1/*
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	.globl _Umul
25.umul:
26_Umul:	/* needed for export */
27	or	%o0, %o1, %o4
28	mov	%o0, %y		! multiplier -> Y
29
30	andncc	%o4, 0xfff, %g0	! test bits 12..31 of *both* args
31	be	Lmul_shortway	! if zero, can do it the short way
32	 andcc	%g0, %g0, %o4	! zero the partial product and clear N and V
33
34	/*
35	 * Long multiply.  32 steps, followed by a final shift step.
36	 */
37	mulscc	%o4, %o1, %o4	! 1
38	mulscc	%o4, %o1, %o4	! 2
39	mulscc	%o4, %o1, %o4	! 3
40	mulscc	%o4, %o1, %o4	! 4
41	mulscc	%o4, %o1, %o4	! 5
42	mulscc	%o4, %o1, %o4	! 6
43	mulscc	%o4, %o1, %o4	! 7
44	mulscc	%o4, %o1, %o4	! 8
45	mulscc	%o4, %o1, %o4	! 9
46	mulscc	%o4, %o1, %o4	! 10
47	mulscc	%o4, %o1, %o4	! 11
48	mulscc	%o4, %o1, %o4	! 12
49	mulscc	%o4, %o1, %o4	! 13
50	mulscc	%o4, %o1, %o4	! 14
51	mulscc	%o4, %o1, %o4	! 15
52	mulscc	%o4, %o1, %o4	! 16
53	mulscc	%o4, %o1, %o4	! 17
54	mulscc	%o4, %o1, %o4	! 18
55	mulscc	%o4, %o1, %o4	! 19
56	mulscc	%o4, %o1, %o4	! 20
57	mulscc	%o4, %o1, %o4	! 21
58	mulscc	%o4, %o1, %o4	! 22
59	mulscc	%o4, %o1, %o4	! 23
60	mulscc	%o4, %o1, %o4	! 24
61	mulscc	%o4, %o1, %o4	! 25
62	mulscc	%o4, %o1, %o4	! 26
63	mulscc	%o4, %o1, %o4	! 27
64	mulscc	%o4, %o1, %o4	! 28
65	mulscc	%o4, %o1, %o4	! 29
66	mulscc	%o4, %o1, %o4	! 30
67	mulscc	%o4, %o1, %o4	! 31
68	mulscc	%o4, %o1, %o4	! 32
69	mulscc	%o4, %g0, %o4	! final shift
70
71
72	/*
73	 * Normally, with the shift-and-add approach, if both numbers are
74	 * positive you get the correct result.  With 32-bit two's-complement
75	 * numbers, -x is represented as
76	 *
77	 *		  x		    32
78	 *	( 2  -  ------ ) mod 2  *  2
79	 *		   32
80	 *		  2
81	 *
82	 * (the `mod 2' subtracts 1 from 1.bbbb).  To avoid lots of 2^32s,
83	 * we can treat this as if the radix point were just to the left
84	 * of the sign bit (multiply by 2^32), and get
85	 *
86	 *	-x  =  (2 - x) mod 2
87	 *
88	 * Then, ignoring the `mod 2's for convenience:
89	 *
90	 *   x *  y	= xy
91	 *  -x *  y	= 2y - xy
92	 *   x * -y	= 2x - xy
93	 *  -x * -y	= 4 - 2x - 2y + xy
94	 *
95	 * For signed multiplies, we subtract (x << 32) from the partial
96	 * product to fix this problem for negative multipliers (see mul.s).
97	 * Because of the way the shift into the partial product is calculated
98	 * (N xor V), this term is automatically removed for the multiplicand,
99	 * so we don't have to adjust.
100	 *
101	 * But for unsigned multiplies, the high order bit wasn't a sign bit,
102	 * and the correction is wrong.  So for unsigned multiplies where the
103	 * high order bit is one, we end up with xy - (y << 32).  To fix it
104	 * we add y << 32.
105	 */
106#if 0
107	tst	%o1
108	bl,a	1f		! if %o1 < 0 (high order bit = 1),
109	 add	%o4, %o0, %o4	! %o4 += %o0 (add y to upper half)
110
1111:
112	rd	%y, %o0		! get lower half of product
113	retl
114	 addcc	%o4, %g0, %o1	! put upper half in place and set Z for %o1==0
115#else
116	/* Faster code from tege@sics.se.  */
117	sra	%o1, 31, %o2	! make mask from sign bit
118	and	%o0, %o2, %o2	! %o2 = 0 or %o0, depending on sign of %o1
119	rd	%y, %o0		! get lower half of product
120	retl
121	 addcc	%o4, %o2, %o1	! add compensation and put upper half in place
122#endif
123
124Lmul_shortway:
125	/*
126	 * Short multiply.  12 steps, followed by a final shift step.
127	 * The resulting bits are off by 12 and (32-12) = 20 bit positions,
128	 * but there is no problem with %o0 being negative (unlike above),
129	 * and overflow is impossible (the answer is at most 24 bits long).
130	 */
131	mulscc	%o4, %o1, %o4	! 1
132	mulscc	%o4, %o1, %o4	! 2
133	mulscc	%o4, %o1, %o4	! 3
134	mulscc	%o4, %o1, %o4	! 4
135	mulscc	%o4, %o1, %o4	! 5
136	mulscc	%o4, %o1, %o4	! 6
137	mulscc	%o4, %o1, %o4	! 7
138	mulscc	%o4, %o1, %o4	! 8
139	mulscc	%o4, %o1, %o4	! 9
140	mulscc	%o4, %o1, %o4	! 10
141	mulscc	%o4, %o1, %o4	! 11
142	mulscc	%o4, %o1, %o4	! 12
143	mulscc	%o4, %g0, %o4	! final shift
144
145	/*
146	 * %o4 has 20 of the bits that should be in the result; %y has
147	 * the bottom 12 (as %y's top 12).  That is:
148	 *
149	 *	  %o4		    %y
150	 * +----------------+----------------+
151	 * | -12- |   -20-  | -12- |   -20-  |
152	 * +------(---------+------)---------+
153	 *	   -----result-----
154	 *
155	 * The 12 bits of %o4 left of the `result' area are all zero;
156	 * in fact, all top 20 bits of %o4 are zero.
157	 */
158
159	rd	%y, %o5
160	sll	%o4, 12, %o0	! shift middle bits left 12
161	srl	%o5, 20, %o5	! shift low bits right 20
162	or	%o5, %o0, %o0
163	retl
164	 addcc	%g0, %g0, %o1	! %o1 = zero, and set Z
165
166	.globl	.umul_patch
167.umul_patch:
168	umul	%o0, %o1, %o0
169	retl
170	 rd	%y, %o1
171	nop
172