CheckingGroups.clear();
+ // If we need to check two pointers to the same underlying object
+ // with a non-constant difference, we shouldn't perform any pointer
+ // grouping with those pointers. This is because we can easily get
+ // into cases where the resulting check would return false, even when
+ // the accesses are safe.
+ //
+ // The following example shows this:
+ // for (i = 0; i < 1000; ++i)
+ // a[5000 + i * m] = a[i] + a[i + 9000]
+ //
+ // Here grouping gives a check of (5000, 5000 + 1000 * m) against
+ // (0, 10000) which is always false. However, if m is 1, there is no
+ // dependence. Not grouping the checks for a[i] and a[i + 9000] allows
+ // us to perform an accurate check in this case.
+ //
+ // The above case requires that we have an UnknownDependence between
+ // accesses to the same underlying object. This cannot happen unless
+ // ShouldRetryWithRuntimeCheck is set, and therefore UseDependencies
+ // is also false. In this case we will use the fallback path and create
+ // separate checking groups for all pointers.
+
// If we don't have the dependency partitions, construct a new
- // checking pointer group for each pointer.
+ // checking pointer group for each pointer. This is also required
+ // for correctness, because in this case we can have checking between
+ // pointers to the same underlying object.
if (!UseDependencies) {
for (unsigned I = 0; I < Pointers.size(); ++I)
CheckingGroups.push_back(CheckingPtrGroup(I, *this));