mirror of https://github.com/rust-lang/nomicon
Alexis Beingessner
9 years ago
committed by
Manish Goregaokar
parent
df793ee850
commit
745dcebe39
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% Drop Check
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We have seen how lifetimes provide us some fairly simple rules for ensuring
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that never read dangling references. However up to this point we have only ever
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interacted with the *outlives* relationship in an inclusive manner. That is,
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when we talked about `'a: 'b`, it was ok for `'a` to live *exactly* as long as
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`'b`. At first glance, this seems to be a meaningless distinction. Nothing ever
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gets dropped at the same time as another, right? This is why we used the
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following desugarring of `let` statements:
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```rust
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let x;
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let y;
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```
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```rust
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{
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let x;
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{
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let y;
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}
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}
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```
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Each creates its own scope, clearly establishing that one drops before the
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other. However, what if we do the following?
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```rust
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let (x, y) = (vec![], vec![]);
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```
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Does either value strictly outlive the other? The answer is in fact *no*,
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neither value strictly outlives the other. Of course, one of x or y will be
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dropped before the other, but the actual order is not specified. Tuples aren't
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special in this regard; composite structures just don't guarantee their
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destruction order as of Rust 1.0.
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We *could* specify this for the fields of built-in composites like tuples and
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structs. However, what about something like Vec? Vec has to manually drop its
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elements via pure-library code. In general, anything that implements Drop has
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a chance to fiddle with its innards during its final death knell. Therefore
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the compiler can't sufficiently reason about the actual destruction order
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of the contents of any type that implements Drop.
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So why do we care? We care because if the type system isn't careful, it could
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accidentally make dangling pointers. Consider the following simple program:
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```rust
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struct Inspector<'a>(&'a u8);
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fn main() {
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let (days, inspector);
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days = Box::new(1);
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inspector = Inspector(&days);
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}
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```
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This program is totally sound and compiles today. The fact that `days` does
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not *strictly* outlive `inspector` doesn't matter. As long as the `inspector`
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is alive, so is days.
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However if we add a destructor, the program will no longer compile!
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```rust,ignore
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struct Inspector<'a>(&'a u8);
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impl<'a> Drop for Inspector<'a> {
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fn drop(&mut self) {
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println!("I was only {} days from retirement!", self.0);
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}
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}
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fn main() {
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let (days, inspector);
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days = Box::new(1);
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inspector = Inspector(&days);
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// Let's say `days` happens to get dropped first.
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// Then when Inspector is dropped, it will try to read free'd memory!
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}
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```
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```text
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<anon>:12:28: 12:32 error: `days` does not live long enough
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<anon>:12 inspector = Inspector(&days);
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^~~~
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<anon>:9:11: 15:2 note: reference must be valid for the block at 9:10...
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<anon>:9 fn main() {
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<anon>:10 let (days, inspector);
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<anon>:11 days = Box::new(1);
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<anon>:12 inspector = Inspector(&days);
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<anon>:13 // Let's say `days` happens to get dropped first.
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<anon>:14 // Then when Inspector is dropped, it will try to read free'd memory!
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...
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<anon>:10:27: 15:2 note: ...but borrowed value is only valid for the block suffix following statement 0 at 10:26
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<anon>:10 let (days, inspector);
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<anon>:11 days = Box::new(1);
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<anon>:12 inspector = Inspector(&days);
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<anon>:13 // Let's say `days` happens to get dropped first.
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<anon>:14 // Then when Inspector is dropped, it will try to read free'd memory!
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<anon>:15 }
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```
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Implementing Drop lets the Inspector execute some arbitrary code *during* its
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death. This means it can potentially observe that types that are supposed to
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live as long as it does actually were destroyed first.
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Interestingly, only *generic* types need to worry about this. If they aren't
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generic, then the only lifetimes they can harbor are `'static`, which will truly
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live *forever*. This is why this problem is referred to as *sound generic drop*.
