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+/*
+ * This m4 code has been taken from The SPARC Architecture Manual Version 8.
+ */
+
+/*
+ * Division/Remainder
+ *
+ * Input is:
+ * dividend -- the thing being divided
+ * divisor -- how many ways to divide it
+ * Important parameters:
+ * N -- how many bits per iteration we try to get
+ * as our current guess: define(N, 4) define(TWOSUPN, 16)
+ * WORDSIZE -- how many bits altogether we're talking about:
+ * obviously: define(WORDSIZE, 32)
+ * A derived constant:
+ * TOPBITS -- how many bits are in the top "decade" of a number:
+ * define(TOPBITS, eval( WORDSIZE - N*((WORDSIZE-1)/N) ) )
+ * Important variables are:
+ * Q -- the partial quotient under development -- initially 0
+ * R -- the remainder so far -- initially == the dividend
+ * ITER -- number of iterations of the main division loop which will
+ * be required. Equal to CEIL( lg2(quotient)/N )
+ * Note that this is log_base_(2ˆN) of the quotient.
+ * V -- the current comparand -- initially divisor*2ˆ(ITER*N-1)
+ * Cost:
+ * current estimate for non-large dividend is
+ * CEIL( lg2(quotient) / N ) x ( 10 + 7N/2 ) + C
+ * a large dividend is one greater than 2ˆ(31-TOPBITS) and takes a
+ * different path, as the upper bits of the quotient must be developed
+ * one bit at a time.
+ * This uses the m4 and cpp macro preprocessors.
+ */
+
+define(dividend, `%o0')
+define(divisor,`%o1')
+define(Q, `%o2')
+define(R, `%o3')
+define(ITER, `%o4')
+define(V, `%o5')
+define(SIGN, `%g3')
+define(T, `%g1')
+define(SC,`%g2')
+/*
+ * This is the recursive definition of how we develop quotient digits.
+ * It takes three important parameters:
+ * $1 -- the current depth, 1<=$1<=N
+ * $2 -- the current accumulation of quotient bits
+ * N -- max depth
+ * We add a new bit to $2 and either recurse or insert the bits in the quotient.
+ * Dynamic input:
+ * R -- current remainder
+ * Q -- current quotient
+ * V -- current comparand
+ * cc -- set on current value of R
+ * Dynamic output:
+ * R', Q', V', cc'
+ */
+
+#include "../assembly.h"
+
+define(DEVELOP_QUOTIENT_BITS,
+` !depth $1, accumulated bits $2
+ bl L.$1.eval(TWOSUPN+$2)
+ srl V,1,V
+ ! remainder is nonnegative
+ 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(TWOSUPN+$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:')
+')
+ifelse( ANSWER, `quotient', `
+.text
+ .align 32
+DEFINE_COMPILERRT_FUNCTION(__udivsi3)
+ b divide
+ mov 0,SIGN ! result always nonnegative
+.text
+ .align 32
+DEFINE_COMPILERRT_FUNCTION(__divsi3)
+ orcc divisor,dividend,%g0 ! are either dividend or divisor negative
+ bge divide ! if not, skip this junk
+ xor divisor,dividend,SIGN ! record sign of result in sign of SIGN
+ tst divisor
+ bge 2f
+ tst dividend
+ ! divisor < 0
+ bge divide
+ neg divisor
+ 2:
+ ! dividend < 0
+ neg dividend
+ ! FALL THROUGH
+',`
+.text
+ .align 32
+DEFINE_COMPILERRT_FUNCTION(__umodsi3)
+ b divide
+ mov 0,SIGN ! result always nonnegative
+.text
+ .align 32
+DEFINE_COMPILERRT_FUNCTION(__modsi3)
+ orcc divisor,dividend,%g0 ! are either dividend or divisor negative
+ bge divide ! if not, skip this junk
+ mov dividend,SIGN ! record sign of result in sign of SIGN
+ tst divisor
+ bge 2f
+ tst dividend
+ ! divisor < 0
+ bge divide
+ neg divisor
+ 2:
+ ! dividend < 0
+ neg dividend
+ ! FALL THROUGH
+')
+
+divide:
+ ! Compute size of quotient, scale comparand.
+ orcc divisor,%g0,V ! movcc divisor,V
+ te 2 ! if divisor = 0
+ mov dividend,R
+ mov 0,Q
+ sethi %hi(1<<(WORDSIZE-TOPBITS-1)),T
+ cmp R,T
+ blu not_really_big
+ mov 0,ITER
+ !
+ ! Here, the dividend is >= 2ˆ(31-N) or so. We must be careful here,
+ ! as our usual N-at-a-shot divide step will cause overflow and havoc.
+ ! The total 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 don't even bother to compare to R.
+1:
+ cmp V,T
+ bgeu 3f
+ mov 1,SC
+ sll V,N,V
+ b 1b
+ inc ITER
+! Now compute SC
+2: addcc V,V,V
+ bcc not_too_big
+ add SC,1,SC
+ ! We're here if the divisor overflowed when 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 do_single_div
+ dec SC
+not_too_big:
+3: cmp V,R
+ blu 2b
+ nop
+ be do_single_div
+ nop
+! 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...
+do_single_div:
+ deccc SC
+ bl end_regular_divide
+ nop
+ sub R,V,R
+ mov 1,Q
+ b,a end_single_divloop
+ ! EMPTY
+single_divloop:
+ sll Q,1,Q
+ bl 1f
+ srl V,1,V
+ ! R >= 0
+ sub R,V,R
+ b 2f
+ inc Q
+ 1: ! R < 0
+ add R,V,R
+ dec Q
+ 2:
+ end_single_divloop:
+ deccc SC
+ bge single_divloop
+ tst R
+ b,a end_regular_divide
+ ! EMPTY
+
+not_really_big:
+1:
+ sll V,N,V
+ cmp V,R
+ bleu 1b
+ inccc ITER
+ be got_result
+ dec ITER
+do_regular_divide:
+ ! Do the main division iteration
+ tst R
+ ! Fall through into divide loop
+divloop:
+ sll Q,N,Q
+ DEVELOP_QUOTIENT_BITS( 1, 0 )
+end_regular_divide:
+ deccc ITER
+ bge divloop
+ tst R
+ bl,a got_result
+ ! non-restoring fixup if remainder < 0, otherwise annulled
+ifelse( ANSWER, `quotient',
+` dec Q
+',` add R,divisor,R
+')
+
+got_result:
+ tst SIGN
+ bl,a 1f
+ ! negate for answer < 0, otherwise annulled
+ifelse( ANSWER, `quotient',
+` neg %o2,%o2 ! Q <- -Q
+',` neg %o3,%o3 ! R <- -R
+')
+1:
+ retl ! leaf-routine return
+ifelse( ANSWER, `quotient',
+` mov %o2,%o0 ! quotient <- Q
+',` mov %o3,%o0 ! remainder <- R
+')