// This is the marker algorithm from "Simple and Efficient Construction of
// Static Single Assignment Form"
// The simple, non-marker algorithm places phi nodes at any join
// Here, we place markers, and only place phi nodes if they end up necessary.
// They are only necessary if they break a cycle (IE we recursively visit
// ourselves again), or we discover, while getting the value of the operands,
// that there are two or more definitions needing to be merged.
// This still will leave non-minimal form in the case of irreducible control
// flow, where phi nodes may be in cycles with themselves, but unnecessary.
MemoryAccess *MemorySSAUpdater::getPreviousDefRecursive(BasicBlock *BB) {
// Single predecessor case, just recurse, we can only have one definition.
if (BasicBlock *Pred = BB->getSinglePredecessor()) {
return getPreviousDefFromEnd(Pred);
} else if (VisitedBlocks.count(BB)) {
// We hit our node again, meaning we had a cycle, we must insert a phi
// node to break it so we have an operand. The only case this will
// insert useless phis is if we have irreducible control flow.
return MSSA->createMemoryPhi(BB);
} else if (VisitedBlocks.insert(BB).second) {
// Mark us visited so we can detect a cycle
SmallVector<MemoryAccess *, 8> PhiOps;
// Recurse to get the values in our predecessors for placement of a
// potential phi node. This will insert phi nodes if we cycle in order to
// break the cycle and have an operand.
for (auto *Pred : predecessors(BB))
PhiOps.push_back(getPreviousDefFromEnd(Pred));
// Now try to simplify the ops to avoid placing a phi.
// This may return null if we never created a phi yet, that's okay
MemoryPhi *Phi = dyn_cast_or_null<MemoryPhi>(MSSA->getMemoryAccess(BB));
bool PHIExistsButNeedsUpdate = false;
// See if the existing phi operands match what we need.
// Unlike normal SSA, we only allow one phi node per block, so we can't just
// create a new one.
if (Phi && Phi->getNumOperands() != 0)
if (!std::equal(Phi->op_begin(), Phi->op_end(), PhiOps.begin())) {
PHIExistsButNeedsUpdate = true;
}
// See if we can avoid the phi by simplifying it.
auto *Result = tryRemoveTrivialPhi(Phi, PhiOps);
// If we couldn't simplify, we may have to create a phi
if (Result == Phi) {
if (!Phi)
Phi = MSSA->createMemoryPhi(BB);
// These will have been filled in by the recursive read we did above.
if (PHIExistsButNeedsUpdate) {
std::copy(PhiOps.begin(), PhiOps.end(), Phi->op_begin());
std::copy(pred_begin(BB), pred_end(BB), Phi->block_begin());
} else {
unsigned i = 0;
for (auto *Pred : predecessors(BB))
Phi->addIncoming(PhiOps[i++], Pred);
}
Result = Phi;
}
if (MemoryPhi *MP = dyn_cast<MemoryPhi>(Result))
InsertedPHIs.push_back(MP);
// Set ourselves up for the next variable by resetting visited state.
VisitedBlocks.erase(BB);
return Result;
}
llvm_unreachable("Should have hit one of the three cases above");
}
// This is the marker algorithm from "Simple and Efficient Construction of
// Static Single Assignment Form"
// The simple, non-marker algorithm places phi nodes at any join
// Here, we place markers, and only place phi nodes if they end up necessary.
// They are only necessary if they break a cycle (IE we recursively visit
// ourselves again), or we discover, while getting the value of the operands,
// that there are two or more definitions needing to be merged.
// This still will leave non-minimal form in the case of irreducible control
// flow, where phi nodes may be in cycles with themselves, but unnecessary.
MemoryAccess *MemorySSAUpdater::getPreviousDefRecursive(
BasicBlock *BB,
DenseMap<BasicBlock *, TrackingVH<MemoryAccess>> &CachedPreviousDef) {
// First, do a cache lookup. Without this cache, certain CFG structures
// (like a series of if statements) take exponential time to visit.
auto Cached = CachedPreviousDef.find(BB);
if (Cached != CachedPreviousDef.end()) {
return Cached->second;
}
if (BasicBlock *Pred = BB->getSinglePredecessor()) {
// Single predecessor case, just recurse, we can only have one definition.
MemoryAccess *Result = getPreviousDefFromEnd(Pred, CachedPreviousDef);
CachedPreviousDef.insert({BB, Result});
return Result;
}
if (VisitedBlocks.count(BB)) {
// We hit our node again, meaning we had a cycle, we must insert a phi
// node to break it so we have an operand. The only case this will
// insert useless phis is if we have irreducible control flow.
MemoryAccess *Result = MSSA->createMemoryPhi(BB);
CachedPreviousDef.insert({BB, Result});
return Result;
}
if (VisitedBlocks.insert(BB).second) {
// Mark us visited so we can detect a cycle
SmallVector<TrackingVH<MemoryAccess>, 8> PhiOps;
// Recurse to get the values in our predecessors for placement of a
// potential phi node. This will insert phi nodes if we cycle in order to
// break the cycle and have an operand.
for (auto *Pred : predecessors(BB))
PhiOps.push_back(getPreviousDefFromEnd(Pred, CachedPreviousDef));
// Now try to simplify the ops to avoid placing a phi.
// This may return null if we never created a phi yet, that's okay
MemoryPhi *Phi = dyn_cast_or_null<MemoryPhi>(MSSA->getMemoryAccess(BB));
// See if we can avoid the phi by simplifying it.
auto *Result = tryRemoveTrivialPhi(Phi, PhiOps);
// If we couldn't simplify, we may have to create a phi
if (Result == Phi) {
if (!Phi)
Phi = MSSA->createMemoryPhi(BB);
// See if the existing phi operands match what we need.
// Unlike normal SSA, we only allow one phi node per block, so we can't just
// create a new one.
if (Phi->getNumOperands() != 0) {
// FIXME: Figure out whether this is dead code and if so remove it.
if (!std::equal(Phi->op_begin(), Phi->op_end(), PhiOps.begin())) {
// These will have been filled in by the recursive read we did above.
std::copy(PhiOps.begin(), PhiOps.end(), Phi->op_begin());
std::copy(pred_begin(BB), pred_end(BB), Phi->block_begin());
}
} else {
unsigned i = 0;
for (auto *Pred : predecessors(BB))
Phi->addIncoming(&*PhiOps[i++], Pred);
InsertedPHIs.push_back(Phi);
}
Result = Phi;
}
// Set ourselves up for the next variable by resetting visited state.
VisitedBlocks.erase(BB);
CachedPreviousDef.insert({BB, Result});
return Result;
}
llvm_unreachable("Should have hit one of the three cases above");
}
