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1 /* -*- Mode: C++; tab-width: 8; indent-tabs-mode: nil; c-basic-offset: 4 -*- |
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2 * vim: set ts=8 sts=4 et sw=4 tw=99: |
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3 * This Source Code Form is subject to the terms of the Mozilla Public |
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4 * License, v. 2.0. If a copy of the MPL was not distributed with this |
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5 * file, You can obtain one at http://mozilla.org/MPL/2.0/. */ |
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6 |
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7 #include "jit/MoveResolver.h" |
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8 |
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9 using namespace js; |
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10 using namespace js::jit; |
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11 |
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12 MoveResolver::MoveResolver() |
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13 : hasCycles_(false) |
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14 { |
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15 } |
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16 |
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17 void |
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18 MoveResolver::resetState() |
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19 { |
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20 hasCycles_ = false; |
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21 } |
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22 |
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23 bool |
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24 MoveResolver::addMove(const MoveOperand &from, const MoveOperand &to, MoveOp::Type type) |
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25 { |
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26 // Assert that we're not doing no-op moves. |
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27 JS_ASSERT(!(from == to)); |
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28 PendingMove *pm = movePool_.allocate(); |
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29 if (!pm) |
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30 return false; |
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31 new (pm) PendingMove(from, to, type); |
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32 pending_.pushBack(pm); |
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33 return true; |
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34 } |
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35 |
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36 // Given move (A -> B), this function attempts to find any move (B -> *) in the |
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37 // pending move list, and returns the first one. |
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38 MoveResolver::PendingMove * |
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39 MoveResolver::findBlockingMove(const PendingMove *last) |
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40 { |
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41 for (PendingMoveIterator iter = pending_.begin(); iter != pending_.end(); iter++) { |
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42 PendingMove *other = *iter; |
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43 |
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44 if (other->from() == last->to()) { |
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45 // We now have pairs in the form (A -> X) (X -> y). The second pair |
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46 // blocks the move in the first pair, so return it. |
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47 return other; |
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48 } |
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49 } |
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50 |
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51 // No blocking moves found. |
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52 return nullptr; |
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53 } |
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54 |
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55 bool |
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56 MoveResolver::resolve() |
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57 { |
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58 resetState(); |
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59 orderedMoves_.clear(); |
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60 |
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61 InlineList<PendingMove> stack; |
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62 |
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63 // This is a depth-first-search without recursion, which tries to find |
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64 // cycles in a list of moves. The output is not entirely optimal for cases |
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65 // where a source has multiple destinations, i.e.: |
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66 // [stack0] -> A |
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67 // [stack0] -> B |
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68 // |
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69 // These moves may not occur next to each other in the list, making it |
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70 // harder for the emitter to optimize memory to memory traffic. However, we |
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71 // expect duplicate sources to be rare in greedy allocation, and indicative |
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72 // of an error in LSRA. |
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73 // |
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74 // Algorithm. |
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75 // |
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76 // S = Traversal stack. |
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77 // P = Pending move list. |
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78 // O = Ordered list of moves. |
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79 // |
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80 // As long as there are pending moves in P: |
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81 // Let |root| be any pending move removed from P |
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82 // Add |root| to the traversal stack. |
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83 // As long as S is not empty: |
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84 // Let |L| be the most recent move added to S. |
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85 // |
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86 // Find any pending move M whose source is L's destination, thus |
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87 // preventing L's move until M has completed. |
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88 // |
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89 // If a move M was found, |
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90 // Remove M from the pending list. |
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91 // If M's destination is |root|, |
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92 // Annotate M and |root| as cycles. |
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93 // Add M to S. |
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94 // do not Add M to O, since M may have other conflictors in P that have not yet been processed. |
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95 // Otherwise, |
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96 // Add M to S. |
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97 // Otherwise, |
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98 // Remove L from S. |
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99 // Add L to O. |
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100 // |
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101 while (!pending_.empty()) { |
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102 PendingMove *pm = pending_.popBack(); |
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103 |
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104 // Add this pending move to the cycle detection stack. |
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105 stack.pushBack(pm); |
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106 |
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107 while (!stack.empty()) { |
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108 PendingMove *blocking = findBlockingMove(stack.peekBack()); |
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109 |
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110 if (blocking) { |
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111 if (blocking->to() == pm->from()) { |
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112 // We annotate cycles at each move in the cycle, and |
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113 // assert that we do not find two cycles in one move chain |
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114 // traversal (which would indicate two moves to the same |
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115 // destination). |
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116 pm->setCycleEnd(); |
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117 blocking->setCycleBegin(pm->type()); |
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118 hasCycles_ = true; |
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119 pending_.remove(blocking); |
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120 stack.pushBack(blocking); |
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121 } else { |
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122 // This is a new link in the move chain, so keep |
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123 // searching for a cycle. |
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124 pending_.remove(blocking); |
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125 stack.pushBack(blocking); |
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126 } |
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127 } else { |
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128 // Otherwise, pop the last move on the search stack because it's |
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129 // complete and not participating in a cycle. The resulting |
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130 // move can safely be added to the ordered move list. |
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131 PendingMove *done = stack.popBack(); |
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132 if (!orderedMoves_.append(*done)) |
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133 return false; |
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134 movePool_.free(done); |
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135 } |
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136 } |
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137 } |
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138 |
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139 return true; |
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140 } |