blob: 96fe2a17e6081897bfcbf798b588ceec8a229061 [file] [log] [blame]
 /* * Copyright (C) 2015 The Android Open Source Project * * Licensed under the Apache License, Version 2.0 (the "License"); * you may not use this file except in compliance with the License. * You may obtain a copy of the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. * See the License for the specific language governing permissions and * limitations under the License. */ #include "code_generator_utils.h" #include "nodes.h" #include "base/logging.h" namespace art { void CalculateMagicAndShiftForDivRem(int64_t divisor, bool is_long, int64_t* magic, int* shift) { // It does not make sense to calculate magic and shift for zero divisor. DCHECK_NE(divisor, 0); /* Implementation according to H.S.Warren's "Hacker's Delight" (Addison Wesley, 2002) * Chapter 10 and T.Grablund, P.L.Montogomery's "Division by Invariant Integers Using * Multiplication" (PLDI 1994). * The magic number M and shift S can be calculated in the following way: * Let nc be the most positive value of numerator(n) such that nc = kd - 1, * where divisor(d) >= 2. * Let nc be the most negative value of numerator(n) such that nc = kd + 1, * where divisor(d) <= -2. * Thus nc can be calculated like: * nc = exp + exp % d - 1, where d >= 2 and exp = 2^31 for int or 2^63 for long * nc = -exp + (exp + 1) % d, where d >= 2 and exp = 2^31 for int or 2^63 for long * * So the shift p is the smallest p satisfying * 2^p > nc * (d - 2^p % d), where d >= 2 * 2^p > nc * (d + 2^p % d), where d <= -2. * * The magic number M is calculated by * M = (2^p + d - 2^p % d) / d, where d >= 2 * M = (2^p - d - 2^p % d) / d, where d <= -2. * * Notice that p is always bigger than or equal to 32 (resp. 64), so we just return 32 - p * (resp. 64 - p) as the shift number S. */ int64_t p = is_long ? 63 : 31; const uint64_t exp = is_long ? (UINT64_C(1) << 63) : (UINT32_C(1) << 31); // Initialize the computations. uint64_t abs_d = (divisor >= 0) ? divisor : -divisor; uint64_t sign_bit = is_long ? static_cast(divisor) >> 63 : static_cast(divisor) >> 31; uint64_t tmp = exp + sign_bit; uint64_t abs_nc = tmp - 1 - (tmp % abs_d); uint64_t quotient1 = exp / abs_nc; uint64_t remainder1 = exp % abs_nc; uint64_t quotient2 = exp / abs_d; uint64_t remainder2 = exp % abs_d; /* * To avoid handling both positive and negative divisor, "Hacker's Delight" * introduces a method to handle these 2 cases together to avoid duplication. */ uint64_t delta; do { p++; quotient1 = 2 * quotient1; remainder1 = 2 * remainder1; if (remainder1 >= abs_nc) { quotient1++; remainder1 = remainder1 - abs_nc; } quotient2 = 2 * quotient2; remainder2 = 2 * remainder2; if (remainder2 >= abs_d) { quotient2++; remainder2 = remainder2 - abs_d; } delta = abs_d - remainder2; } while (quotient1 < delta || (quotient1 == delta && remainder1 == 0)); *magic = (divisor > 0) ? (quotient2 + 1) : (-quotient2 - 1); if (!is_long) { *magic = static_cast(*magic); } *shift = is_long ? p - 64 : p - 32; } bool IsBooleanValueOrMaterializedCondition(HInstruction* cond_input) { return !cond_input->IsCondition() || !cond_input->IsEmittedAtUseSite(); } } // namespace art