Martin Schwidefsky | d9f7a74 | 2006-09-28 16:55:39 +0200 | [diff] [blame] | 1 | /* |
Martin Schwidefsky | d9f7a74 | 2006-09-28 16:55:39 +0200 | [diff] [blame] | 2 | * __div64_32 implementation for 31 bit. |
| 3 | * |
Heiko Carstens | a53c8fa | 2012-07-20 11:15:04 +0200 | [diff] [blame] | 4 | * Copyright IBM Corp. 2006 |
Martin Schwidefsky | d9f7a74 | 2006-09-28 16:55:39 +0200 | [diff] [blame] | 5 | * Author(s): Martin Schwidefsky (schwidefsky@de.ibm.com), |
| 6 | */ |
| 7 | |
| 8 | #include <linux/types.h> |
| 9 | #include <linux/module.h> |
| 10 | |
| 11 | #ifdef CONFIG_MARCH_G5 |
| 12 | |
| 13 | /* |
| 14 | * Function to divide an unsigned 64 bit integer by an unsigned |
| 15 | * 31 bit integer using signed 64/32 bit division. |
| 16 | */ |
| 17 | static uint32_t __div64_31(uint64_t *n, uint32_t base) |
| 18 | { |
| 19 | register uint32_t reg2 asm("2"); |
| 20 | register uint32_t reg3 asm("3"); |
| 21 | uint32_t *words = (uint32_t *) n; |
| 22 | uint32_t tmp; |
| 23 | |
| 24 | /* Special case base==1, remainder = 0, quotient = n */ |
| 25 | if (base == 1) |
| 26 | return 0; |
| 27 | /* |
| 28 | * Special case base==0 will cause a fixed point divide exception |
| 29 | * on the dr instruction and may not happen anyway. For the |
| 30 | * following calculation we can assume base > 1. The first |
| 31 | * signed 64 / 32 bit division with an upper half of 0 will |
| 32 | * give the correct upper half of the 64 bit quotient. |
| 33 | */ |
| 34 | reg2 = 0UL; |
| 35 | reg3 = words[0]; |
| 36 | asm volatile( |
| 37 | " dr %0,%2\n" |
| 38 | : "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" ); |
| 39 | words[0] = reg3; |
| 40 | reg3 = words[1]; |
| 41 | /* |
| 42 | * To get the lower half of the 64 bit quotient and the 32 bit |
| 43 | * remainder we have to use a little trick. Since we only have |
| 44 | * a signed division the quotient can get too big. To avoid this |
| 45 | * the 64 bit dividend is halved, then the signed division will |
| 46 | * work. Afterwards the quotient and the remainder are doubled. |
| 47 | * If the last bit of the dividend has been one the remainder |
| 48 | * is increased by one then checked against the base. If the |
| 49 | * remainder has overflown subtract base and increase the |
| 50 | * quotient. Simple, no ? |
| 51 | */ |
| 52 | asm volatile( |
| 53 | " nr %2,%1\n" |
| 54 | " srdl %0,1\n" |
| 55 | " dr %0,%3\n" |
| 56 | " alr %0,%0\n" |
| 57 | " alr %1,%1\n" |
| 58 | " alr %0,%2\n" |
| 59 | " clr %0,%3\n" |
| 60 | " jl 0f\n" |
| 61 | " slr %0,%3\n" |
Martin Schwidefsky | 4fa81ed2 | 2009-03-18 13:27:32 +0100 | [diff] [blame] | 62 | " ahi %1,1\n" |
Martin Schwidefsky | d9f7a74 | 2006-09-28 16:55:39 +0200 | [diff] [blame] | 63 | "0:\n" |
| 64 | : "+d" (reg2), "+d" (reg3), "=d" (tmp) |
| 65 | : "d" (base), "2" (1UL) : "cc" ); |
| 66 | words[1] = reg3; |
| 67 | return reg2; |
| 68 | } |
| 69 | |
| 70 | /* |
| 71 | * Function to divide an unsigned 64 bit integer by an unsigned |
| 72 | * 32 bit integer using the unsigned 64/31 bit division. |
| 73 | */ |
| 74 | uint32_t __div64_32(uint64_t *n, uint32_t base) |
| 75 | { |
| 76 | uint32_t r; |
| 77 | |
| 78 | /* |
| 79 | * If the most significant bit of base is set, divide n by |
| 80 | * (base/2). That allows to use 64/31 bit division and gives a |
| 81 | * good approximation of the result: n = (base/2)*q + r. The |
| 82 | * result needs to be corrected with two simple transformations. |
| 83 | * If base is already < 2^31-1 __div64_31 can be used directly. |
| 84 | */ |
| 85 | r = __div64_31(n, ((signed) base < 0) ? (base/2) : base); |
| 86 | if ((signed) base < 0) { |
| 87 | uint64_t q = *n; |
| 88 | /* |
| 89 | * First transformation: |
| 90 | * n = (base/2)*q + r |
| 91 | * = ((base/2)*2)*(q/2) + ((q&1) ? (base/2) : 0) + r |
| 92 | * Since r < (base/2), r + (base/2) < base. |
| 93 | * With q1 = (q/2) and r1 = r + ((q&1) ? (base/2) : 0) |
| 94 | * n = ((base/2)*2)*q1 + r1 with r1 < base. |
| 95 | */ |
| 96 | if (q & 1) |
| 97 | r += base/2; |
| 98 | q >>= 1; |
| 99 | /* |
| 100 | * Second transformation. ((base/2)*2) could have lost the |
| 101 | * last bit. |
| 102 | * n = ((base/2)*2)*q1 + r1 |
| 103 | * = base*q1 - ((base&1) ? q1 : 0) + r1 |
| 104 | */ |
| 105 | if (base & 1) { |
| 106 | int64_t rx = r - q; |
| 107 | /* |
| 108 | * base is >= 2^31. The worst case for the while |
| 109 | * loop is n=2^64-1 base=2^31+1. That gives a |
| 110 | * maximum for q=(2^64-1)/2^31 = 0x1ffffffff. Since |
| 111 | * base >= 2^31 the loop is finished after a maximum |
| 112 | * of three iterations. |
| 113 | */ |
| 114 | while (rx < 0) { |
| 115 | rx += base; |
| 116 | q--; |
| 117 | } |
| 118 | r = rx; |
| 119 | } |
| 120 | *n = q; |
| 121 | } |
| 122 | return r; |
| 123 | } |
| 124 | |
| 125 | #else /* MARCH_G5 */ |
| 126 | |
| 127 | uint32_t __div64_32(uint64_t *n, uint32_t base) |
| 128 | { |
| 129 | register uint32_t reg2 asm("2"); |
| 130 | register uint32_t reg3 asm("3"); |
| 131 | uint32_t *words = (uint32_t *) n; |
| 132 | |
| 133 | reg2 = 0UL; |
| 134 | reg3 = words[0]; |
| 135 | asm volatile( |
| 136 | " dlr %0,%2\n" |
| 137 | : "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" ); |
| 138 | words[0] = reg3; |
| 139 | reg3 = words[1]; |
| 140 | asm volatile( |
| 141 | " dlr %0,%2\n" |
| 142 | : "+d" (reg2), "+d" (reg3) : "d" (base) : "cc" ); |
| 143 | words[1] = reg3; |
| 144 | return reg2; |
| 145 | } |
| 146 | |
| 147 | #endif /* MARCH_G5 */ |