Convert And+Sub into bic if possible

Convert
 i1: AND a, b
     SUB a, i1
into:
     BIC a, a, b

Also works if `i1` is AND b, a.

Bug: 304972513
Fixes: 304972513
Test: art/test/testrunner/testrunner.py --host --64 --target
Change-Id: I22218c263f52b58d87431186588ac166dc93246a

1 files changed, 30 insertions, 0 deletions

diff --git a/compiler/optimizing/instruction_simplifier_shared.cc b/compiler/optimizing/instruction_simplifier_shared.cc
index 50ea2b929b..deb8f93492 100644
--- a/compiler/optimizing/instruction_simplifier_shared.cc
+++ b/compiler/optimizing/instruction_simplifier_shared.cc
@@ -18,6 +18,7 @@
 
 #include "code_generator.h"
 #include "mirror/array-inl.h"
+#include "nodes.h"
 
 namespace art HIDDEN {
 
@@ -229,6 +230,35 @@ bool TryMergeNegatedInput(HBinaryOperation* op) {
   return false;
 }
 
+bool TryMergeWithAnd(HSub* instruction) {
+  HAnd* and_instr = instruction->GetRight()->AsAndOrNull();
+  if (and_instr == nullptr) {
+    return false;
+  }
+
+  HInstruction* value = instruction->GetLeft();
+
+  HInstruction* left = and_instr->GetLeft();
+  const bool left_is_equal = left == value;
+  HInstruction* right = and_instr->GetRight();
+  const bool right_is_equal = right == value;
+  if (!left_is_equal && !right_is_equal) {
+    return false;
+  }
+
+  HBitwiseNegatedRight* bnr = new (instruction->GetBlock()->GetGraph()->GetAllocator())
+      HBitwiseNegatedRight(instruction->GetType(),
+                           HInstruction::InstructionKind::kAnd,
+                           value,
+                           left_is_equal ? right : left,
+                           instruction->GetDexPc());
+  instruction->GetBlock()->ReplaceAndRemoveInstructionWith(instruction, bnr);
+  // Since we don't run DCE after this phase, try to manually remove the And instruction.
+  if (!and_instr->HasUses()) {
+    and_instr->GetBlock()->RemoveInstruction(and_instr);
+  }
+  return true;
+}
 
 bool TryExtractArrayAccessAddress(CodeGenerator* codegen,
                                   HInstruction* access,