Lines Matching refs:predecessors
94 "predecessors (i.e. don't add nops that are executed). In log2 "
306 /// Count of predecessors of any block within the chain which have not
308 /// until those predecessors are scheduled (or we find a sufficiently good
528 /// predecessors. Filters based on loop.
813 for (MachineBasicBlock *SuccPred : Succ->predecessors()) {
904 /// lower part of a trellis if all of the predecessors of S are either in S or
920 // To avoid reviewing the same predecessors twice.
925 for (auto SuccPred : Succ->predecessors()) {
965 // predecessor. If the best incoming predecessors aren't the same,
1024 for (MachineBasicBlock *SuccPred : Succ->predecessors()) {
1025 // Skip any placed predecessors that are not BB
1088 /// into all of its unplaced, unfiltered predecessors, that are not BB.
1103 for (MachineBasicBlock *Pred : Succ->predecessors()) {
1104 // Make sure all unplaced and unfiltered predecessors can be
1132 // whether Succ can be duplicated into all its unplaced predecessors, we
1135 // D. TODO(iteratee): ignore sufficiently cold predecessors for
1163 // fallthrough from predecessors from Succ to its successors. We may need
1171 // If the duplication candidate has more unplaced predecessors than
1183 // In this example Dup has 2 successors and 3 predecessors, duplication of Dup
1262 // If PDom can't tail-duplicate into it's non-BB predecessors, then this
1264 for (MachineBasicBlock* Pred : PDom->predecessors()) {
1272 // If we can't tail-duplicate PDom to its predecessors, then skip this
1362 // There isn't a better layout when there are no unscheduled predecessors.
1486 for (MachineBasicBlock *Pred : Succ->predecessors()) {
1747 "Attempting to place block with unscheduled predecessors in worklist.");
1751 for (MachineBasicBlock *Pred : ChainBB->predecessors()) {
1879 for (MachineBasicBlock *Pred : Top->predecessors()) {
1947 for (MachineBasicBlock *Pred : NewTop->predecessors()) {
2000 /// from predecessors of old top.
2011 /// and it has more than one predecessors
2013 /// If it is below one of its predecessors P, only P can fall through to
2014 /// it, all other predecessors need a jump to it, and another conditional
2016 /// predecessors jump to it, then fall through to loop header. So all its
2017 /// predecessors except P can reduce one taken branch.
2038 for (MachineBasicBlock *Pred : OldTop->predecessors()) {
2074 // Walk backwards through any straight line of predecessors.
2126 // header has be pre-merged into a chain due to predecessors not having
2245 for (MachineBasicBlock *Pred : Top->predecessors()) {
2383 for (auto *Pred : ChainHeaderBB->predecessors()) {
2513 for (auto LoopPred : L.getHeader()->predecessors())
2557 // predecessors to the header if there is one which will result in strictly
2579 "LoopChain should not have unscheduled predecessors.");
2882 // Force alignment if all the predecessors are jumps. We already checked
2890 // cold relative to the block. When this is true, other predecessors make up
2901 /// Tail duplicate \p BB into (some) predecessors if profitable, repeating if
2954 // of unscheduled predecessors.
2961 /// Tail duplicate \p BB into (some) predecessors if profitable.
3056 // We're only looking for unscheduled predecessors that match the filter.
3128 // Find out the predecessors of BB and BB can be beneficially duplicated into
3155 // For each predecessors of BB, compute the benefit of duplicating BB,
3195 // predecessors.
3231 // No predecessors can optimally fallthrough to BB.
3357 // Align all of the blocks that have no fall-through predecessors to a