path: root/lib/Transforms/Utils/SimplifyCFG.cpp

//===- SimplifyCFG.cpp - Code to perform CFG simplification ---------------===//
// 
//                     The LLVM Compiler Infrastructure
//
// This file was developed by the LLVM research group and is distributed under
// the University of Illinois Open Source License. See LICENSE.TXT for details.
// 
//===----------------------------------------------------------------------===//
//
// Peephole optimize the CFG.
//
//===----------------------------------------------------------------------===//

#include "llvm/Transforms/Utils/Local.h"
#include "llvm/Constants.h"
#include "llvm/Instructions.h"
#include "llvm/Support/CFG.h"
#include <algorithm>
#include <functional>
using namespace llvm;

// PropagatePredecessors - This gets "Succ" ready to have the predecessors from
// "BB".  This is a little tricky because "Succ" has PHI nodes, which need to
// have extra slots added to them to hold the merge edges from BB's
// predecessors, and BB itself might have had PHI nodes in it.  This function
// returns true (failure) if the Succ BB already has a predecessor that is a
// predecessor of BB and incoming PHI arguments would not be discernible.
//
// Assumption: Succ is the single successor for BB.
//
static bool PropagatePredecessorsForPHIs(BasicBlock *BB, BasicBlock *Succ) {
  assert(*succ_begin(BB) == Succ && "Succ is not successor of BB!");

  if (!isa<PHINode>(Succ->front()))
    return false;  // We can make the transformation, no problem.

  // If there is more than one predecessor, and there are PHI nodes in
  // the successor, then we need to add incoming edges for the PHI nodes
  //
  const std::vector<BasicBlock*> BBPreds(pred_begin(BB), pred_end(BB));

  // Check to see if one of the predecessors of BB is already a predecessor of
  // Succ.  If so, we cannot do the transformation if there are any PHI nodes
  // with incompatible values coming in from the two edges!
  //
  for (pred_iterator PI = pred_begin(Succ), PE = pred_end(Succ); PI != PE; ++PI)
    if (find(BBPreds.begin(), BBPreds.end(), *PI) != BBPreds.end()) {
      // Loop over all of the PHI nodes checking to see if there are
      // incompatible values coming in.
      for (BasicBlock::iterator I = Succ->begin();
           PHINode *PN = dyn_cast<PHINode>(I); ++I) {
        // Loop up the entries in the PHI node for BB and for *PI if the values
        // coming in are non-equal, we cannot merge these two blocks (instead we
        // should insert a conditional move or something, then merge the
        // blocks).
        int Idx1 = PN->getBasicBlockIndex(BB);
        int Idx2 = PN->getBasicBlockIndex(*PI);
        assert(Idx1 != -1 && Idx2 != -1 &&
               "Didn't have entries for my predecessors??");
        if (PN->getIncomingValue(Idx1) != PN->getIncomingValue(Idx2))
          return true;  // Values are not equal...
      }
    }

  // Loop over all of the PHI nodes in the successor BB
  for (BasicBlock::iterator I = Succ->begin();
       PHINode *PN = dyn_cast<PHINode>(I); ++I) {
    Value *OldVal = PN->removeIncomingValue(BB, false);
    assert(OldVal && "No entry in PHI for Pred BB!");

    // If this incoming value is one of the PHI nodes in BB...
    if (isa<PHINode>(OldVal) && cast<PHINode>(OldVal)->getParent() == BB) {
      PHINode *OldValPN = cast<PHINode>(OldVal);
      for (std::vector<BasicBlock*>::const_iterator PredI = BBPreds.begin(), 
             End = BBPreds.end(); PredI != End; ++PredI) {
        PN->addIncoming(OldValPN->getIncomingValueForBlock(*PredI), *PredI);
      }
    } else {
      for (std::vector<BasicBlock*>::const_iterator PredI = BBPreds.begin(), 
             End = BBPreds.end(); PredI != End; ++PredI) {
        // Add an incoming value for each of the new incoming values...
        PN->addIncoming(OldVal, *PredI);
      }
    }
  }
  return false;
}