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Sound generic drop is enforced by the *drop checker*. As of this writing, some
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of the finer details of how the drop checker validates types is totally up in
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the air. However The Big Rule is the subtlety that we have focused on this whole
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section:
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**For a generic type to soundly implement drop, it must strictly outlive all of
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its generic arguments.**
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This rule is sufficient but not necessary to satisfy the drop checker. That is,
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if your type obeys this rule then it's *definitely* sound to drop. However
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there are special cases where you can fail to satisfy this, but still
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successfully pass the borrow checker. These are the precise rules that are
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currently up in the air.
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It turns out that when writing unsafe code, we generally don't need to
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worry at all about doing the right thing for the drop checker. However there
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is *one* special case that you need to worry about, which we will look at in
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the next section.
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@ -0,0 +1,140 @@
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% Splitting Lifetimes
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The mutual exclusion property of mutable references can be very limiting when
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working with a composite structure. The borrow checker understands some basic stuff, but
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will fall over pretty easily. It *does* understand structs sufficiently to
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know that it's possible to borrow disjoint fields of a struct simultaneously.
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So this works today:
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```rust
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struct Foo {
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a: i32,
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b: i32,
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c: i32,
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}
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let mut x = Foo {a: 0, b: 0, c: 0};
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let a = &mut x.a;
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let b = &mut x.b;
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let c = &x.c;
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*b += 1;
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let c2 = &x.c;
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*a += 10;
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println!("{} {} {} {}", a, b, c, c2);
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```
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However borrowck doesn't understand arrays or slices in any way, so this doesn't
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work:
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```rust,ignore
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let x = [1, 2, 3];
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let a = &mut x[0];
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let b = &mut x[1];
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println!("{} {}", a, b);
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```
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```text
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<anon>:3:18: 3:22 error: cannot borrow immutable indexed content `x[..]` as mutable
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<anon>:3 let a = &mut x[0];
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^~~~
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<anon>:4:18: 4:22 error: cannot borrow immutable indexed content `x[..]` as mutable
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<anon>:4 let b = &mut x[1];
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^~~~
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error: aborting due to 2 previous errors
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```
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While it was plausible that borrowck could understand this simple case, it's
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pretty clearly hopeless for borrowck to understand disjointness in general
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container types like a tree, especially if distinct keys actually *do* map
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to the same value.
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In order to "teach" borrowck that what we're doing is ok, we need to drop down
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to unsafe code. For instance, mutable slices expose a `split_at_mut` function that
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consumes the slice and returns *two* mutable slices. One for everything to the
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left of the index, and one for everything to the right. Intuitively we know this
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is safe because the slices don't alias. However the implementation requires some
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unsafety:
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```rust,ignore
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fn split_at_mut(&mut self, mid: usize) -> (&mut [T], &mut [T]) {
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unsafe {
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let self2: &mut [T] = mem::transmute_copy(&self);
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(ops::IndexMut::index_mut(self, ops::RangeTo { end: mid } ),
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ops::IndexMut::index_mut(self2, ops::RangeFrom { start: mid } ))
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}
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}
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```
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This is pretty plainly dangerous. We use transmute to duplicate the slice with an
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*unbounded* lifetime, so that it can be treated as disjoint from the other until
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we unify them when we return.
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However more subtle is how iterators that yield mutable references work.
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The iterator trait is defined as follows:
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```rust
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trait Iterator {
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type Item;
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fn next(&mut self) -> Option<Self::Item>;
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}
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```
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Given this definition, Self::Item has *no* connection to `self`. This means
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that we can call `next` several times in a row, and hold onto all the results
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*concurrently*. This is perfectly fine for by-value iterators, which have exactly
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these semantics. It's also actually fine for shared references, as they admit
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arbitrarily many references to the same thing (although the
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iterator needs to be a separate object from the thing being shared). But mutable
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references make this a mess. At first glance, they might seem completely
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incompatible with this API, as it would produce multiple mutable references to
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the same object!
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However it actually *does* work, exactly because iterators are one-shot objects.
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Everything an IterMut yields will be yielded *at most* once, so we don't *actually*
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ever yield multiple mutable references to the same piece of data.
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In general all mutable iterators require *some* unsafe code *somewhere*, though.
