// this many instructions in common.
unsigned minCommonTailLength = TailMergeSize;
- // If there's a successor block, there are some cases which don't require
- // new branching and as such are very likely to be profitable.
- if (SuccBB) {
- if (SuccBB->pred_size() == MergePotentials.size() &&
- !MergePotentials[0].getBlock()->empty()) {
- // If all the predecessors have at least one tail instruction in common,
- // merging is very likely to be a win since it won't require an increase
- // in static branches, and it will decrease the static instruction count.
- bool AllPredsMatch = true;
- MachineBasicBlock::iterator FirstNonTerm;
- unsigned MinNumTerms = CountTerminators(MergePotentials[0].getBlock(),
- FirstNonTerm);
- if (FirstNonTerm != MergePotentials[0].getBlock()->end()) {
- for (unsigned i = 1, e = MergePotentials.size(); i != e; ++i) {
- MachineBasicBlock::iterator OtherFirstNonTerm;
- unsigned NumTerms = CountTerminators(MergePotentials[0].getBlock(),
- OtherFirstNonTerm);
- if (NumTerms < MinNumTerms)
- MinNumTerms = NumTerms;
- if (OtherFirstNonTerm == MergePotentials[i].getBlock()->end() ||
- OtherFirstNonTerm->isIdenticalTo(FirstNonTerm)) {
- AllPredsMatch = false;
- break;
- }
- }
-
- // If they all have an instruction in common, do any amount of merging.
- if (AllPredsMatch)
- minCommonTailLength = MinNumTerms + 1;
- }
- }
- }
-
DEBUG(errs() << "\nTryTailMergeBlocks: ";
for (unsigned i = 0, e = MergePotentials.size(); i != e; ++i)
errs() << "BB#" << MergePotentials[i].getBlock()->getNumber()