On Mon, Apr 15, 2024 at 11:37 AM Benno Lossin <benno.lossin@xxxxxxxxx> wrote:
>
> On 15.04.24 09:13, Alice Ryhl wrote:
> > +impl UserSlice {
> > + /// Constructs a user slice from a raw pointer and a length in bytes.
> > + ///
> > + /// Constructing a [`UserSlice`] performs no checks on the provided address and length, it can
> > + /// safely be constructed inside a kernel thread with no current userspace process. Reads and
> > + /// writes wrap the kernel APIs `copy_from_user` and `copy_to_user`, which check the memory map
> > + /// of the current process and enforce that the address range is within the user range (no
> > + /// additional calls to `access_ok` are needed).
> > + ///
> > + /// Callers must be careful to avoid time-of-check-time-of-use (TOCTOU) issues. The simplest way
> > + /// is to create a single instance of [`UserSlice`] per user memory block as it reads each byte
> > + /// at most once.
> > + pub fn new(ptr: *mut c_void, length: usize) -> Self {
>
> What would happen if I call this with a kernel pointer and then
> read/write to it? For example
>
> let mut arr = [MaybeUninit::uninit(); 64];
> let ptr: *mut [MaybeUninit<u8>] = &mut arr;
> let ptr = ptr.cast::<c_void>();
>
> let slice = UserSlice::new(ptr, 64);
> let (mut r, mut w) = slice.reader_writer();
>
> r.read_raw(&mut arr)?;
> // SAFETY: `arr` was initialized above.
> w.write_slice(unsafe { MaybeUninit::slice_assume_init_ref(&arr) })?;
>
> I think this would violate the exclusivity of `&mut` without any
> `unsafe` code. (the `unsafe` block at the end cannot possibly be wrong)
This will fail with an EFAULT error. There is a check on the C side
that verifies that the address is in userspace. (The access_ok call.)
Alice
