nomicon/dropck.md

% Drop Check

We have seen how lifetimes provide us some fairly simple rules for ensuring that we never read dangling references. However up to this point we have only ever interacted with the outlives relationship in an inclusive manner. That is, when we talked about 'a: 'b, it was ok for 'a to live exactly as long as 'b. At first glance, this seems to be a meaningless distinction. Nothing ever gets dropped at the same time as another, right? This is why we used the following desugarring of let statements:

let x;
let y;

{
    let x;
    {
        let y;
    }
}

Each creates its own scope, clearly establishing that one drops before the other. However, what if we do the following?

let (x, y) = (vec![], vec![]);

Does either value strictly outlive the other? The answer is in fact no, neither value strictly outlives the other. Of course, one of x or y will be dropped before the other, but the actual order is not specified. Tuples aren't special in this regard; composite structures just don't guarantee their destruction order as of Rust 1.0.

We could specify this for the fields of built-in composites like tuples and structs. However, what about something like Vec? Vec has to manually drop its elements via pure-library code. In general, anything that implements Drop has a chance to fiddle with its innards during its final death knell. Therefore the compiler can't sufficiently reason about the actual destruction order of the contents of any type that implements Drop.

So why do we care? We care because if the type system isn't careful, it could accidentally make dangling pointers. Consider the following simple program:

struct Inspector<'a>(&'a u8);

fn main() {
    let (inspector, days);
    days = Box::new(1);
    inspector = Inspector(&days);
}

This program is totally sound and compiles today. The fact that days does not strictly outlive inspector doesn't matter. As long as the inspector is alive, so is days.

However if we add a destructor, the program will no longer compile!

struct Inspector<'a>(&'a u8);

impl<'a> Drop for Inspector<'a> {
    fn drop(&mut self) {
        println!("I was only {} days from retirement!", self.0);
    }
}

fn main() {
    let (inspector, days);
    days = Box::new(1);
    inspector = Inspector(&days);
    // Let's say `days` happens to get dropped first.
    // Then when Inspector is dropped, it will try to read free'd memory!
}

<anon>:12:28: 12:32 error: `days` does not live long enough
<anon>:12     inspector = Inspector(&days);
                                     ^~~~
<anon>:9:11: 15:2 note: reference must be valid for the block at 9:10...
<anon>:9 fn main() {
<anon>:10     let (inspector, days);
<anon>:11     days = Box::new(1);
<anon>:12     inspector = Inspector(&days);
<anon>:13     // Let's say `days` happens to get dropped first.
<anon>:14     // Then when Inspector is dropped, it will try to read free'd memory!
          ...
<anon>:10:27: 15:2 note: ...but borrowed value is only valid for the block suffix following statement 0 at 10:26
<anon>:10     let (inspector, days);
<anon>:11     days = Box::new(1);
<anon>:12     inspector = Inspector(&days);
<anon>:13     // Let's say `days` happens to get dropped first.
<anon>:14     // Then when Inspector is dropped, it will try to read free'd memory!
<anon>:15 }

Implementing Drop lets the Inspector execute some arbitrary code during its death. This means it can potentially observe that types that are supposed to live as long as it does actually were destroyed first.

Interestingly, only generic types need to worry about this. If they aren't generic, then the only lifetimes they can harbor are 'static, which will truly live forever. This is why this problem is referred to as sound generic drop. Sound generic drop is enforced by the drop checker. As of this writing, some of the finer details of how the drop checker validates types is totally up in the air. However The Big Rule is the subtlety that we have focused on this whole section:

For a generic type to soundly implement drop, its generics arguments must strictly outlive it.

This rule is sufficient but not necessary to satisfy the drop checker. That is, if your type obeys this rule then it's definitely sound to drop. However there are special cases where you can fail to satisfy this, but still successfully pass the borrow checker. These are the precise rules that are currently up in the air.

It turns out that when writing unsafe code, we generally don't need to worry at all about doing the right thing for the drop checker. However there is one special case that you need to worry about, which we will look at in the next section.