From: Catherine Moore <clm@gcc.gnu.org>
Date: Wed, 27 May 1998 12:40:31 +0000 (-0400)
Subject: Replace .urem and .rem routines.
X-Git-Url: https://git.libre-soc.org/?a=commitdiff_plain;h=9718d50bb51c94b253272d70bd2249153926217b;p=gcc.git

Replace .urem and .rem routines.

Replace .urem and .rem routines.  Fix a problem
with signed operands and uses leaf procedure calling
convention.

From-SVN: r20090
---

diff --git a/gcc/config/sparc/lb1spc.asm b/gcc/config/sparc/lb1spc.asm
index ec9532d3ec0..d4dea1ae765 100644
--- a/gcc/config/sparc/lb1spc.asm
+++ b/gcc/config/sparc/lb1spc.asm
@@ -282,205 +282,347 @@ got_result:
 #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:
-	save	%sp, -64, %sp
 	b	divide
-	mov	0, %l2		! result always positive
+	mov	0, %g3		! result always positive
+
+        .align 4
 	.global .rem
 	.proc 4
 .rem:
-	save	%sp, -64, %sp
-	orcc	%i1, %i0, %g0	! is either operand negative
-	bge	divide		! if not, skip this junk
-	mov	%i0, %l2	! record sign of result in sign of %i2
-	tst	%i1
-	bge	2f
-	tst	%i0
-	! %i1 < 0
-	bge	divide
-	neg	%i1
-2:	! %i0 < 0
-	neg	%i0
-	!	FALL THROUGH
+	! 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:
-	! Compute size of quotient, scale comparand.
-	orcc	%i1, %g0, %l1		! movcc %i1, %l1
-	te	2			! if %i1 = 0
-	mov	%i0, %i3
-	mov	0, %i2
-	sethi	%hi(1<<(32-4-1)), %l3
-	cmp 	%i3, %l3
+	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
-	mov	0, %l0
-	!
-	! Here, the %i0 is >= 2^(31-3) or so.  We must be careful here,
-	! as our usual 3-at-a-shot divide step will cause overflow and havoc.
-	! The total number of bits in the result here is 3*%l0+%l4, where
-	! %l4 <= 3.
-	! Compute %l0 in an unorthodox manner: know we need to Shift %l1 into
-	! the top decade: so do not even bother to compare to %i3.
-1:	cmp	%l1, %l3
-	bgeu	3f
-	mov	1, %l4
-	sll	%l1, 3, %l1
-	b	1b
-	inc	%l0
-	!
-	! Now compute %l4
-	!
-2:	addcc	%l1, %l1, %l1
-	bcc	not_too_big
-	add	%l4, 1, %l4
-	!
-	! We are here if the %i1 overflowed when Shifting.
-	! This means that %i3 has the high-order bit set.
-	! Restore %l1 and subtract from %i3.
-	sll	%l3, 4, %l3
-	srl	%l1, 1, %l1
-	add	%l1, %l3, %l1
-	b	do_single_div
-	dec	%l4
-not_too_big:
-3:	cmp	%l1, %i3
-	blu	2b
-	nop
-	be	do_single_div
-	nop
-	! %l1 > %i3: went too far: back up 1 step
-	! 	srl	%l1, 1, %l1
-	!	dec	%l4
+	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 %i3 >= %l1, so we can do the
+	! 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 %i3 >= 0.  Because both %i3 and %l1 may have the
-	! high-order bit set in the first step, just falling into the regular
+	! 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:
-	deccc	%l4
-	bl	end_regular_divide
-	nop
-	sub	%i3, %l1, %i3
-	mov	1, %i2
-	b	end_single_divloop
-	nop
-single_divloop:
-	sll	%i2, 1, %i2
-	bl	1f
-	srl	%l1, 1, %l1
-	! %i3 >= 0
-	sub	%i3, %l1, %i3
-	b	2f
-	inc	%i2
-1:	! %i3 < 0
-	add	%i3, %l1, %i3
-	dec	%i2
-end_single_divloop:
-2:	deccc	%l4
-	bge	single_divloop
-	tst	%i3
-	b	end_regular_divide
-	nop
+	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	%l1, 3, %l1
-	cmp	%l1, %i3
+1:
