Fix debug output to include a newline after printing a Value, now

that Value's operator<< doesn't include one.


git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@79240 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Dan Gohman
2009-08-17 15:25:05 +00:00
parent fa6a1cf1ed
commit 8732577230
3 changed files with 18 additions and 18 deletions

View File

@ -222,7 +222,7 @@ void Reassociate::LinearizeExpr(BinaryOperator *I) {
isReassociableOp(RHS, I->getOpcode()) &&
"Not an expression that needs linearization?");
DOUT << "Linear" << *LHS << *RHS << *I;
DOUT << "Linear" << *LHS << '\n' << *RHS << '\n' << *I << '\n';
// Move the RHS instruction to live immediately before I, avoiding breaking
// dominator properties.
@ -235,7 +235,7 @@ void Reassociate::LinearizeExpr(BinaryOperator *I) {
++NumLinear;
MadeChange = true;
DOUT << "Linearized: " << *I;
DOUT << "Linearized: " << *I << '\n';
// If D is part of this expression tree, tail recurse.
if (isReassociableOp(I->getOperand(1), I->getOpcode()))
@ -334,10 +334,10 @@ void Reassociate::RewriteExprTree(BinaryOperator *I,
if (I->getOperand(0) != Ops[i].Op ||
I->getOperand(1) != Ops[i+1].Op) {
Value *OldLHS = I->getOperand(0);
DOUT << "RA: " << *I;
DOUT << "RA: " << *I << '\n';
I->setOperand(0, Ops[i].Op);
I->setOperand(1, Ops[i+1].Op);
DOUT << "TO: " << *I;
DOUT << "TO: " << *I << '\n';
MadeChange = true;
++NumChanged;
@ -350,9 +350,9 @@ void Reassociate::RewriteExprTree(BinaryOperator *I,
assert(i+2 < Ops.size() && "Ops index out of range!");
if (I->getOperand(1) != Ops[i].Op) {
DOUT << "RA: " << *I;
DOUT << "RA: " << *I << '\n';
I->setOperand(1, Ops[i].Op);
DOUT << "TO: " << *I;
DOUT << "TO: " << *I << '\n';
MadeChange = true;
++NumChanged;
}
@ -450,7 +450,7 @@ static Instruction *BreakUpSubtract(LLVMContext &Context, Instruction *Sub,
Sub->replaceAllUsesWith(New);
Sub->eraseFromParent();
DOUT << "Negated: " << *New;
DOUT << "Negated: " << *New << '\n';
return New;
}