1// SPDX-License-Identifier: GPL-2.0
2/*
3 * Copyright (C) 2003 Bernardo Innocenti <bernie@develer.com>
4 *
5 * Based on former do_div() implementation from asm-parisc/div64.h:
6 *	Copyright (C) 1999 Hewlett-Packard Co
7 *	Copyright (C) 1999 David Mosberger-Tang <davidm@hpl.hp.com>
8 *
9 *
10 * Generic C version of 64bit/32bit division and modulo, with
11 * 64bit result and 32bit remainder.
12 *
13 * The fast case for (n>>32 == 0) is handled inline by do_div().
14 *
15 * Code generated for this function might be very inefficient
16 * for some CPUs. __div64_32() can be overridden by linking arch-specific
17 * assembly versions such as arch/ppc/lib/div64.S and arch/sh/lib/div64.S
18 * or by defining a preprocessor macro in arch/include/asm/div64.h.
19 */
20
21#include <linux/bitops.h>
22#include <linux/export.h>
23#include <linux/math.h>
24#include <linux/math64.h>
25#include <linux/minmax.h>
26#include <linux/log2.h>
27
28/* Not needed on 64bit architectures */
29#if BITS_PER_LONG == 32
30
31#ifndef __div64_32
32uint32_t __attribute__((weak)) __div64_32(uint64_t *n, uint32_t base)
33{
34	uint64_t rem = *n;
35	uint64_t b = base;
36	uint64_t res, d = 1;
37	uint32_t high = rem >> 32;
38
39	/* Reduce the thing a bit first */
40	res = 0;
41	if (high >= base) {
42		high /= base;
43		res = (uint64_t) high << 32;
44		rem -= (uint64_t) (high*base) << 32;
45	}
46
47	while ((int64_t)b > 0 && b < rem) {
48		b = b+b;
49		d = d+d;
50	}
51
52	do {
53		if (rem >= b) {
54			rem -= b;
55			res += d;
56		}
57		b >>= 1;
58		d >>= 1;
59	} while (d);
60
61	*n = res;
62	return rem;
63}
64EXPORT_SYMBOL(__div64_32);
65#endif
66
67#ifndef div_s64_rem
68s64 div_s64_rem(s64 dividend, s32 divisor, s32 *remainder)
69{
70	u64 quotient;
71
72	if (dividend < 0) {
73		quotient = div_u64_rem(-dividend, abs(divisor), (u32 *)remainder);
74		*remainder = -*remainder;
75		if (divisor > 0)
76			quotient = -quotient;
77	} else {
78		quotient = div_u64_rem(dividend, abs(divisor), (u32 *)remainder);
79		if (divisor < 0)
80			quotient = -quotient;
81	}
82	return quotient;
83}
84EXPORT_SYMBOL(div_s64_rem);
85#endif
86
87/*
88 * div64_u64_rem - unsigned 64bit divide with 64bit divisor and remainder
89 * @dividend:	64bit dividend
90 * @divisor:	64bit divisor
91 * @remainder:  64bit remainder
92 *
93 * This implementation is a comparable to algorithm used by div64_u64.
94 * But this operation, which includes math for calculating the remainder,
95 * is kept distinct to avoid slowing down the div64_u64 operation on 32bit
96 * systems.
97 */
98#ifndef div64_u64_rem
99u64 div64_u64_rem(u64 dividend, u64 divisor, u64 *remainder)
100{
101	u32 high = divisor >> 32;
102	u64 quot;
103
104	if (high == 0) {
105		u32 rem32;
106		quot = div_u64_rem(dividend, divisor, &rem32);
107		*remainder = rem32;
108	} else {
109		int n = fls(high);
110		quot = div_u64(dividend >> n, divisor >> n);
111
112		if (quot != 0)
113			quot--;
114
115		*remainder = dividend - quot * divisor;
116		if (*remainder >= divisor) {
117			quot++;
118			*remainder -= divisor;
119		}
120	}
121
122	return quot;
123}
124EXPORT_SYMBOL(div64_u64_rem);
125#endif
126
127/*
128 * div64_u64 - unsigned 64bit divide with 64bit divisor
129 * @dividend:	64bit dividend
130 * @divisor:	64bit divisor
131 *
132 * This implementation is a modified version of the algorithm proposed
133 * by the book 'Hacker's Delight'.  The original source and full proof
134 * can be found here and is available for use without restriction.
135 *
136 * 'http://www.hackersdelight.org/hdcodetxt/divDouble.c.txt'
137 */
138#ifndef div64_u64
139u64 div64_u64(u64 dividend, u64 divisor)
140{
141	u32 high = divisor >> 32;
142	u64 quot;
143
144	if (high == 0) {
145		quot = div_u64(dividend, divisor);
146	} else {
147		int n = fls(high);
148		quot = div_u64(dividend >> n, divisor >> n);
149
150		if (quot != 0)
151			quot--;
152		if ((dividend - quot * divisor) >= divisor)
153			quot++;
154	}
155
156	return quot;
157}
158EXPORT_SYMBOL(div64_u64);
159#endif
160
161#ifndef div64_s64
162s64 div64_s64(s64 dividend, s64 divisor)
163{
164	s64 quot, t;
165
166	quot = div64_u64(abs(dividend), abs(divisor));
167	t = (dividend ^ divisor) >> 63;
168
169	return (quot ^ t) - t;
170}
171EXPORT_SYMBOL(div64_s64);
172#endif
173
174#endif /* BITS_PER_LONG == 32 */
175
176/*
177 * Iterative div/mod for use when dividend is not expected to be much
178 * bigger than divisor.
179 */
180u32 iter_div_u64_rem(u64 dividend, u32 divisor, u64 *remainder)
181{
182	return __iter_div_u64_rem(dividend, divisor, remainder);
183}
184EXPORT_SYMBOL(iter_div_u64_rem);
185
186#ifndef mul_u64_u64_div_u64
187u64 mul_u64_u64_div_u64(u64 a, u64 b, u64 c)
188{
189	u64 res = 0, div, rem;
190	int shift;
191
192	/* can a * b overflow ? */
193	if (ilog2(a) + ilog2(b) > 62) {
194		/*
195		 * Note that the algorithm after the if block below might lose
196		 * some precision and the result is more exact for b > a. So
197		 * exchange a and b if a is bigger than b.
198		 *
199		 * For example with a = 43980465100800, b = 100000000, c = 1000000000
200		 * the below calculation doesn't modify b at all because div == 0
201		 * and then shift becomes 45 + 26 - 62 = 9 and so the result
202		 * becomes 4398035251080. However with a and b swapped the exact
203		 * result is calculated (i.e. 4398046510080).
204		 */
205		if (a > b)
206			swap(a, b);
207
208		/*
209		 * (b * a) / c is equal to
210		 *
211		 *      (b / c) * a +
212		 *      (b % c) * a / c
213		 *
214		 * if nothing overflows. Can the 1st multiplication
215		 * overflow? Yes, but we do not care: this can only
216		 * happen if the end result can't fit in u64 anyway.
217		 *
218		 * So the code below does
219		 *
220		 *      res = (b / c) * a;
221		 *      b = b % c;
222		 */
223		div = div64_u64_rem(b, c, &rem);
224		res = div * a;
225		b = rem;
226
227		shift = ilog2(a) + ilog2(b) - 62;
228		if (shift > 0) {
229			/* drop precision */
230			b >>= shift;
231			c >>= shift;
232			if (!c)
233				return res;
234		}
235	}
236
237	return res + div64_u64(a * b, c);
238}
239EXPORT_SYMBOL(mul_u64_u64_div_u64);
240#endif
241