1 /*
2 * Copyright (C) 2015 The Android Open Source Project
3 *
4 * Licensed under the Apache License, Version 2.0 (the "License");
5 * you may not use this file except in compliance with the License.
6 * You may obtain a copy of the License at
7 *
8 * http://www.apache.org/licenses/LICENSE-2.0
9 *
10 * Unless required by applicable law or agreed to in writing, software
11 * distributed under the License is distributed on an "AS IS" BASIS,
12 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
13 * See the License for the specific language governing permissions and
14 * limitations under the License.
15 */
16
17 #include "code_generator_utils.h"
18 #include "nodes.h"
19
20 #include "base/logging.h"
21
22 namespace art {
23
CalculateMagicAndShiftForDivRem(int64_t divisor,bool is_long,int64_t * magic,int * shift)24 void CalculateMagicAndShiftForDivRem(int64_t divisor, bool is_long,
25 int64_t* magic, int* shift) {
26 // It does not make sense to calculate magic and shift for zero divisor.
27 DCHECK_NE(divisor, 0);
28
29 /* Implementation according to H.S.Warren's "Hacker's Delight" (Addison Wesley, 2002)
30 * Chapter 10 and T.Grablund, P.L.Montogomery's "Division by Invariant Integers Using
31 * Multiplication" (PLDI 1994).
32 * The magic number M and shift S can be calculated in the following way:
33 * Let nc be the most positive value of numerator(n) such that nc = kd - 1,
34 * where divisor(d) >= 2.
35 * Let nc be the most negative value of numerator(n) such that nc = kd + 1,
36 * where divisor(d) <= -2.
37 * Thus nc can be calculated like:
38 * nc = exp + exp % d - 1, where d >= 2 and exp = 2^31 for int or 2^63 for long
39 * nc = -exp + (exp + 1) % d, where d >= 2 and exp = 2^31 for int or 2^63 for long
40 *
41 * So the shift p is the smallest p satisfying
42 * 2^p > nc * (d - 2^p % d), where d >= 2
43 * 2^p > nc * (d + 2^p % d), where d <= -2.
44 *
45 * The magic number M is calculated by
46 * M = (2^p + d - 2^p % d) / d, where d >= 2
47 * M = (2^p - d - 2^p % d) / d, where d <= -2.
48 *
49 * Notice that p is always bigger than or equal to 32 (resp. 64), so we just return 32 - p
50 * (resp. 64 - p) as the shift number S.
51 */
52
53 int64_t p = is_long ? 63 : 31;
54 const uint64_t exp = is_long ? (UINT64_C(1) << 63) : (UINT32_C(1) << 31);
55
56 // Initialize the computations.
57 uint64_t abs_d = (divisor >= 0) ? divisor : -divisor;
58 uint64_t sign_bit = is_long ? static_cast<uint64_t>(divisor) >> 63 :
59 static_cast<uint32_t>(divisor) >> 31;
60 uint64_t tmp = exp + sign_bit;
61 uint64_t abs_nc = tmp - 1 - (tmp % abs_d);
62 uint64_t quotient1 = exp / abs_nc;
63 uint64_t remainder1 = exp % abs_nc;
64 uint64_t quotient2 = exp / abs_d;
65 uint64_t remainder2 = exp % abs_d;
66
67 /*
68 * To avoid handling both positive and negative divisor, "Hacker's Delight"
69 * introduces a method to handle these 2 cases together to avoid duplication.
70 */
71 uint64_t delta;
72 do {
73 p++;
74 quotient1 = 2 * quotient1;
75 remainder1 = 2 * remainder1;
76 if (remainder1 >= abs_nc) {
77 quotient1++;
78 remainder1 = remainder1 - abs_nc;
79 }
80 quotient2 = 2 * quotient2;
81 remainder2 = 2 * remainder2;
82 if (remainder2 >= abs_d) {
83 quotient2++;
84 remainder2 = remainder2 - abs_d;
85 }
86 delta = abs_d - remainder2;
87 } while (quotient1 < delta || (quotient1 == delta && remainder1 == 0));
88
89 *magic = (divisor > 0) ? (quotient2 + 1) : (-quotient2 - 1);
90
91 if (!is_long) {
92 *magic = static_cast<int>(*magic);
93 }
94
95 *shift = is_long ? p - 64 : p - 32;
96 }
97
IsBooleanValueOrMaterializedCondition(HInstruction * cond_input)98 bool IsBooleanValueOrMaterializedCondition(HInstruction* cond_input) {
99 return !cond_input->IsCondition() || !cond_input->IsEmittedAtUseSite();
100 }
101
102 } // namespace art
103