Bug fix in the transfer function for compound assignments: if the value

of the LHS expressions is Unknown, the value of the entire expression
does not evaluate to the RHS (as is the case with normal assignments).

llvm-svn: 48102
This commit is contained in:
Ted Kremenek 2008-03-09 08:12:37 +00:00
parent c4c9dde04c
commit cfd8b19ef7
1 changed files with 9 additions and 13 deletions

View File

@ -1162,21 +1162,13 @@ void GRExprEngine::VisitBinaryOperator(BinaryOperator* B,
}
if (LeftV.isUnknown()) {
// While we do not know the location to store RightV,
// the entire expression does evaluate to RightV.
if (RightV.isUnknown()) {
Dst.Add(N2);
continue;
}
St = SetRVal(St, B, RightV);
break;
assert (isa<UnknownVal>(GetRVal(St, B)));
Dst.Add(N2);
continue;
}
// At this pointer we know that the LHS evaluates to an LVal
// that is neither "Unknown" or "Unintialized."
// that is neither "Unknown" or "Undefined."
LVal LeftLV = cast<LVal>(LeftV);
@ -1196,12 +1188,16 @@ void GRExprEngine::VisitBinaryOperator(BinaryOperator* B,
// Propagate unknown values.
if (V.isUnknown()) {
// The value bound to LeftV is unknown. Thus we just
// propagate the current node (as "B" is already bound to nothing).
assert (isa<UnknownVal>(GetRVal(St, B)));
Dst.Add(N2);
continue;
}
if (RightV.isUnknown()) {
St = SetRVal(SetRVal(St, LeftLV, RightV), B, RightV);
assert (isa<UnknownVal>(GetRVal(St, B)));
St = SetRVal(St, LeftLV, UnknownVal());
break;
}