 : 2^32 <= d < 2^64
* =========================
*
* We first express d as
*
* d = dh.2^k + dl 2^31 <= dh < 2^32,
* 0 <= dl < 2^k, 1 <= k <= 32 (1)
*
* i.e. find the highest bit set in d, subtract 32, and split d into
* dh and dl at that point.
*
* We then express x as
*
* x = xh.2^k + xl 0 <= xl < 2^k (2)
*
* giving
*
* xh.2^k = x - xl from (2)
* xh.2^k <= x since xl >= 0 from (1)
* xh.2^k < 2^64 since xh < 2^64 from (1)
* xh < 2^(64-k) (3)
*
* We then solve the division
*
* xh = dh.q' + r' 0 <= r' < dh (4)
*
* which we can do using a divl instruction since
*
* 0 <= xh < 2^64 since x < 2^64 and xh < x
*
* and
*
* 1 <= dh < 2^32 from (1)
*
* and
*
* dh.q' = xh - r' from (4)
* dh.q' <= xh since r' >= 0 from (4)
* dh.q' < 2^(64-k) from (3) (5)
* q'.2^31 <= dh.q' since dh >= 2^31 from (1) (6)
* q'.2^31 < 2^(64-k) from (5) and (6)
* q' < 2^(33-k)
* q' < 2^32 since k >= 1 from (1) (7)
*
* This gives us
*
* xh.2^k = dh.q'.2^k + r'.2^k from (4)
* x - xl = ( d - dl ).q' + r'.2^k from (1) and (2)
* x = d.q' + ( r'.2^k + xl ) - dl.q' (8)
*
* Now
*
* r'.2^k + xl < r'.2^k + 2^k since xl < 2^k from (2)
* r'.2^k + xl < ( r' + 1 ).2^k
* r'.2^k + xl < dh.2^k since r' < dh from (4)
* r'.2^k + xl < ( d - dl ) from (1) (9)
*
*
* (missing)
*
*
* This gives us two cases to consider:
*
* case (a):
*
* dl.q' <= ( r'.2^k + xl ) (15a)
*
* in which case
*
* x = d.q' + ( r'.2^k + xl - dl.q' )
*
* is a direct solution to the division, since
*
* r'.2^k + xl < d from (9)
* ( r'.2^k + xl - dl.q' ) < d since dl >= 0 and q' >= 0
*
* and
*
* 0 <= ( r'.2^k + xl - dl.q' ) from (15a)
*
* case (b):
*
* dl.q' > ( r'.2^k + xl ) (15b)
*
* Express
*
* x = d.(q'-1) + ( r'.2^k + xl ) + ( d - dl.q' )
*
*
* (missing)
*
*
* special case: k = 32 cannot be handled with shifts
*
* (missing)
*
*/
#include <stdint.h>
#include <assert.h>
typedef uint64_t UDItype;
struct uint64 {
uint32_t l;
uint32_t h;
};
static inline void udivmod64_lo ( const struct uint64 *x,
const struct uint64 *d,
struct uint64 *q,
struct uint64 *r ) {
uint32_t r_dash;
q->h = 0;
r->h = 0;
r_dash = x->h;
if ( x->h >= d->l ) {
__asm__ ( "divl %2"
: "=&a" ( q->h ), "=&d" ( r_dash )
: "g" ( d->l ), "0" ( x->h ), "1" ( 0 ) );
}
__asm__ ( "divl %2"
: "=&a" ( q->l ), "=&d" ( r->l )
: "g" ( d->l ), "0" ( x->l ), "1" ( r_dash ) );
}
static void udivmod64 ( const struct uint64 *x,
const struct uint64 *d,
struct uint64 *q,
struct uint64 *r ) {
if ( d->h == 0 ) {
udivmod64_lo ( x, d, q, r );
} else {
assert ( 0 );
while ( 1 ) {};
}
}
/**
* 64-bit division with remainder
*
* @v x Dividend
* @v d Divisor
* @ret r Remainder
* @ret q Quotient
*/
UDItype __udivmoddi4 ( UDItype x, UDItype d, UDItype *r ) {
UDItype q;
UDItype *_x = &x;
UDItype *_d = &d;
UDItype *_q = &q;
UDItype *_r = r;
udivmod64 ( ( struct uint64 * ) _x, ( struct uint64 * ) _d,
( struct uint64 * ) _q, ( struct uint64 * ) _r );
return q;
}
/**
* 64-bit division
*
* @v x Dividend
* @v d Divisor
* @ret q Quotient
*/
UDItype __udivdi3 ( UDItype x, UDItype d ) {
UDItype r;
return __udivmoddi4 ( x, d, &r );
}