* Copyright (c) 1992, 1993
* The Regents of the University of California. All rights reserved.
* This software was developed by the Computer Systems Engineering group
* at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
* contributed to Berkeley.
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* from: $Header: divrem.m4,v 1.4 92/06/25 13:23:57 torek Exp $
* Division and remainder, from Appendix E of the Sparc Version 8
* Architecture Manual, with fixes from Gordon Irlam.
#if defined(LIBC_SCCS) && !defined(lint)
.asciz "@(#)divrem.m4 8.1 (Berkeley) 6/4/93"
#endif /* LIBC_SCCS and not lint */
* Input: dividend and divisor in %o0 and %o1 respectively.
* .div name of function to generate
* div div=div => %o0 / %o1; div=rem => %o0 % %o1
* true true=true => signed; true=false => unsigned
* N how many bits per iteration we try to get (4)
* WORDSIZE total number of bits (32)
* TWOSUPN 2^N, for label generation (m4 exponentiation currently broken)
* TOPBITS number of bits in the top decade of a number
* 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)
* 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
/* m4 reminder: d => if a is b, then c, else d */
* This is the recursive definition for developing quotient digits.
* $1 the current depth, 1 <= $1 <= 4
* $2 the current accumulation of quotient bits
* We add a new bit to $2 and either recurse or insert the bits in
* the quotient. %o3, %o2, and %o5 are inputs and outputs as defined above;
* the condition codes are expected to reflect the input %o3, and are
* modified to reflect the output %o3.
#include <machine/trap.h>
! compute sign of result; if neither is negative, no problem
orcc %o1, %o0, %g0 ! either negative?
bge 2f ! no, go do the divide
xor %o1, %o0, %g6 ! compute sign in any case
! %o1 is definitely negative; %o0 might also be negative
bge 2f ! if %o0 not negative...
neg %o1 ! in any case, make %o1 nonneg
1: ! %o0 is negative, %o1 is nonnegative
neg %o0 ! make %o0 nonnegative
! Ready to divide. Compute size of quotient; scale comparand.
! Divide by zero trap. If it returns, return 0 (about as
! wrong as possible, but that is what SunOS does...).
cmp %o3, %o5 ! if %o1 exceeds %o0, done
blu Lgot_result ! (and algorithm fails otherwise)
sethi %hi(1 << (32 - 4 - 1)), %g1
! 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.
! We get here if the %o1 overflowed while shifting.
! This means that %o3 has the high-order bit set.
! Restore %o5 and subtract from %o3.
sll %g1, 4, %g1 ! high order bit
srl %o5, 1, %o5 ! rest of %o5
/* NB: these are commented out in the V8-Sparc manual as well */
/* (I do not understand this) */
! %o5 > %o3: went too far: back up 1 step
! do single-bit divide steps
! We have to be careful here. We know that %o3 >= %o5, so we can do the
! first divide step without thinking. BUT, the others are conditional,
! and are only done if %o3 >= 0. Because both %o3 and %o5 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...
tst %o3 ! set up for initial iteration
! depth 1, accumulated bits 0
! depth 2, accumulated bits 1
! depth 3, accumulated bits 3
! depth 4, accumulated bits 7
! depth 4, accumulated bits 5
! depth 3, accumulated bits 1
! depth 4, accumulated bits 3
! depth 4, accumulated bits 1
! depth 2, accumulated bits -1
! depth 3, accumulated bits -1
! depth 4, accumulated bits -1
! depth 4, accumulated bits -3
! depth 3, accumulated bits -3
! depth 4, accumulated bits -5
! depth 4, accumulated bits -7
! non-restoring fixup here (one instruction only!)
! check to see if answer should be < 0