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Whether it's raw pointers, or safely composing on top of *another* IterMut.
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For instance, VecDeque's IterMut:
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```rust,ignore
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struct IterMut<'a, T:'a> {
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// The whole backing array. Some of these indices are initialized!
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ring: &'a mut [T],
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tail: usize,
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head: usize,
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}
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impl<'a, T> Iterator for IterMut<'a, T> {
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type Item = &'a mut T;
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fn next(&mut self) -> Option<&'a mut T> {
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if self.tail == self.head {
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return None;
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}
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let tail = self.tail;
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self.tail = wrap_index(self.tail.wrapping_add(1), self.ring.len());
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unsafe {
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// might as well do unchecked indexing since wrap_index has us
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// in-bounds, and many of the "middle" indices are uninitialized
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// anyway.
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let elem = self.ring.get_unchecked_mut(tail);
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// round-trip through a raw pointer to unbound the lifetime from
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// ourselves
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Some(&mut *(elem as *mut _))
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}
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}
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}
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```
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A very subtle but interesting detail in this design is that it *relies on
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privacy to be sound*. Borrowck works on some very simple rules. One of those rules
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is that if we have a live &mut Foo and Foo contains an &mut Bar, then that &mut
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Bar is *also* live. Since IterMut is always live when `next` can be called, if
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`ring` were public then we could mutate `ring` while outstanding mutable borrows
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to it exist!
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@ -0,0 +1,87 @@
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% PhantomData
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When working with unsafe code, we can often end up in a situation where
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types or lifetimes are logically associated with a struct, but not actually
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part of a field. This most commonly occurs with lifetimes. For instance, the
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`Iter` for `&'a [T]` is (approximately) defined as follows:
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```rust,ignore
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struct Iter<'a, T: 'a> {
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ptr: *const T,
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end: *const T,
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}
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```
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However because `'a` is unused within the struct's body, it's *unbounded*.
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Because of the troubles this has historically caused, unbounded lifetimes and
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types are *illegal* in struct definitions. Therefore we must somehow refer
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to these types in the body. Correctly doing this is necessary to have
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correct variance and drop checking.
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We do this using *PhantomData*, which is a special marker type. PhantomData
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consumes no space, but simulates a field of the given type for the purpose of
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static analysis. This was deemed to be less error-prone than explicitly telling
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the type-system the kind of variance that you want, while also providing other
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useful such as the information needed by drop check.
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Iter logically contains a bunch of `&'a T`s, so this is exactly what we tell
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the PhantomData to simulate:
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```
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use std::marker;
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struct Iter<'a, T: 'a> {
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ptr: *const T,
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end: *const T,
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_marker: marker::PhantomData<&'a T>,
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}
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```
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and that's it. The lifetime will be bounded, and your iterator will be variant
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over `'a` and `T`. Everything Just Works.
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Another important example is Vec, which is (approximately) defined as follows:
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```
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struct Vec<T> {
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data: *const T, // *const for variance!
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len: usize,
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cap: usize,
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}
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```
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Unlike the previous example it *appears* that everything is exactly as we
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want. Every generic argument to Vec shows up in the at least one field.
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Good to go!
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Nope.
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The drop checker will generously determine that Vec<T> does not own any values
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of type T. This will in turn make it conclude that it does *not* need to worry
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about Vec dropping any T's in its destructor for determining drop check
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soundness. This will in turn allow people to create unsoundness using
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Vec's destructor.
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In order to tell dropck that we *do* own values of type T, and therefore may
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drop some T's when *we* drop, we must add an extra PhantomData saying exactly
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that:
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```
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use std::marker;
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struct Vec<T> {
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data: *const T, // *const for covariance!
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len: usize,
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cap: usize,
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_marker: marker::PhantomData<T>,
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}
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```
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Raw pointers that own an allocation is such a pervasive pattern that the
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standard library made a utility for itself called `Unique<T>` which:
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* wraps a `*const T` for variance
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* includes a `PhantomData<T>`,
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* auto-derives Send/Sync as if T was contained
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* marks the pointer as NonZero for the null-pointer optimization
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