1 files changed, 169 insertions, 0 deletions
diff --git a/arch/sparc/lib/umul.S b/arch/sparc/lib/umul.S
new file mode 100644
index 000000000000..a784720a8a22
--- /dev/null
+++ b/arch/sparc/lib/umul.S
@@ -0,0 +1,169 @@
+/* $Id: umul.S,v 1.4 1996/09/30 02:22:39 davem Exp $
+ * umul.S:      This routine was taken from glibc-1.09 and is covered
+ *              by the GNU Library General Public License Version 2.
+ */
+/*
+ * Unsigned multiply.  Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the
+ * upper 32 bits of the 64-bit product).
+ *
+ * This code optimizes short (less than 13-bit) multiplies.  Short
+ * multiplies require 25 instruction cycles, and long ones require
+ * 45 instruction cycles.
+ *
+ * On return, overflow has occurred (%o1 is not zero) if and only if
+ * the Z condition code is clear, allowing, e.g., the following:
+ *
+ *      call    .umul
+ *      nop
+ *      bnz     overflow        (or tnz)
+ */
+        .globl .umul
+.umul:
+        or      %o0, %o1, %o4
+        mov     %o0, %y         ! multiplier -> Y
+        andncc  %o4, 0xfff, %g0 ! test bits 12..31 of *both* args
+        be      Lmul_shortway   ! if zero, can do it the short way
+         andcc  %g0, %g0, %o4   ! zero the partial product and clear N and V
+        /*
+         * Long multiply.  32 steps, followed by a final shift step.
+         */
+        mulscc  %o4, %o1, %o4   ! 1
+        mulscc  %o4, %o1, %o4   ! 2
+        mulscc  %o4, %o1, %o4   ! 3
+        mulscc  %o4, %o1, %o4   ! 4
+        mulscc  %o4, %o1, %o4   ! 5
+        mulscc  %o4, %o1, %o4   ! 6
+        mulscc  %o4, %o1, %o4   ! 7
+        mulscc  %o4, %o1, %o4   ! 8
+        mulscc  %o4, %o1, %o4   ! 9
+        mulscc  %o4, %o1, %o4   ! 10
+        mulscc  %o4, %o1, %o4   ! 11
+        mulscc  %o4, %o1, %o4   ! 12
+        mulscc  %o4, %o1, %o4   ! 13
+        mulscc  %o4, %o1, %o4   ! 14
+        mulscc  %o4, %o1, %o4   ! 15
+        mulscc  %o4, %o1, %o4   ! 16
+        mulscc  %o4, %o1, %o4   ! 17
+        mulscc  %o4, %o1, %o4   ! 18
+        mulscc  %o4, %o1, %o4   ! 19
+        mulscc  %o4, %o1, %o4   ! 20
+        mulscc  %o4, %o1, %o4   ! 21
+        mulscc  %o4, %o1, %o4   ! 22
+        mulscc  %o4, %o1, %o4   ! 23
+        mulscc  %o4, %o1, %o4   ! 24
+        mulscc  %o4, %o1, %o4   ! 25
+        mulscc  %o4, %o1, %o4   ! 26
+        mulscc  %o4, %o1, %o4   ! 27
+        mulscc  %o4, %o1, %o4   ! 28
+        mulscc  %o4, %o1, %o4   ! 29
+        mulscc  %o4, %o1, %o4   ! 30
+        mulscc  %o4, %o1, %o4   ! 31
+        mulscc  %o4, %o1, %o4   ! 32
+        mulscc  %o4, %g0, %o4   ! final shift
+        /*
+         * Normally, with the shift-and-add approach, if both numbers are
+         * positive you get the correct result.  With 32-bit two's-complement
+         * numbers, -x is represented as
+         *
+         *                x                 32
+         *      ( 2  -  ------ ) mod 2  *  2
+         *                 32
+         *                2
+         *
+         * (the `mod 2' subtracts 1 from 1.bbbb).  To avoid lots of 2^32s,
+         * we can treat this as if the radix point were just to the left
+         * of the sign bit (multiply by 2^32), and get
+         *
+         *      -x  =  (2 - x) mod 2
+         *
+         * Then, ignoring the `mod 2's for convenience:
+         *
+         *   x *  y     = xy
+         *  -x *  y     = 2y - xy
+         *   x * -y     = 2x - xy
+         *  -x * -y     = 4 - 2x - 2y + xy
+         *
+         * For signed multiplies, we subtract (x << 32) from the partial
+         * product to fix this problem for negative multipliers (see mul.s).
