linux/lib/math/div64.c
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   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/export.h>
  22#include <linux/kernel.h>
  23#include <linux/math64.h>
  24
  25/* Not needed on 64bit architectures */
  26#if BITS_PER_LONG == 32
  27
  28#ifndef __div64_32
  29uint32_t __attribute__((weak)) __div64_32(uint64_t *n, uint32_t base)
  30{
  31        uint64_t rem = *n;
  32        uint64_t b = base;
  33        uint64_t res, d = 1;
  34        uint32_t high = rem >> 32;
  35
  36        /* Reduce the thing a bit first */
  37        res = 0;
  38        if (high >= base) {
  39                high /= base;
  40                res = (uint64_t) high << 32;
  41                rem -= (uint64_t) (high*base) << 32;
  42        }
  43
  44        while ((int64_t)b > 0 && b < rem) {
  45                b = b+b;
  46                d = d+d;
  47        }
  48
  49        do {
  50                if (rem >= b) {
  51                        rem -= b;
  52                        res += d;
  53                }
  54                b >>= 1;
  55                d >>= 1;
  56        } while (d);
  57
  58        *n = res;
  59        return rem;
  60}
  61EXPORT_SYMBOL(__div64_32);
  62#endif
  63
  64/**
  65 * div_s64_rem - signed 64bit divide with 64bit divisor and remainder
  66 * @dividend:   64bit dividend
  67 * @divisor:    64bit divisor
  68 * @remainder:  64bit remainder
  69 */
  70#ifndef div_s64_rem
  71s64 div_s64_rem(s64 dividend, s32 divisor, s32 *remainder)
  72{
  73        u64 quotient;
  74
  75        if (dividend < 0) {
  76                quotient = div_u64_rem(-dividend, abs(divisor), (u32 *)remainder);
  77                *remainder = -*remainder;
  78                if (divisor > 0)
  79                        quotient = -quotient;
  80        } else {
  81                quotient = div_u64_rem(dividend, abs(divisor), (u32 *)remainder);
  82                if (divisor < 0)
  83                        quotient = -quotient;
  84        }
  85        return quotient;
  86}
  87EXPORT_SYMBOL(div_s64_rem);
  88#endif
  89
  90/**
  91 * div64_u64_rem - unsigned 64bit divide with 64bit divisor and remainder
  92 * @dividend:   64bit dividend
  93 * @divisor:    64bit divisor
  94 * @remainder:  64bit remainder
  95 *
  96 * This implementation is a comparable to algorithm used by div64_u64.
  97 * But this operation, which includes math for calculating the remainder,
  98 * is kept distinct to avoid slowing down the div64_u64 operation on 32bit
  99 * systems.
 100 */
 101#ifndef div64_u64_rem
 102u64 div64_u64_rem(u64 dividend, u64 divisor, u64 *remainder)
 103{
 104        u32 high = divisor >> 32;
 105        u64 quot;
 106
 107        if (high == 0) {
 108                u32 rem32;
 109                quot = div_u64_rem(dividend, divisor, &rem32);
 110                *remainder = rem32;
 111        } else {
 112                int n = 1 + fls(high);
 113                quot = div_u64(dividend >> n, divisor >> n);
 114
 115                if (quot != 0)
 116                        quot--;
 117
 118                *remainder = dividend - quot * divisor;
 119                if (*remainder >= divisor) {
 120                        quot++;
 121                        *remainder -= divisor;
 122                }
 123        }
 124
 125        return quot;
 126}
 127EXPORT_SYMBOL(div64_u64_rem);
 128#endif
 129
 130/**
 131 * div64_u64 - unsigned 64bit divide with 64bit divisor
 132 * @dividend:   64bit dividend
 133 * @divisor:    64bit divisor
 134 *
 135 * This implementation is a modified version of the algorithm proposed
 136 * by the book 'Hacker's Delight'.  The original source and full proof
 137 * can be found here and is available for use without restriction.
 138 *
 139 * 'http://www.hackersdelight.org/hdcodetxt/divDouble.c.txt'
 140 */
 141#ifndef div64_u64
 142u64 div64_u64(u64 dividend, u64 divisor)
 143{
 144        u32 high = divisor >> 32;
 145        u64 quot;
 146
 147        if (high == 0) {
 148                quot = div_u64(dividend, divisor);
 149        } else {
 150                int n = 1 + fls(high);
 151                quot = div_u64(dividend >> n, divisor >> n);
 152
 153                if (quot != 0)
 154                        quot--;
 155                if ((dividend - quot * divisor) >= divisor)
 156                        quot++;
 157        }
 158
 159        return quot;
 160}
 161EXPORT_SYMBOL(div64_u64);
 162#endif
 163
 164/**
 165 * div64_s64 - signed 64bit divide with 64bit divisor
 166 * @dividend:   64bit dividend
 167 * @divisor:    64bit divisor
 168 */
 169#ifndef div64_s64
 170s64 div64_s64(s64 dividend, s64 divisor)
 171{
 172        s64 quot, t;
 173
 174        quot = div64_u64(abs(dividend), abs(divisor));
 175        t = (dividend ^ divisor) >> 63;
 176
 177        return (quot ^ t) - t;
 178}
 179EXPORT_SYMBOL(div64_s64);
 180#endif
 181
 182#endif /* BITS_PER_LONG == 32 */
 183
 184/*
 185 * Iterative div/mod for use when dividend is not expected to be much
 186 * bigger than divisor.
 187 */
 188u32 iter_div_u64_rem(u64 dividend, u32 divisor, u64 *remainder)
 189{
 190        return __iter_div_u64_rem(dividend, divisor, remainder);
 191}
 192EXPORT_SYMBOL(iter_div_u64_rem);
 193
 194#ifndef mul_u64_u64_div_u64
 195u64 mul_u64_u64_div_u64(u64 a, u64 b, u64 c)
 196{
 197        u64 res = 0, div, rem;
 198        int shift;
 199
 200        /* can a * b overflow ? */
 201        if (ilog2(a) + ilog2(b) > 62) {
 202                /*
 203                 * (b * a) / c is equal to
 204                 *
 205                 *      (b / c) * a +
 206                 *      (b % c) * a / c
 207                 *
 208                 * if nothing overflows. Can the 1st multiplication
 209                 * overflow? Yes, but we do not care: this can only
 210                 * happen if the end result can't fit in u64 anyway.
 211                 *
 212                 * So the code below does
 213                 *
 214                 *      res = (b / c) * a;
 215                 *      b = b % c;
 216                 */
 217                div = div64_u64_rem(b, c, &rem);
 218                res = div * a;
 219                b = rem;
 220
 221                shift = ilog2(a) + ilog2(b) - 62;
 222                if (shift > 0) {
 223                        /* drop precision */
 224                        b >>= shift;
 225                        c >>= shift;
 226                        if (!c)
 227                                return res;
 228                }
 229        }
 230
 231        return res + div64_u64(a * b, c);
 232}
 233#endif
 234