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/*
* Division and remainder, from Appendix E of the Sparc Version 8
* Architecture Manual, with fixes from Gordon Irlam.
*/
/*
* Input: dividend and divisor in %o0 and %o1 respectively.
*
* m4 parameters:
* NAME name of function to generate
* OP OP=div => %o0 / %o1; OP=rem => %o0 % %o1
* S S=true => signed; S=false => unsigned
*
* Algorithm parameters:
* N how many bits per iteration we try to get (4)
* WORDSIZE total number of bits (32)
*
* Derived constants:
* TOPBITS number of bits in the top `decade' of a number
*
* Important variables:
* Q the partial quotient under development (initially 0)
* R the remainder so far, initially the dividend
* ITER number of main division loop iterations required;
* equal to ceil(log2(quotient) / N). Note that this
* is the log base (2^N) of the quotient.
* V the current comparand, initially divisor*2^(ITER*N-1)
*
* Cost:
* Current estimate for non-large dividend is
* ceil(log2(quotient) / N) * (10 + 7N/2) + C
* A large dividend is one greater than 2^(31-TOPBITS) and takes a
* different path, as the upper bits of the quotient must be developed
* one bit at a time.
*/
define(N, `4')dnl
define(WORDSIZE, `32')dnl
define(TOPBITS, eval(WORDSIZE - N*((WORDSIZE-1)/N)))dnl
dnl
define(dividend, `%o0')dnl
define(divisor, `%o1')dnl
define(Q, `%o2')dnl
define(R, `%o3')dnl
define(ITER, `%o4')dnl
define(V, `%o5')dnl
dnl
dnl m4 reminder: ifelse(a,b,c,d) => if a is b, then c, else d
define(T, `%g1')dnl
define(SC, `%g7')dnl
ifelse(S, `true', `define(SIGN, `%g6')')dnl
dnl
dnl This is the recursive definition for developing quotient digits.
dnl
dnl Parameters:
dnl $1 the current depth, 1 <= $1 <= N
dnl $2 the current accumulation of quotient bits
dnl N max depth
dnl
dnl We add a new bit to $2 and either recurse or insert the bits in
dnl the quotient. R, Q, and V are inputs and outputs as defined above;
dnl the condition codes are expected to reflect the input R, and are
dnl modified to reflect the output R.
dnl
define(DEVELOP_QUOTIENT_BITS,
` ! depth $1, accumulated bits $2
bl L.$1.eval(2**N+$2)
srl V,1,V
! remainder is positive
subcc R,V,R
ifelse($1, N,
` b 9f
add Q, ($2*2+1), Q
', ` DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2+1)')')
L.$1.eval(2**N+$2):
! remainder is negative
addcc R,V,R
ifelse($1, N,
` b 9f
add Q, ($2*2-1), Q
', ` DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2-1)')')
ifelse($1, 1, `9:')')dnl
#include "sysdep.h"
#ifdef __linux__
#include <asm/traps.h>
#else
#ifdef __svr4__
#include <sys/trap.h>
#else
#include <machine/trap.h>
#endif
#endif
FUNC(NAME)
ifelse(S, `true',
` ! compute sign of result; if neither is negative, no problem
orcc divisor, dividend, %g0 ! either negative?
bge 2f ! no, go do the divide
ifelse(OP, `div',
` xor divisor, dividend, SIGN ! compute sign in any case',
` mov dividend, SIGN ! sign of remainder matches dividend')
tst divisor
bge 1f
tst dividend
! divisor is definitely negative; dividend might also be negative
bge 2f ! if dividend not negative...
sub %g0, divisor, divisor ! in any case, make divisor nonneg
1: ! dividend is negative, divisor is nonnegative
sub %g0, dividend, dividend ! make dividend nonnegative
2:
')
! Ready to divide. Compute size of quotient; scale comparand.
orcc divisor, %g0, V
bne 1f
mov dividend, R
! Divide by zero trap. If it returns, return 0 (about as
! wrong as possible, but that is what SunOS does...).
ta ST_DIV0
retl
clr %o0
1:
cmp R, V ! if divisor exceeds dividend, done
blu Lgot_result ! (and algorithm fails otherwise)
clr Q
sethi %hi(1 << (WORDSIZE - TOPBITS - 1)), T
cmp R, T
blu Lnot_really_big
clr ITER
! `Here the dividend is >= 2**(31-N) or so. We must be careful here,
! as our usual N-at-a-shot divide step will cause overflow and havoc.
! The number of bits in the result here is N*ITER+SC, where SC <= N.
! Compute ITER in an unorthodox manner: know we need to shift V into
! the top decade: so do not even bother to compare to R.'
1:
cmp V, T
bgeu 3f
mov 1, SC
sll V, N, V
b 1b
add ITER, 1, ITER
! Now compute SC.
2: addcc V, V, V
bcc Lnot_too_big
add SC, 1, SC
! We get here if the divisor overflowed while shifting.
! This means that R has the high-order bit set.
! Restore V and subtract from R.
sll T, TOPBITS, T ! high order bit
srl V, 1, V ! rest of V
add V, T, V
b Ldo_single_div
sub SC, 1, SC
Lnot_too_big:
3: cmp V, R
blu 2b
nop
be Ldo_single_div
nop
/* NB: these are commented out in the V8-Sparc manual as well */
/* (I do not understand this) */
! V > R: went too far: back up 1 step
! srl V, 1, V
! dec SC
! do single-bit divide steps
!
! We have to be careful here. We know that R >= V, so we can do the
! first divide step without thinking. BUT, the others are conditional,
! and are only done if R >= 0. Because both R and V may have the high-
! order bit set in the first step, just falling into the regular
! division loop will mess up the first time around.
! So we unroll slightly...
Ldo_single_div:
subcc SC, 1, SC
bl Lend_regular_divide
nop
sub R, V, R
mov 1, Q
b Lend_single_divloop
nop
Lsingle_divloop:
sll Q, 1, Q
bl 1f
srl V, 1, V
! R >= 0
sub R, V, R
b 2f
add Q, 1, Q
1: ! R < 0
add R, V, R
sub Q, 1, Q
2:
Lend_single_divloop:
subcc SC, 1, SC
bge Lsingle_divloop
tst R
b,a Lend_regular_divide
Lnot_really_big:
1:
sll V, N, V
cmp V, R
bleu 1b
addcc ITER, 1, ITER
be Lgot_result
sub ITER, 1, ITER
tst R ! set up for initial iteration
Ldivloop:
sll Q, N, Q
DEVELOP_QUOTIENT_BITS(1, 0)
Lend_regular_divide:
subcc ITER, 1, ITER
bge Ldivloop
tst R
bl,a Lgot_result
! non-restoring fixup here (one instruction only!)
ifelse(OP, `div',
` sub Q, 1, Q
', ` add R, divisor, R
')
Lgot_result:
ifelse(S, `true',
` ! check to see if answer should be < 0
tst SIGN
bl,a 1f
ifelse(OP, `div', `sub %g0, Q, Q', `sub %g0, R, R')
1:')
retl
ifelse(OP, `div', `mov Q, %o0', `mov R, %o0')
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