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 desugaring 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.

Obeying this rule is (usually) necessary to satisfy the borrow checker; obeying it is sufficient but not necessary to be sound. That is, if your type obeys this rule then it's definitely sound to drop.

The reason that it is not always necessary to satisfy the above rule is that some Drop implementations will not access borrowed data even though their type gives them the capability for such access.

For example, this variant of the above Inspector example will never access borrowed data:

struct Inspector<'a>(&'a u8, &'static str);

impl<'a> Drop for Inspector<'a> {
    fn drop(&mut self) {
        println!("Inspector(_, {}) knows when *not* to inspect.", self.1);
    }
}

fn main() {
    let (inspector, days);
    days = Box::new(1);
    inspector = Inspector(&days, "gadget");
    // Let's say `days` happens to get dropped first.
    // Even when Inspector is dropped, its destructor will not access the
    // borrowed `days`.
}

Likewise, this variant will also never access borrowed data:

use std::fmt;

struct Inspector<T: fmt::Display>(T, &'static str);

impl<T: fmt::Display> Drop for Inspector<T> {
    fn drop(&mut self) {
        println!("Inspector(_, {}) knows when *not* to inspect.", self.1);
    }
}

fn main() {
    let (inspector, days): (Inspector<&u8>, Box<u8>);
    days = Box::new(1);
    inspector = Inspector(&days, "gadget");
    // Let's say `days` happens to get dropped first.
    // Even when Inspector is dropped, its destructor will not access the
    // borrowed `days`.
}

However, both of the above variants are rejected by the borrow checker during the analysis of fn main, saying that days does not live long enough.

The reason is that the borrow checking analysis of main does not know about the internals of each Inspector's Drop implementation. As far as the borrow checker knows while it is analyzing main, the body of an inspector's destructor might access that borrowed data.

Therefore, the drop checker forces all borrowed data in a value to strictly outlive that value.

An Escape Hatch

The precise rules that govern drop checking may be less restrictive in the future.

The current analysis is deliberately conservative and trivial; it forces all borrowed data in a value to outlive that value, which is certainly sound.

Future versions of the language may make the analysis more precise, to reduce the number of cases where sound code is rejected as unsafe. This would help address cases such as the two Inspectors above that know not to inspect during destruction.

In the meantime, there is an unstable attribute that one can use to assert (unsafely) that a generic type's destructor is guaranteed to not access any expired data, even if its type gives it the capability to do so.

That attribute is called may_dangle and was introduced in [RFC 1327] (https://github.com/rust-lang/rfcs/blob/master/text/1327-dropck-param-eyepatch.md). To deploy it on the Inspector example from above, we would write:

struct Inspector<'a>(&'a u8, &'static str);

unsafe impl<#[may_dangle] 'a> Drop for Inspector<'a> {
    fn drop(&mut self) {
        println!("Inspector(_, {}) knows when *not* to inspect.", self.1);
    }
}

Use of this attribute requires the Drop impl to be marked unsafe because the compiler is not checking the implicit assertion that no potentially expired data (e.g. self.0 above) is accessed.

The attribute can be applied to any number of lifetime and type parameters. In the following example, we assert that we access no data behind a reference of lifetime 'b and that the only uses of T will be moves or drops, but omit the attribute from 'a and U, because we do access data with that lifetime and that type:

use std::fmt::Display;

struct Inspector<'a, 'b, T, U: Display>(&'a u8, &'b u8, T, U);

unsafe impl<'a, #[may_dangle] 'b, #[may_dangle] T, U: Display> Drop for Inspector<'a, 'b, T, U> {
    fn drop(&mut self) {
        println!("Inspector({}, _, _, {})", self.0, self.3);
    }
}

It is sometimes obvious that no such access can occur, like the case above. However, when dealing with a generic type parameter, such access can occur indirectly. Examples of such indirect access are:

• invoking a callback,
• via a trait method call.

(Future changes to the language, such as impl specialization, may add other avenues for such indirect access.)

Here is an example of invoking a callback:

struct Inspector<T>(T, &'static str, Box<for <'r> fn(&'r T) -> String>);

impl<T> Drop for Inspector<T> {
    fn drop(&mut self) {
        // The `self.2` call could access a borrow e.g. if `T` is `&'a _`.
        println!("Inspector({}, {}) unwittingly inspects expired data.",
                 (self.2)(&self.0), self.1);
    }
}

Here is an example of a trait method call:

use std::fmt;

struct Inspector<T: fmt::Display>(T, &'static str);

impl<T: fmt::Display> Drop for Inspector<T> {
    fn drop(&mut self) {
        // There is a hidden call to `<T as Display>::fmt` below, which
        // could access a borrow e.g. if `T` is `&'a _`
        println!("Inspector({}, {}) unwittingly inspects expired data.",
                 self.0, self.1);
    }
}

And of course, all of these accesses could be further hidden within some other method invoked by the destructor, rather than being written directly within it.

In all of the above cases where the &'a u8 is accessed in the destructor, adding the #[may_dangle] attribute makes the type vulnerable to misuse that the borrower checker will not catch, inviting havoc. It is better to avoid adding the attribute.

Is that all about drop checker?

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.