+         * Because of the way the shift into the partial product is calculated
+         * (N xor V), this term is automatically removed for the multiplicand,
+         * so we don't have to adjust.
+         *
+         * But for unsigned multiplies, the high order bit wasn't a sign bit,
+         * and the correction is wrong.  So for unsigned multiplies where the
+         * high order bit is one, we end up with xy - (y << 32).  To fix it
+         * we add y << 32.
+         */
+#if 0
+        tst     %o1
+        bl,a    1f              ! if %o1 < 0 (high order bit = 1),
+         add    %o4, %o0, %o4   ! %o4 += %o0 (add y to upper half)
+1:
+        rd      %y, %o0         ! get lower half of product
+        retl
+         addcc  %o4, %g0, %o1   ! put upper half in place and set Z for %o1==0
+#else
+        /* Faster code from tege@sics.se.  */
+        sra     %o1, 31, %o2    ! make mask from sign bit
+        and     %o0, %o2, %o2   ! %o2 = 0 or %o0, depending on sign of %o1
+        rd      %y, %o0         ! get lower half of product
+        retl
+         addcc  %o4, %o2, %o1   ! add compensation and put upper half in place
+#endif
+Lmul_shortway:
+        /*
+         * Short multiply.  12 steps, followed by a final shift step.
+         * The resulting bits are off by 12 and (32-12) = 20 bit positions,
+         * but there is no problem with %o0 being negative (unlike above),
+         * and overflow is impossible (the answer is at most 24 bits long).
+         */
+        mulscc  %o4, %o1, %o4   ! 1
+        mulscc  %o4, %o1, %o4   ! 2
+        mulscc  %o4, %o1, %o4   ! 3
+        mulscc  %o4, %o1, %o4   ! 4
+        mulscc  %o4, %o1, %o4   ! 5
+        mulscc  %o4, %o1, %o4   ! 6
+        mulscc  %o4, %o1, %o4   ! 7
+        mulscc  %o4, %o1, %o4   ! 8
+        mulscc  %o4, %o1, %o4   ! 9
+        mulscc  %o4, %o1, %o4   ! 10
+        mulscc  %o4, %o1, %o4   ! 11
+        mulscc  %o4, %o1, %o4   ! 12
+        mulscc  %o4, %g0, %o4   ! final shift
+        /*
+         * %o4 has 20 of the bits that should be in the result; %y has
+         * the bottom 12 (as %y's top 12).  That is:
+         *
+         *        %o4               %y
+         * +----------------+----------------+
+         * | -12- |   -20-  | -12- |   -20-  |
+         * +------(---------+------)---------+
+         *         -----result-----
+         *
+         * The 12 bits of %o4 left of the `result' area are all zero;
+         * in fact, all top 20 bits of %o4 are zero.
+         */
+        rd      %y, %o5
+        sll     %o4, 12, %o0    ! shift middle bits left 12
+        srl     %o5, 20, %o5    ! shift low bits right 20
+        or      %o5, %o0, %o0
+        retl
+         addcc  %g0, %g0, %o1   ! %o1 = zero, and set Z
+        .globl  .umul_patch
+.umul_patch:
+        umul    %o0, %o1, %o0
+        retl
+         rd     %y, %o1
+        nop

diff --git a/arch/sparc/lib/umul.S b/arch/sparc/lib/umul.S new file mode 100644 index 000000000000..a784720a8a22 --- /dev/null +++ b/arch/sparc/lib/umul.S
@@ -0,0 +1,169 @@
	1	/* $Id: umul.S,v 1.4 1996/09/30 02:22:39 davem Exp $
	2	* umul.S: This routine was taken from glibc-1.09 and is covered
	3	* by the GNU Library General Public License Version 2.
