Comment by AlotOfReading

4 hours ago

Under NLL, reference lifetime is defined by the places where the reference is used, and noaccess types can't be used. So then reference lifetime has to devolve to the spans where the reference is live, which is only defined lexically.

So for example:

    let mut data = vec!['a', 'b', 'c'];
    let b: &noaccess [char] = &data[..];
    data.push('d'); // Does this error?

Hence the question

1 comment

AlotOfReading

amluto 3 hours ago

Aha, gotcha. I assume that any such reference would either do nothing or would be used by unsafe code (presumably by conversion through a raw pointer but maybe direct unsafe conversion to regular references could also be allowed).

Your example is sneaky, though. Lifetime issues aside (suppose the next line of code uses b), that’s a noaccess reference to memory (an object? a place? I’m not sure what the current term is) that is only guaranteed to exist so long as data is not mutated. So the code with a subsequent use of data would error.

But if it were instead:

    let b: &noaccess = data;

Then it would not error.