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Remove dead code
llvm-svn: 4952
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b8ec19c4a1
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@ -25,9 +25,6 @@ namespace {
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class GCSE : public FunctionPass {
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std::set<Instruction*> WorkList;
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DominatorSet *DomSetInfo;
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#if 0
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ImmediateDominators *ImmDominator;
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#endif
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ValueNumbering *VN;
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public:
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virtual bool runOnFunction(Function &F);
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@ -61,9 +58,6 @@ bool GCSE::runOnFunction(Function &F) {
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// Get pointers to the analysis results that we will be using...
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DomSetInfo = &getAnalysis<DominatorSet>();
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#if 0
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ImmDominator = &getAnalysis<ImmediateDominators>();
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#endif
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VN = &getAnalysis<ValueNumbering>();
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// Step #1: Add all instructions in the function to the worklist for
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@ -254,32 +248,6 @@ Instruction *GCSE::EliminateCSE(Instruction *I, Instruction *Other) {
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// PRE than GCSE.
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//
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return 0;
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#if 0
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// Handle the most general case now. In this case, neither I dom Other nor
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// Other dom I. Because we are in SSA form, we are guaranteed that the
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// operands of the two instructions both dominate the uses, so we _know_
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// that there must exist a block that dominates both instructions (if the
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// operands of the instructions are globals or constants, worst case we
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// would get the entry node of the function). Search for this block now.
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//
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// Search up the immediate dominator chain of BB1 for the shared dominator
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BasicBlock *SharedDom = (*ImmDominator)[BB1];
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while (!DomSetInfo->dominates(SharedDom, BB2))
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SharedDom = (*ImmDominator)[SharedDom];
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// At this point, shared dom must dominate BOTH BB1 and BB2...
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assert(SharedDom && DomSetInfo->dominates(SharedDom, BB1) &&
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DomSetInfo->dominates(SharedDom, BB2) && "Dominators broken!");
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// Rip 'I' out of BB1, and move it to the end of SharedDom.
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BB1->getInstList().remove(I);
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SharedDom->getInstList().insert(--SharedDom->end(), I);
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// Eliminate 'Other' now.
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ReplaceInstWithInst(I, Other);
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#endif
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}
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if (isa<LoadInst>(Ret))
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