	4	*/
	5
	6
	7	/*
	8	* Unsigned multiply. Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the
	9	* upper 32 bits of the 64-bit product).
	10	*
	11	* This code optimizes short (less than 13-bit) multiplies. Short
	12	* multiplies require 25 instruction cycles, and long ones require
	13	* 45 instruction cycles.
	14	*
	15	* On return, overflow has occurred (%o1 is not zero) if and only if
	16	* the Z condition code is clear, allowing, e.g., the following:
	17	*
	18	* call .umul
	19	* nop
	20	* bnz overflow (or tnz)
	21	*/
	22
	23	.globl .umul
	24	.umul:
	25	or %o0, %o1, %o4
	26	mov %o0, %y ! multiplier -> Y
	27
	28	andncc %o4, 0xfff, %g0 ! test bits 12..31 of both args
	29	be Lmul_shortway ! if zero, can do it the short way
	30	andcc %g0, %g0, %o4 ! zero the partial product and clear N and V
	31
	32	/*
	33	* Long multiply. 32 steps, followed by a final shift step.
	34	*/
	35	mulscc %o4, %o1, %o4 ! 1
	36	mulscc %o4, %o1, %o4 ! 2
	37	mulscc %o4, %o1, %o4 ! 3
	38	mulscc %o4, %o1, %o4 ! 4
	39	mulscc %o4, %o1, %o4 ! 5
	40	mulscc %o4, %o1, %o4 ! 6
	41	mulscc %o4, %o1, %o4 ! 7
	42	mulscc %o4, %o1, %o4 ! 8
	43	mulscc %o4, %o1, %o4 ! 9
	44	mulscc %o4, %o1, %o4 ! 10
	45	mulscc %o4, %o1, %o4 ! 11
	46	mulscc %o4, %o1, %o4 ! 12
	47	mulscc %o4, %o1, %o4 ! 13
	48	mulscc %o4, %o1, %o4 ! 14
	49	mulscc %o4, %o1, %o4 ! 15
	50	mulscc %o4, %o1, %o4 ! 16
	51	mulscc %o4, %o1, %o4 ! 17
	52	mulscc %o4, %o1, %o4 ! 18
	53	mulscc %o4, %o1, %o4 ! 19
	54	mulscc %o4, %o1, %o4 ! 20
	55	mulscc %o4, %o1, %o4 ! 21
	56	mulscc %o4, %o1, %o4 ! 22
	57	mulscc %o4, %o1, %o4 ! 23
	58	mulscc %o4, %o1, %o4 ! 24
	59	mulscc %o4, %o1, %o4 ! 25
	60	mulscc %o4, %o1, %o4 ! 26
	61	mulscc %o4, %o1, %o4 ! 27
	62	mulscc %o4, %o1, %o4 ! 28
	63	mulscc %o4, %o1, %o4 ! 29
	64	mulscc %o4, %o1, %o4 ! 30
	65	mulscc %o4, %o1, %o4 ! 31
	66	mulscc %o4, %o1, %o4 ! 32
	67	mulscc %o4, %g0, %o4 ! final shift
	68
	69
	70	/*
	71	* Normally, with the shift-and-add approach, if both numbers are
	72	* positive you get the correct result. With 32-bit two's-complement
	73	* numbers, -x is represented as
	74	*
	75	* x 32
	76	* ( 2 - ------ ) mod 2 * 2
	77	* 32
	78	* 2
	79	*
	80	* (the `mod 2' subtracts 1 from 1.bbbb). To avoid lots of 2^32s,
	81	* we can treat this as if the radix point were just to the left
	82	* of the sign bit (multiply by 2^32), and get
	83	*
	84	* -x = (2 - x) mod 2
	85	*
	86	* Then, ignoring the `mod 2's for convenience:
	87	*
	88	* x * y = xy
	89	* -x * y = 2y - xy
	90	* x * -y = 2x - xy
	91	* -x * -y = 4 - 2x - 2y + xy
	92	*
	93	* For signed multiplies, we subtract (x << 32) from the partial
	94	* product to fix this problem for negative multipliers (see mul.s).
