https://gcc.gnu.org/bugzilla/show_bug.cgi?id=118461
Bug ID: 118461
Summary: constexpr lifetime tracking allows access to
out-of-lifetime const variables
Product: gcc
Version: unknown
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
Assignee: unassigned at gcc dot gnu.org
Reporter: richard-gccbugzilla at metafoo dot co.uk
Target Milestone: ---
Even after the fix for PR96630, GCC still accepts this invalid code that
accesses a variable outside its lifetime:
constexpr int f() {
const int *p = 0;
for (int i = 0; i != 2; ++i) {
if (i == 1) { return *p; }
const int n = 0;
p = &n;
}
}
static_assert(f() == 0);
(Live: https://godbolt.org/z/3fEEoq3T6)
Even though `*p` denotes `n`, which is usable in constant expressions, the
evaluation of `*p` still has undefined behavior because `n` is not within its
lifetime, so this is non-constant by [expr.const]/10.8.
Note in particular that the special provision in [expr.const]/17 that
effectively says we get to ignore the lifetime rules in some cases does not
apply to variables whose lifetime started within the evaluation. So this is
valid:
void g() {
const int n = 0;
static_assert([] { return n; }() == 0);
}
... because we get to assume that n is alive throughout the evaluation, but the
same doesn't apply in the testcase above, because the lifetime of the variable
started within the evaluation of `f() == 0`.
