102 lines
3.5 KiB
C++
102 lines
3.5 KiB
C++
/*
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* Copyright (C) 2015 The Android Open Source Project
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*
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* Licensed under the Apache License, Version 2.0 (the "License");
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* you may not use this file except in compliance with the License.
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* You may obtain a copy of the License at
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*
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* http://www.apache.org/licenses/LICENSE-2.0
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*
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* Unless required by applicable law or agreed to in writing, software
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* distributed under the License is distributed on an "AS IS" BASIS,
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* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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* See the License for the specific language governing permissions and
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* limitations under the License.
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*/
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#include "code_generator_utils.h"
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#include "nodes.h"
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#include "base/logging.h"
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namespace art {
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void CalculateMagicAndShiftForDivRem(int64_t divisor, bool is_long,
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int64_t* magic, int* shift) {
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// It does not make sense to calculate magic and shift for zero divisor.
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DCHECK_NE(divisor, 0);
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/* Implementation according to H.S.Warren's "Hacker's Delight" (Addison Wesley, 2002)
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* Chapter 10 and T.Grablund, P.L.Montogomery's "Division by Invariant Integers Using
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* Multiplication" (PLDI 1994).
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* The magic number M and shift S can be calculated in the following way:
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* Let nc be the most positive value of numerator(n) such that nc = kd - 1,
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* where divisor(d) >= 2.
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* Let nc be the most negative value of numerator(n) such that nc = kd + 1,
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* where divisor(d) <= -2.
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* Thus nc can be calculated like:
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* nc = exp + exp % d - 1, where d >= 2 and exp = 2^31 for int or 2^63 for long
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* nc = -exp + (exp + 1) % d, where d >= 2 and exp = 2^31 for int or 2^63 for long
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*
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* So the shift p is the smallest p satisfying
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* 2^p > nc * (d - 2^p % d), where d >= 2
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* 2^p > nc * (d + 2^p % d), where d <= -2.
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*
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* The magic number M is calculated by
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* M = (2^p + d - 2^p % d) / d, where d >= 2
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* M = (2^p - d - 2^p % d) / d, where d <= -2.
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*
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* Notice that p is always bigger than or equal to 32 (resp. 64), so we just return 32 - p
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* (resp. 64 - p) as the shift number S.
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*/
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int64_t p = is_long ? 63 : 31;
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const uint64_t exp = is_long ? (UINT64_C(1) << 63) : (UINT32_C(1) << 31);
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// Initialize the computations.
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uint64_t abs_d = (divisor >= 0) ? divisor : -divisor;
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uint64_t sign_bit = is_long ? static_cast<uint64_t>(divisor) >> 63 :
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static_cast<uint32_t>(divisor) >> 31;
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uint64_t tmp = exp + sign_bit;
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uint64_t abs_nc = tmp - 1 - (tmp % abs_d);
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uint64_t quotient1 = exp / abs_nc;
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uint64_t remainder1 = exp % abs_nc;
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uint64_t quotient2 = exp / abs_d;
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uint64_t remainder2 = exp % abs_d;
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/*
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* To avoid handling both positive and negative divisor, "Hacker's Delight"
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* introduces a method to handle these 2 cases together to avoid duplication.
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*/
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uint64_t delta;
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do {
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p++;
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quotient1 = 2 * quotient1;
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remainder1 = 2 * remainder1;
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if (remainder1 >= abs_nc) {
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quotient1++;
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remainder1 = remainder1 - abs_nc;
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}
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quotient2 = 2 * quotient2;
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remainder2 = 2 * remainder2;
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if (remainder2 >= abs_d) {
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quotient2++;
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remainder2 = remainder2 - abs_d;
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}
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delta = abs_d - remainder2;
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} while (quotient1 < delta || (quotient1 == delta && remainder1 == 0));
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*magic = (divisor > 0) ? (quotient2 + 1) : (-quotient2 - 1);
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if (!is_long) {
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*magic = static_cast<int>(*magic);
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}
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*shift = is_long ? p - 64 : p - 32;
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}
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bool IsBooleanValueOrMaterializedCondition(HInstruction* cond_input) {
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return !cond_input->IsCondition() || !cond_input->IsEmittedAtUseSite();
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}
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} // namespace art
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