	95	* Because of the way the shift into the partial product is calculated
	96	* (N xor V), this term is automatically removed for the multiplicand,
	97	* so we don't have to adjust.
	98	*
	99	* But for unsigned multiplies, the high order bit wasn't a sign bit,
	100	* and the correction is wrong. So for unsigned multiplies where the
	101	* high order bit is one, we end up with xy - (y << 32). To fix it
	102	* we add y << 32.
	103	*/
	104	#if 0
	105	tst %o1
	106	bl,a 1f ! if %o1 < 0 (high order bit = 1),
	107	add %o4, %o0, %o4 ! %o4 += %o0 (add y to upper half)
	108
	109	1:
	110	rd %y, %o0 ! get lower half of product
	111	retl
	112	addcc %o4, %g0, %o1 ! put upper half in place and set Z for %o1==0
	113	#else
	114	/* Faster code from tege@sics.se. */
	115	sra %o1, 31, %o2 ! make mask from sign bit
	116	and %o0, %o2, %o2 ! %o2 = 0 or %o0, depending on sign of %o1
	117	rd %y, %o0 ! get lower half of product
	118	retl
	119	addcc %o4, %o2, %o1 ! add compensation and put upper half in place
	120	#endif
	121
	122	Lmul_shortway:
	123	/*
	124	* Short multiply. 12 steps, followed by a final shift step.
	125	* The resulting bits are off by 12 and (32-12) = 20 bit positions,
	126	* but there is no problem with %o0 being negative (unlike above),
	127	* and overflow is impossible (the answer is at most 24 bits long).
	128	*/
	129	mulscc %o4, %o1, %o4 ! 1
	130	mulscc %o4, %o1, %o4 ! 2
	131	mulscc %o4, %o1, %o4 ! 3
	132	mulscc %o4, %o1, %o4 ! 4
	133	mulscc %o4, %o1, %o4 ! 5
	134	mulscc %o4, %o1, %o4 ! 6
	135	mulscc %o4, %o1, %o4 ! 7
	136	mulscc %o4, %o1, %o4 ! 8
	137	mulscc %o4, %o1, %o4 ! 9
	138	mulscc %o4, %o1, %o4 ! 10
	139	mulscc %o4, %o1, %o4 ! 11
	140	mulscc %o4, %o1, %o4 ! 12
	141	mulscc %o4, %g0, %o4 ! final shift
	142
	143	/*
	144	* %o4 has 20 of the bits that should be in the result; %y has
	145	* the bottom 12 (as %y's top 12). That is:
	146	*
	147	* %o4 %y
	148	* +----------------+----------------+
	149	* \| -12- \| -20- \| -12- \| -20- \|
	150	* +------(---------+------)---------+
	151	* -----result-----
	152	*
	153	* The 12 bits of %o4 left of the `result' area are all zero;
	154	* in fact, all top 20 bits of %o4 are zero.
	155	*/
	156
	157	rd %y, %o5
	158	sll %o4, 12, %o0 ! shift middle bits left 12
	159	srl %o5, 20, %o5 ! shift low bits right 20
	160	or %o5, %o0, %o0
	161	retl
	162	addcc %g0, %g0, %o1 ! %o1 = zero, and set Z
	163
	164	.globl .umul_patch
	165	.umul_patch:
	166	umul %o0, %o1, %o0
	167	retl
	168	rd %y, %o1
	169	nop