/// GetIfCondition - Given a basic block (BB) with two predecessors (and
/// presumably PHI nodes in it), check to see if the merge at this block is due
/// to an "if condition".  If so, return the boolean condition that determines
/// which entry into BB will be taken.  Also, return by references the block
/// that will be entered from if the condition is true, and the block that will
/// be entered if the condition is false.
/// 
///
static Value *GetIfCondition(BasicBlock *BB,
                             BasicBlock *&IfTrue, BasicBlock *&IfFalse) {
  assert(std::distance(pred_begin(BB), pred_end(BB)) == 2 &&
         "Function can only handle blocks with 2 predecessors!");
  BasicBlock *Pred1 = *pred_begin(BB);
  BasicBlock *Pred2 = *++pred_begin(BB);

  // We can only handle branches.  Other control flow will be lowered to
  // branches if possible anyway.
  if (!isa<BranchInst>(Pred1->getTerminator()) ||
      !isa<BranchInst>(Pred2->getTerminator()))
    return 0;
  BranchInst *Pred1Br = cast<BranchInst>(Pred1->getTerminator());
  BranchInst *Pred2Br = cast<BranchInst>(Pred2->getTerminator());

  // Eliminate code duplication by ensuring that Pred1Br is conditional if
  // either are.
  if (Pred2Br->isConditional()) {
    // If both branches are conditional, we don't have an "if statement".  In
    // reality, we could transform this case, but since the condition will be
    // required anyway, we stand no chance of eliminating it, so the xform is
    // probably not profitable.
    if (Pred1Br->isConditional())
      return 0;

    std::swap(Pred1, Pred2);
    std::swap(Pred1Br, Pred2Br);
  }

  if (Pred1Br->isConditional()) {
    // If we found a conditional branch predecessor, make sure that it branches
    // to BB and Pred2Br.  If it doesn't, this isn't an "if statement".
    if (Pred1Br->getSuccessor(0) == BB &&
        Pred1Br->getSuccessor(1) == Pred2) {
      IfTrue = Pred1;
      IfFalse = Pred2;
    } else if (Pred1Br->getSuccessor(0) == Pred2 &&
               Pred1Br->getSuccessor(1) == BB) {
      IfTrue = Pred2;
      IfFalse = Pred1;
    } else {
      // We know that one arm of the conditional goes to BB, so the other must
      // go somewhere unrelated, and this must not be an "if statement".
      return 0;
    }

    // The only thing we have to watch out for here is to make sure that Pred2
    // doesn't have incoming edges from other blocks.  If it does, the condition
    // doesn't dominate BB.
    if (++pred_begin(Pred2) != pred_end(Pred2))
      return 0;

    return Pred1Br->getCondition();
  }

  // Ok, if we got here, both predecessors end with an unconditional branch to
  // BB.  Don't panic!  If both blocks only have a single (identical)
  // predecessor, and THAT is a conditional branch, then we're all ok!
  if (pred_begin(Pred1) == pred_end(Pred1) ||
      ++pred_begin(Pred1) != pred_end(Pred1) ||
      pred_begin(Pred2) == pred_end(Pred2) ||
      ++pred_begin(Pred2) != pred_end(Pred2) ||
      *pred_begin(Pred1) != *pred_begin(Pred2))
    return 0;

  // Otherwise, if this is a conditional branch, then we can use it!
  BasicBlock *CommonPred = *pred_begin(Pred1);
  if (BranchInst *BI = dyn_cast<BranchInst>(CommonPred->getTerminator())) {
    assert(BI->isConditional() && "Two successors but not conditional?");
    if (BI->getSuccessor(0) == Pred1) {
      IfTrue = Pred1;
      IfFalse = Pred2;
    } else {
      IfTrue = Pred2;
      IfFalse = Pred1;
    }
    return BI->getCondition();
  }
  return 0;
}


// If we have a merge point of an "if condition" as accepted above, return true
// if the specified value dominates the block.  We don't handle the true
// generality of domination here, just a special case which works well enough
// for us.
static bool DominatesMergePoint(Value *V, BasicBlock *BB) {
  if (Instruction *I = dyn_cast<Instruction>(V)) {
    BasicBlock *PBB = I->getParent();
    // If this instruction is defined i