+	sll	%o5, 4, %o5
+	cmp	%o5, %o3
 	bleu	1b
-	inccc	%l0
+	addcc	%o4, 1, %o4
 	be	got_result
-	dec	%l0
-do_regular_divide:
-	! Do the main division iteration
-	tst	%i3
-	! Fall through into divide loop
+	sub	%o4, 1, %o4
+
+	tst	%o3	! set up for initial iteration
 divloop:
-	sll	%i2, 3, %i2
-	! depth 1, accumulated bits 0
-	bl	L.1.8
-	srl	%l1,1,%l1
+	sll	%o2, 4, %o2
+		! depth 1, accumulated bits 0
+	bl	L1.16
+	srl	%o5,1,%o5
 	! remainder is positive
-	subcc	%i3,%l1,%i3
+	subcc	%o3,%o5,%o3
 	! depth 2, accumulated bits 1
-	bl	L.2.9
-	srl	%l1,1,%l1
+	bl	L2.17
+	srl	%o5,1,%o5
 	! remainder is positive
-	subcc	%i3,%l1,%i3
+	subcc	%o3,%o5,%o3
 	! depth 3, accumulated bits 3
-	bl	L.3.11
-	srl	%l1,1,%l1
+	bl	L3.19
+	srl	%o5,1,%o5
 	! remainder is positive
-	subcc	%i3,%l1,%i3
+	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	%i2, (3*2+1), %i2
-L.3.11:	! remainder is negative
-	addcc	%i3,%l1,%i3
+	add	%o2, (7*2+1), %o2
+L4.23:
+	! remainder is negative
+	addcc	%o3,%o5,%o3
 	b	9f
-	add	%i2, (3*2-1), %i2
-L.2.9:	! remainder is negative
-	addcc	%i3,%l1,%i3
+	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	L.3.9
-	srl	%l1,1,%l1
+	bl	L3.17
+	srl	%o5,1,%o5
 	! remainder is positive
-	subcc	%i3,%l1,%i3
+	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	%i2, (1*2+1), %i2
-L.3.9:	! remainder is negative
-	addcc	%i3,%l1,%i3
+	add	%o2, (3*2+1), %o2
+	
+L4.19:
+	! remainder is negative
+	addcc	%o3,%o5,%o3
 	b	9f
-	add	%i2, (1*2-1), %i2
-L.1.8:	! remainder is negative
-	addcc	%i3,%l1,%i3
+	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	L.2.7
-	srl	%l1,1,%l1
+	bl	L2.15
+	srl	%o5,1,%o5
 	! remainder is positive
-	subcc	%i3,%l1,%i3
+	subcc	%o3,%o5,%o3
 	! depth 3, accumulated bits -1
-	bl	L.3.7
-	srl	%l1,1,%l1
+	bl	L3.15
+	srl	%o5,1,%o5
 	! remainder is positive
-	subcc	%i3,%l1,%i3
+	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	%i2, (-1*2+1), %i2
-L.3.7:	! remainder is negative
-	addcc	%i3,%l1,%i3
+	add	%o2, (-1*2+1), %o2
+	
+L4.15:
+	! remainder is negative
+	addcc	%o3,%o5,%o3
 	b	9f
-	add	%i2, (-1*2-1), %i2
-L.2.7:	! remainder is negative
-	addcc	%i3,%l1,%i3
+	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	L.3.5
-	srl	%l1,1,%l1
+	bl	L3.13
+	srl	%o5,1,%o5
 	! remainder is positive
-	subcc	%i3,%l1,%i3
+	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	%i2, (-3*2+1), %i2
-L.3.5:	! remainder is negative
-	addcc	%i3,%l1,%i3
+	add	%o2, (-5*2+1), %o2
+	
+L4.11:
+	! remainder is negative
+	addcc	%o3,%o5,%o3
 	b	9f
-	add	%i2, (-3*2-1), %i2
+	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:
-9:	deccc	%l0
+	subcc	%o4, 1, %o4
 	bge	divloop
-	tst	%i3
-	bge	got_result
-	nop
-	! non-restoring fixup here
-	add	%i3, %i1, %i3
+	tst	%o3
+	bl,a	got_result
+	! non-restoring fixup here (one instruction only!)
+	add	%o3, %o1, %o3
+
 got_result:
-	tst	%l2
-	bge	1f
-	restore
-	! answer < 0
-	retl		! leaf-routine return
-	neg	%o3, %o0	! remainder <- -%i3
-1:	retl		! leaf-routine return
-	mov	%o3, %o0	! remainder <- %i3
-#endif
+	! check to see if answer should be < 0
+	tst	%g3
+	bl,a	1f
+	sub %g0, %o3, %o3
+1:
+	retl
+	mov %o3, %o0
 
+#endif