> I didn't find (or could construct) a single C or C++ testcase that
> wasn't fine with the new default, so I switched it (IIRC the default
> is disambiguating the most cases).
Yes, the default is the optimistic answer and a few tests rely on it:
FAIL: gcc.dg/tree-ssa/alias-18.c scan-tree-dump fre "with 1"
FAIL: gcc.dg/tree-ssa/pr13146.c scan-tree-dump optimized "return 0;"
FAIL: gcc.dg/tree-ssa/pr27799.c (test for excess errors)
FAIL: gcc.dg/tree-ssa/pr38895.c scan-tree-dump optimized "return 1;"
For alias-18.c:
struct A {
int i;
int j;
float x;
};
struct B {
struct A a;
int k;
};
int test1 (struct A *p, struct B *q)
{
p->i = 1;
q->k = -1;
return p->i;
}
the 1 is expected to be propagated but isn't if the default is changed.
IIUC the key here is that 'struct A' is a sub-structure of 'struct B' so, if
*p is not in the access path of *q, the references cannot alias. So I can
propose this for the end of nonaliasing_component_refs_p:
/* We haven't found any common base to apply offset-based disambiguation.
There are two cases:
1. The base access types have the same alias set. This can happen
in Ada when a function with an unconstrained parameter passed by
reference is called on a constrained object and inlined: the types
have the same alias set but aren't equivalent. The references may
alias in this case.
2. The base access types don't have the same alias set, i.e. one set
is a subset of the other. We have proved that B1 is not in the
access path B2.path and that B2 is not in the access path B1.path
so the references may not alias. */
return get_alias_set (type1) == get_alias_set (type2);
--
Eric Botcazou
