lb1spc.asm

这是leon3处理器的交叉编译链

	b	9f
	add	%o2, (-5*2+1), %o2
	
L4.11:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-5*2-1), %o2
	
L3.13:
	! remainder is negative
	addcc	%o3,%o5,%o3
	! depth 4, accumulated bits -7
	bl	L4.9
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-7*2+1), %o2

L4.9:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-7*2-1), %o2
	
	9:
end_regular_divide:
	subcc	%o4, 1, %o4
	bge	divloop
	tst	%o3
	bl,a	got_result
	! non-restoring fixup here (one instruction only!)
	sub	%o2, 1, %o2


got_result:
	! check to see if answer should be < 0
	tst	%g3
	bl,a	1f
	sub %g0, %o2, %o2
1:
	retl
	mov %o2, %o0
#endif

#ifdef L_modsi3
/* This implementation was taken from glibc:
 *
 * Input: dividend and divisor in %o0 and %o1 respectively.
 *
 * 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.
 */
.text
	.align 4
	.global	.urem
	.proc 4
.urem:
	b	divide
	mov	0, %g3		! result always positive

        .align 4
	.global .rem
	.proc 4
.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 nonneg
1:	! %o0 is negative, %o1 is nonnegative
	sub	%g0, %o0, %o0	! make %o0 nonnegative
2:

	! Ready to divide.  Compute size of quotient; scale comparand.
divide:
	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	0x2   !ST_DIV0
		retl
		clr	%o0

1:
	cmp	%o3, %o5		! if %o1 exceeds %o0, done
	blu	got_result		! (and algorithm fails otherwise)
	clr	%o2
	sethi	%hi(1 << (32 - 4 - 1)), %g1
	cmp	%o3, %g1
	blu	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	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	do_single_div
		sub	%g2, 1, %g2

	not_too_big:
	3:	cmp	%o5, %o3
		blu	2b
		nop
		be	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...
	do_single_div:
		subcc	%g2, 1, %g2
		bl	end_regular_divide
		nop
		sub	%o3, %o5, %o3
		mov	1, %o2
		b	end_single_divloop
		nop
	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:
	end_single_divloop:
		subcc	%g2, 1, %g2
		bge	single_divloop
		tst	%o3
		b,a	end_regular_divide

not_really_big:
1:
	sll	%o5, 4, %o5
	cmp	%o5, %o3
	bleu	1b
	addcc	%o4, 1, %o4
	be	got_result
	sub	%o4, 1, %o4

	tst	%o3	! set up for initial iteration
divloop:
	sll	%o2, 4, %o2
		! depth 1, accumulated bits 0
	bl	L1.16
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	! depth 2, accumulated bits 1
	bl	L2.17
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	! depth 3, accumulated bits 3
	bl	L3.19
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	! depth 4, accumulated bits 7
	bl	L4.23
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	add	%o2, (7*2+1), %o2
L4.23:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	add	%o2, (7*2-1), %o2
	
L3.19:
	! remainder is negative
	addcc	%o3,%o5,%o3
	! depth 4, accumulated bits 5
	bl	L4.21
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	add	%o2, (5*2+1), %o2
	
L4.21:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	add	%o2, (5*2-1), %o2
	
L2.17:
	! remainder is negative
	addcc	%o3,%o5,%o3
	! depth 3, accumulated bits 1
	bl	L3.17
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	! depth 4, accumulated bits 3
	bl	L4.19
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	add	%o2, (3*2+1), %o2
	
L4.19:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	add	%o2, (3*2-1), %o2
	
L3.17:
	! remainder is negative
	addcc	%o3,%o5,%o3
	! depth 4, accumulated bits 1
	bl	L4.17
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	add	%o2, (1*2+1), %o2
	
L4.17:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	add	%o2, (1*2-1), %o2
	
L1.16:
	! remainder is negative
	addcc	%o3,%o5,%o3
	! depth 2, accumulated bits -1
	bl	L2.15
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	! depth 3, accumulated bits -1
	bl	L3.15
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	! depth 4, accumulated bits -1
	bl	L4.15
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-1*2+1), %o2
	
L4.15:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-1*2-1), %o2
	
L3.15:
	! remainder is negative
	addcc	%o3,%o5,%o3
	! depth 4, accumulated bits -3
	bl	L4.13
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-3*2+1), %o2
	
L4.13:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-3*2-1), %o2
	
L2.15:
	! remainder is negative
	addcc	%o3,%o5,%o3
	! depth 3, accumulated bits -3
	bl	L3.13
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	! depth 4, accumulated bits -5
	bl	L4.11
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-5*2+1), %o2
	
L4.11:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-5*2-1), %o2
	
L3.13:
	! remainder is negative
	addcc	%o3,%o5,%o3
	! depth 4, accumulated bits -7
	bl	L4.9
	srl	%o5,1,%o5
	! remainder is positive
	subcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-7*2+1), %o2
	
L4.9:
	! remainder is negative
	addcc	%o3,%o5,%o3
	b	9f
	add	%o2, (-7*2-1), %o2
	
	9:
end_regular_divide:
	subcc	%o4, 1, %o4
	bge	divloop
	tst	%o3
	bl,a	got_result
	! non-restoring fixup here (one instruction only!)
	add	%o3, %o1, %o3

got_result:
	! check to see if answer should be < 0
	tst	%g3
	bl,a	1f
	sub %g0, %o3, %o3
1:
	retl
	mov %o3, %o0

#endif