/* This file is generated from divrem.m4; DO NOT EDIT! *//* * 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: *  .rem	name of function to generate *  rem		rem=div => %o0 / %o1; rem=rem => %o0 % %o1 *  true		true=true => signed; true=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. */#include <sysdep.h>#include <sys/trap.h>ENTRY(.rem)	! compute sign of result; if neither is negative, no problem	orcc	%o1, %o0, %g0	! either negative?	bge	2f			! no, go do the divide	mov	%o0, %g3		! sign of remainder matches %o0	tst	%o1	bge	1f	tst	%o0	! %o1 is definitely negative; %o0 might also be negative	bge	2f			! if %o0 not negative...	sub	%g0, %o1, %o1	! in any case, make %o1 nonneg1:	! %o0 is negative, %o1 is nonnegative	sub	%g0, %o0, %o0	! make %o0 nonnegative2:	! Ready to divide.  Compute size of quotient; scale comparand.	orcc	%o1, %g0, %o5	bne	1f	mov	%o0, %o3		! 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	%o01:	cmp	%o3, %o5			! if %o1 exceeds %o0, done	blu	LOC(got_result)		! (and algorithm fails otherwise)	clr	%o2	sethi	%hi(1 << (32 - 4 - 1)), %g1	cmp	%o3, %g1	blu	LOC(not_really_big)	clr	%o4	! 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	%o5, %g1		bgeu	3f		mov	1, %g2		sll	%o5, 4, %o5		b	1b		add	%o4, 1, %o4	! Now compute %g2.	2:	addcc	%o5, %o5, %o5		bcc	LOC(not_too_big)		add	%g2, 1, %g2		! 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		add	%o5, %g1, %o5		b	LOC(do_single_div)		sub	%g2, 1, %g2	LOC(not_too_big):	3:	cmp	%o5, %o3		blu	2b		nop		be	LOC(do_single_div)		nop	/* 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	!	srl	%o5, 1, %o5	!	dec	%g2	! 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...	LOC(do_single_div):		subcc	%g2, 1, %g2		bl	LOC(end_regular_divide)		nop		sub	%o3, %o5, %o3		mov	1, %o2		b	LOC(end_single_divloop)		nop	LOC(single_divloop):		sll	%o2, 1, %o2		bl	1f		srl	%o5, 1, %o5		! %o3 >= 0		sub	%o3, %o5, %o3		b	2f		add	%o2, 1, %o2	1:	! %o3 < 0		add	%o3, %o5, %o3		sub	%o2, 1, %o2	2:	LOC(end_single_divloop):		subcc	%g2, 1, %g2		bge	LOC(single_divloop)		tst	%o3		b,a	LOC(end_regular_divide)LOC(not_really_big):1:	sll	%o5, 4, %o5	cmp	%o5, %o3	bleu	1b	addcc	%o4, 1, %o4	be	LOC(got_result)	sub	%o4, 1, %o4	tst	%o3	! set up for initial iterationLOC(divloop):	sll	%o2, 4, %o2		! depth 1, accumulated bits 0	bl	LOC(1.16)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3			! depth 2, accumulated bits 1	bl	LOC(2.17)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3			! depth 3, accumulated bits 3	bl	LOC(3.19)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3			! depth 4, accumulated bits 7	bl	LOC(4.23)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3		b	9f		add	%o2, (7*2+1), %o2	LOC(4.23):	! remainder is negative	addcc	%o3,%o5,%o3		b	9f		add	%o2, (7*2-1), %o2		LOC(3.19):	! remainder is negative	addcc	%o3,%o5,%o3			! depth 4, accumulated bits 5	bl	LOC(4.21)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3		b	9f		add	%o2, (5*2+1), %o2	LOC(4.21):	! remainder is negative	addcc	%o3,%o5,%o3		b	9f		add	%o2, (5*2-1), %o2			LOC(2.17):	! remainder is negative	addcc	%o3,%o5,%o3			! depth 3, accumulated bits 1	bl	LOC(3.17)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3			! depth 4, accumulated bits 3	bl	LOC(4.19)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3		b	9f		add	%o2, (3*2+1), %o2	LOC(4.19):	! remainder is negative	addcc	%o3,%o5,%o3		b	9f		add	%o2, (3*2-1), %o2		LOC(3.17):	! remainder is negative	addcc	%o3,%o5,%o3			! depth 4, accumulated bits 1	bl	LOC(4.17)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3		b	9f		add	%o2, (1*2+1), %o2	LOC(4.17):	! remainder is negative	addcc	%o3,%o5,%o3		b	9f		add	%o2, (1*2-1), %o2				LOC(1.16):	! remainder is negative	addcc	%o3,%o5,%o3			! depth 2, accumulated bits -1	bl	LOC(2.15)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3			! depth 3, accumulated bits -1	bl	LOC(3.15)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3			! depth 4, accumulated bits -1	bl	LOC(4.15)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3		b	9f		add	%o2, (-1*2+1), %o2	LOC(4.15):	! remainder is negative	addcc	%o3,%o5,%o3		b	9f		add	%o2, (-1*2-1), %o2		LOC(3.15):	! remainder is negative	addcc	%o3,%o5,%o3			! depth 4, accumulated bits -3	bl	LOC(4.13)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3		b	9f		add	%o2, (-3*2+1), %o2	LOC(4.13):	! remainder is negative	addcc	%o3,%o5,%o3		b	9f		add	%o2, (-3*2-1), %o2			LOC(2.15):	! remainder is negative	addcc	%o3,%o5,%o3			! depth 3, accumulated bits -3	bl	LOC(3.13)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3			! depth 4, accumulated bits -5	bl	LOC(4.11)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3		b	9f		add	%o2, (-5*2+1), %o2	LOC(4.11):	! remainder is negative	addcc	%o3,%o5,%o3		b	9f		add	%o2, (-5*2-1), %o2		LOC(3.13):	! remainder is negative	addcc	%o3,%o5,%o3			! depth 4, accumulated bits -7	bl	LOC(4.9)	srl	%o5,1,%o5	! remainder is positive	subcc	%o3,%o5,%o3		b	9f		add	%o2, (-7*2+1), %o2	LOC(4.9):	! remainder is negative	addcc	%o3,%o5,%o3		b	9f		add	%o2, (-7*2-1), %o2					9:LOC(end_regular_divide):	subcc	%o4, 1, %o4	bge	LOC(divloop)	tst	%o3	bl,a	LOC(got_result)	! non-restoring fixup here (one instruction only!)	add	%o3, %o1, %o3LOC(got_result):	! check to see if answer should be < 0	tst	%g3	bl,a	1f	sub %g0, %o3, %o31:	retl	mov %o3, %o0END(.rem)