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Merge pull request #113 from vorner/defined-drop-order

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dropck: The drop order is now defined
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Alexis Beingessner 6 years ago committed by GitHub
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@ -22,25 +22,23 @@ let y;
}
```
Each creates its own scope, clearly establishing that one drops before the
other. However, what if we do the following?
There are some more complex situations which are not possible to desugar using
scopes, but the order is still defined ‒ variables are dropped in the reverse
order of their definition, fields of structs and tuples in order of their
definition. There are some more details about order of drop in [rfc1875].
Let's do this:
```rust,ignore
let (x, y) = (vec![], vec![]);
let tuple = (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.
The left vector is dropped first. But does it mean the right one strictly
outlives it in the eyes of the borrow checker? The answer to this question is
*no*. The borrow checker could track fields of tuples separately, but it would
still be unable to decide what outlives what in case of vector elements, which
are dropped manually via pure-library code the borrow checker doesn't
understand.
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:
@ -48,20 +46,27 @@ accidentally make dangling pointers. Consider the following simple program:
```rust
struct Inspector<'a>(&'a u8);
struct World<'a> {
inspector: Option<Inspector<'a>>,
days: Box<u8>,
}
fn main() {
let (inspector, days);
days = Box::new(1);
inspector = Inspector(&days);
let mut world = World {
inspector: None,
days: Box::new(1),
};
world.inspector = Some(Inspector(&world.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.
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!
```rust,ignore
```rust,compile_fail
struct Inspector<'a>(&'a u8);
impl<'a> Drop for Inspector<'a> {
@ -70,30 +75,38 @@ impl<'a> Drop for Inspector<'a> {
}
}
struct World<'a> {
inspector: Option<Inspector<'a>>,
days: Box<u8>,
}
fn main() {
let (inspector, days);
days = Box::new(1);
inspector = Inspector(&days);
let mut world = World {
inspector: None,
days: Box::new(1),
};
world.inspector = Some(Inspector(&world.days));
// Let's say `days` happens to get dropped first.
// Then when Inspector is dropped, it will try to read free'd memory!
}
```
```text
error[E0597]: `days` does not live long enough
--> src/main.rs:12:28
error[E0597]: `world.days` does not live long enough
--> src/main.rs:20:39
|
12 | inspector = Inspector(&days);
| ^^^^ borrowed value does not live long enough
20 | world.inspector = Some(Inspector(&world.days));
| ^^^^^^^^^^ borrowed value does not live long enough
...
15 | }
| - `days` dropped here while still borrowed
23 | }
| - `world.days` dropped here while still borrowed
|
= note: values in a scope are dropped in the opposite order they are created
error: aborting due to previous error
```
You can try changing the order of fields or use a tuple instead of the struct,
it'll still not compile.
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.
@ -116,12 +129,14 @@ 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.
though their type gives them the capability for such access, or because we know
the specific drop order and the borrowed data is still fine even if the borrow
checker doesn't know that.
For example, this variant of the above `Inspector` example will never
access borrowed data:
```rust,ignore
```rust,compile_fail
struct Inspector<'a>(&'a u8, &'static str);
impl<'a> Drop for Inspector<'a> {
@ -130,10 +145,17 @@ impl<'a> Drop for Inspector<'a> {
}
}
struct World<'a> {
inspector: Option<Inspector<'a>>,
days: Box<u8>,
}
fn main() {
let (inspector, days);
days = Box::new(1);
inspector = Inspector(&days, "gadget");
let mut world = World {
inspector: None,
days: Box::new(1),
};
world.inspector = Some(Inspector(&world.days, "gadget"));
// Let's say `days` happens to get dropped first.
// Even when Inspector is dropped, its destructor will not access the
// borrowed `days`.
@ -142,21 +164,26 @@ fn main() {
Likewise, this variant will also never access borrowed data:
```rust,ignore
use std::fmt;
struct Inspector<T: fmt::Display>(T, &'static str);
```rust,compile_fail
struct Inspector<T>(T, &'static str);
impl<T: fmt::Display> Drop for Inspector<T> {
impl<T> Drop for Inspector<T> {
fn drop(&mut self) {
println!("Inspector(_, {}) knows when *not* to inspect.", self.1);
}
}
struct World<T> {
inspector: Option<Inspector<T>>,
days: Box<u8>,
}
fn main() {
let (inspector, days): (Inspector<&u8>, Box<u8>);
days = Box::new(1);
inspector = Inspector(&days, "gadget");
let mut world = World {
inspector: None,
days: Box::new(1),
};
world.inspector = Some(Inspector(&world.days, "gadget"));
// Let's say `days` happens to get dropped first.
// Even when Inspector is dropped, its destructor will not access the
// borrowed `days`.
@ -194,9 +221,11 @@ 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][rfc1327].
To deploy it on the `Inspector` example from above, we would write:
To deploy it on the `Inspector` from above, we would write:
```rust
#![feature(dropck_eyepatch)]
```rust,ignore
struct Inspector<'a>(&'a u8, &'static str);
unsafe impl<#[may_dangle] 'a> Drop for Inspector<'a> {
@ -204,6 +233,19 @@ unsafe impl<#[may_dangle] 'a> Drop for Inspector<'a> {
println!("Inspector(_, {}) knows when *not* to inspect.", self.1);
}
}
struct World<'a> {
days: Box<u8>,
inspector: Option<Inspector<'a>>,
}
fn main() {
let mut world = World {
inspector: None,
days: Box::new(1),
};
world.inspector = Some(Inspector(&world.days, "gatget"));
}
```
Use of this attribute requires the `Drop` impl to be marked `unsafe` because the
@ -279,6 +321,12 @@ attribute makes the type vulnerable to misuse that the borrower
checker will not catch, inviting havoc. It is better to avoid adding
the attribute.
# A related side note about drop order
While the drop order of fields inside a struct is defined, relying on it is
fragile and subtle. When the order matters, it is better to use the
[`ManuallyDrop`] wrapper.
# Is that all about drop checker?
It turns out that when writing unsafe code, we generally don't need to
@ -286,4 +334,7 @@ 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.
[rfc1327]: https://github.com/rust-lang/rfcs/blob/master/text/1327-dropck-param-eyepatch.md
[rfc1857]: https://github.com/rust-lang/rfcs/blob/master/text/1857-stabilize-drop-order.md
[`ManuallyDrop`]: ../std/mem/struct.ManuallyDrop.html

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