xref: /original-bsd/lib/libc/sparc/gen/umul.s (revision c3e32dec)
1/*
2 * Copyright (c) 1992, 1993
3 *	The Regents of the University of California.  All rights reserved.
4 *
5 * This software was developed by the Computer Systems Engineering group
6 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
7 * contributed to Berkeley.
8 *
9 * %sccs.include.redist.c%
10 *
11 * from: $Header: umul.s,v 1.4 92/06/25 13:24:05 torek Exp $
12 */
13
14#if defined(LIBC_SCCS) && !defined(lint)
15	.asciz "@(#)umul.s	8.1 (Berkeley) 06/04/93"
16#endif /* LIBC_SCCS and not lint */
17
18/*
19 * Unsigned multiply.  Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the
20 * upper 32 bits of the 64-bit product).
21 *
22 * This code optimizes short (less than 13-bit) multiplies.  Short
23 * multiplies require 25 instruction cycles, and long ones require
24 * 45 instruction cycles.
25 *
26 * On return, overflow has occurred (%o1 is not zero) if and only if
27 * the Z condition code is clear, allowing, e.g., the following:
28 *
29 *	call	.umul
30 *	nop
31 *	bnz	overflow	(or tnz)
32 */
33
34#include "DEFS.h"
35FUNC(.umul)
36	or	%o0, %o1, %o4
37	mov	%o0, %y		! multiplier -> Y
38	andncc	%o4, 0xfff, %g0	! test bits 12..31 of *both* args
39	be	Lmul_shortway	! if zero, can do it the short way
40	andcc	%g0, %g0, %o4	! zero the partial product and clear N and V
41
42	/*
43	 * Long multiply.  32 steps, followed by a final shift step.
44	 */
45	mulscc	%o4, %o1, %o4	! 1
46	mulscc	%o4, %o1, %o4	! 2
47	mulscc	%o4, %o1, %o4	! 3
48	mulscc	%o4, %o1, %o4	! 4
49	mulscc	%o4, %o1, %o4	! 5
50	mulscc	%o4, %o1, %o4	! 6
51	mulscc	%o4, %o1, %o4	! 7
52	mulscc	%o4, %o1, %o4	! 8
53	mulscc	%o4, %o1, %o4	! 9
54	mulscc	%o4, %o1, %o4	! 10
55	mulscc	%o4, %o1, %o4	! 11
56	mulscc	%o4, %o1, %o4	! 12
57	mulscc	%o4, %o1, %o4	! 13
58	mulscc	%o4, %o1, %o4	! 14
59	mulscc	%o4, %o1, %o4	! 15
60	mulscc	%o4, %o1, %o4	! 16
61	mulscc	%o4, %o1, %o4	! 17
62	mulscc	%o4, %o1, %o4	! 18
63	mulscc	%o4, %o1, %o4	! 19
64	mulscc	%o4, %o1, %o4	! 20
65	mulscc	%o4, %o1, %o4	! 21
66	mulscc	%o4, %o1, %o4	! 22
67	mulscc	%o4, %o1, %o4	! 23
68	mulscc	%o4, %o1, %o4	! 24
69	mulscc	%o4, %o1, %o4	! 25
70	mulscc	%o4, %o1, %o4	! 26
71	mulscc	%o4, %o1, %o4	! 27
72	mulscc	%o4, %o1, %o4	! 28
73	mulscc	%o4, %o1, %o4	! 29
74	mulscc	%o4, %o1, %o4	! 30
75	mulscc	%o4, %o1, %o4	! 31
76	mulscc	%o4, %o1, %o4	! 32
77	mulscc	%o4, %g0, %o4	! final shift
78
79
80	/*
81	 * Normally, with the shift-and-add approach, if both numbers are
82	 * positive you get the correct result.  WIth 32-bit two's-complement
83	 * numbers, -x is represented as
84	 *
85	 *		  x		    32
86	 *	( 2  -  ------ ) mod 2  *  2
87	 *		   32
88	 *		  2
89	 *
90	 * (the `mod 2' subtracts 1 from 1.bbbb).  To avoid lots of 2^32s,
91	 * we can treat this as if the radix point were just to the left
92	 * of the sign bit (multiply by 2^32), and get
93	 *
94	 *	-x  =  (2 - x) mod 2
95	 *
96	 * Then, ignoring the `mod 2's for convenience:
97	 *
98	 *   x *  y	= xy
99	 *  -x *  y	= 2y - xy
100	 *   x * -y	= 2x - xy
101	 *  -x * -y	= 4 - 2x - 2y + xy
102	 *
103	 * For signed multiplies, we subtract (x << 32) from the partial
104	 * product to fix this problem for negative multipliers (see mul.s).
105	 * Because of the way the shift into the partial product is calculated
106	 * (N xor V), this term is automatically removed for the multiplicand,
107	 * so we don't have to adjust.
108	 *
109	 * But for unsigned multiplies, the high order bit wasn't a sign bit,
110	 * and the correction is wrong.  So for unsigned multiplies where the
111	 * high order bit is one, we end up with xy - (y << 32).  To fix it
112	 * we add y << 32.
113	 */
114	tst	%o1
115	bl,a	1f		! if %o1 < 0 (high order bit = 1),
116	add	%o4, %o0, %o4	! %o4 += %o0 (add y to upper half)
1171:	rd	%y, %o0		! get lower half of product
118	retl
119	addcc	%o4, %g0, %o1	! put upper half in place and set Z for %o1==0
120
121Lmul_shortway:
122	/*
123	 * Short multiply.  12 steps, followed by a final shift step.
124	 * The resulting bits are off by 12 and (32-12) = 20 bit positions,
125	 * but there is no problem with %o0 being negative (unlike above),
126	 * and overflow is impossible (the answer is at most 24 bits long).
127	 */
128	mulscc	%o4, %o1, %o4	! 1
129	mulscc	%o4, %o1, %o4	! 2
130	mulscc	%o4, %o1, %o4	! 3
131	mulscc	%o4, %o1, %o4	! 4
132	mulscc	%o4, %o1, %o4	! 5
133	mulscc	%o4, %o1, %o4	! 6
134	mulscc	%o4, %o1, %o4	! 7
135	mulscc	%o4, %o1, %o4	! 8
136	mulscc	%o4, %o1, %o4	! 9
137	mulscc	%o4, %o1, %o4	! 10
138	mulscc	%o4, %o1, %o4	! 11
139	mulscc	%o4, %o1, %o4	! 12
140	mulscc	%o4, %g0, %o4	! final shift
141
142	/*
143	 * %o4 has 20 of the bits that should be in the result; %y has
144	 * the bottom 12 (as %y's top 12).  That is:
145	 *
146	 *	  %o4		    %y
147	 * +----------------+----------------+
148	 * | -12- |   -20-  | -12- |   -20-  |
149	 * +------(---------+------)---------+
150	 *	   -----result-----
151	 *
152	 * The 12 bits of %o4 left of the `result' area are all zero;
153	 * in fact, all top 20 bits of %o4 are zero.
154	 */
155
156	rd	%y, %o5
157	sll	%o4, 12, %o0	! shift middle bits left 12
158	srl	%o5, 20, %o5	! shift low bits right 20
159	or	%o5, %o0, %o0
160	retl
161	addcc	%g0, %g0, %o1	! %o1 = zero, and set Z
162