The borrow tracker tracks whether there is 1, more than 1, or no references to a pointer at any particular time and rust automatically drops it when that last reference (the owning one) goes away. Sounds like compile time reference counting to me :P
I didn't invent this way of referring to it, though I don't recall who I stole it from. It's not entirely accurate, but it's a close enough description to capture how rust's mostly automatic memory management works from a distance.
So the problem here is that it is almost entirely wrong. There is no reference count anywhere in the borrow checker’s algorithm, and you can’t do the things with borrows that you can do with reference counting.
It’s just not a good mental model.
For example, with reference counting you can convert a shared reference to a unique reference when you can verify that the count is exactly 1. But converting a `&T` to a `&mut T` is always instantaneous UB, no exceptions. It doesn’t matter if it’s actually the only reference.
Borrows are also orthogonal to dropping/destructors. Borrows can extend the lifetime of a value for convenience reasons, but it is not a general rule that values are dropped when the last reference is gone.
There is a reference count in the algorithm in the sense that the algorithm must keep track of the number of live shared borrows derived from a unique borrow or owned value so that it knows when it becomes legal to mutate it again (i.e. to know when that number goes to zero) or if there are still outstanding ones.
Borrow checking is necessary for dropping and destructors in the sense that without borrows we could drop an owned value while we still have references to it and get a use after free. RAII in rust only works safely because we have the borrow checker reference counting for us to tell us when its again safe to mutate (including drop) owned values.
Yes, rust doesn't support going from an &T to an &mut T, but it does support going from an <currently immutable reference to T> to a <mutable reference to T> in the shape of going from an &mut T which is currently immutably borrowed to an &mut T which is not borrowed. It can do this because it keeps track of how many shared references there are derived from the mutable reference.
You're right that it's possible to leak the owning reference so that the object isn't freed when the last reference is gone - but it's possible to leak a reference in runtime reference runtime reference counted language too.
But yes, it's not a perfect analogy, merely a good one. It's most likely that the implementation doesn't just keep a count of references for instance, but a set of them to enable better diagnostics and more efficient computation.
I didn't invent this way of referring to it, though I don't recall who I stole it from. It's not entirely accurate, but it's a close enough description to capture how rust's mostly automatic memory management works from a distance.
If you want a more literal interpretation of compile time reference counting see also: https://docs.rs/static-rc/0.7.0/static_rc/