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authorRoland McGrath <roland@gnu.org>1995-02-18 01:27:10 +0000
committerRoland McGrath <roland@gnu.org>1995-02-18 01:27:10 +0000
commit28f540f45bbacd939bfd07f213bcad2bf730b1bf (patch)
tree15f07c4c43d635959c6afee96bde71fb1b3614ee /sysdeps/sparc/divrem.m4
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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 "DEFS.h"
+#ifdef __svr4__
+#include <sys/trap.h>
+#else
+#include <machine/trap.h>
+#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')