https://gcc.gnu.org/bugzilla/show_bug.cgi?id=91357
--- Comment #2 from Jonathan Wakely <redi at gcc dot gnu.org> ---
More specifically:
v.operator[](1); /* maybe okay -- forms a pointer one past end */
Not OK. **Dereferences** a past-the end iterator. That's UB. Go to jail. Go
directly to jail. Do not pass Go. Do not collect £200.
v[1]; /* highly suspicious but not obviously invalid or UB */
UB for the same reason.
int *ugly_end = &v[1]; /* hmm. is this okay? */
No. v[1] is UB already, &v[1] doesn't make it unundefined.
std::cout << v[1] << std::endl; /* definitely not okay */
v[2]; /* definitely not okay -- merely computing this pointer is UB */
There's no pointer. vector::operator[] is not specified in terms of pointers.
This is not C.
The assertion is entirely correct. Calling operator[] with an invalid value is
UB, it doesn't matter what you do (or don't do) with the result.
