shard out misc section on lifetimes properly

Author: Gankra

src/doc/tarpl/SUMMARY.md

@@ -15,7 +15,9 @@
 	* [Unbounded Lifetimes](unbounded-lifetimes.md)
 	* [Higher-Rank Trait Bounds](hrtb.md)
 	* [Subtyping and Variance](subtyping.md)
-	* [Misc](lifetime-misc.md)
+	* [Drop Check](dropck.md)
+	* [PhantomData](phantom-data.md)
+	* [Splitting Lifetimes](lifetime-splitting.md)
 * [Type Conversions](conversions.md)
 	* [Coercions](coercions.md)
 	* [The Dot Operator](dot-operator.md)

src/doc/tarpl/dropck.md

@@ -0,0 +1,127 @@
+% Drop Check
+
+We have seen how lifetimes provide us some fairly simple rules for ensuring
+that 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:
+
+```rust
+let x;
+let y;
+```
+
+```rust
+{
+    let x;
+    {
+        let y;
+    }
+}
+```
+
+Each creates its own scope, clearly establishing that one drops before the
+other. However, what if we do the following?
+
+```rust
+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:
+
+```rust
+struct Inspector<'a>(&'a u8);
+
+fn main() {
+    let (days, inspector);
+    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!
+
+```rust,ignore
+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 (days, inspector);
+    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!
+}
+```
+
+```text
+<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 (days, inspector);
+<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 (days, inspector);
+<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, it must strictly outlive all of
+its generic arguments.**
+
+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.

src/doc/tarpl/lifetime-splitting.md

@@ -0,0 +1,140 @@
+% Splitting Lifetimes
+
+The mutual exclusion property of mutable references can be very limiting when
+working with a composite structure. The borrow checker understands some basic stuff, but
+will fall over pretty easily. It *does* understand structs sufficiently to
+know that it's possible to borrow disjoint fields of a struct simultaneously.
+So this works today:
+
+```rust
+struct Foo {
+    a: i32,
+    b: i32,
+    c: i32,
+}
+
+let mut x = Foo {a: 0, b: 0, c: 0};
+let a = &mut x.a;
+let b = &mut x.b;
+let c = &x.c;
+*b += 1;
+let c2 = &x.c;
+*a += 10;
+println!("{} {} {} {}", a, b, c, c2);
+```
+
+However borrowck doesn't understand arrays or slices in any way, so this doesn't
+work:
+
+```rust,ignore
+let x = [1, 2, 3];
+let a = &mut x[0];
+let b = &mut x[1];
+println!("{} {}", a, b);
+```
+
+```text
+<anon>:3:18: 3:22 error: cannot borrow immutable indexed content `x[..]` as mutable
+<anon>:3     let a = &mut x[0];
+                          ^~~~
+<anon>:4:18: 4:22 error: cannot borrow immutable indexed content `x[..]` as mutable
+<anon>:4     let b = &mut x[1];
+                          ^~~~
+error: aborting due to 2 previous errors
+```
+
+While it was plausible that borrowck could understand this simple case, it's
+pretty clearly hopeless for borrowck to understand disjointness in general
+container types like a tree, especially if distinct keys actually *do* map
+to the same value.
+
+In order to "teach" borrowck that what we're doing is ok, we need to drop down
+to unsafe code. For instance, mutable slices expose a `split_at_mut` function that
+consumes the slice and returns *two* mutable slices. One for everything to the
+left of the index, and one for everything to the right. Intuitively we know this
+is safe because the slices don't alias. However the implementation requires some
+unsafety:
+
+```rust,ignore
+fn split_at_mut(&mut self, mid: usize) -> (&mut [T], &mut [T]) {
+    unsafe {
+        let self2: &mut [T] = mem::transmute_copy(&self);
+
+        (ops::IndexMut::index_mut(self, ops::RangeTo { end: mid } ),
+         ops::IndexMut::index_mut(self2, ops::RangeFrom { start: mid } ))
+    }
+}
+```
+
+This is pretty plainly dangerous. We use transmute to duplicate the slice with an
+*unbounded* lifetime, so that it can be treated as disjoint from the other until
+we unify them when we return.
+
+However more subtle is how iterators that yield mutable references work.
+The iterator trait is defined as follows:
+
+```rust
+trait Iterator {
+    type Item;
+
+    fn next(&mut self) -> Option<Self::Item>;
+}
+```
+
+Given this definition, Self::Item has *no* connection to `self`. This means
+that we can call `next` several times in a row, and hold onto all the results
+*concurrently*. This is perfectly fine for by-value iterators, which have exactly
+these semantics. It's also actually fine for shared references, as they admit
+arbitrarily many references to the same thing (although the
+iterator needs to be a separate object from the thing being shared). But mutable
+references make this a mess. At first glance, they might seem completely
+incompatible with this API, as it would produce multiple mutable references to
+the same object!
+
+However it actually *does* work, exactly because iterators are one-shot objects.
+Everything an IterMut yields will be yielded *at most* once, so we don't *actually*
+ever yield multiple mutable references to the same piece of data.
+
+In general all mutable iterators require *some* unsafe code *somewhere*, though.
+Whether it's raw pointers, or safely composing on top of *another* IterMut.
+
+For instance, VecDeque's IterMut:
+
+```rust,ignore
+struct IterMut<'a, T:'a> {
+    // The whole backing array. Some of these indices are initialized!
+    ring: &'a mut [T],
+    tail: usize,
+    head: usize,
+}
+
+impl<'a, T> Iterator for IterMut<'a, T> {
+    type Item = &'a mut T;
+
+    fn next(&mut self) -> Option<&'a mut T> {
+        if self.tail == self.head {
+            return None;
+        }
+        let tail = self.tail;
+        self.tail = wrap_index(self.tail.wrapping_add(1), self.ring.len());
+
+        unsafe {
+            // might as well do unchecked indexing since wrap_index has us
+            // in-bounds, and many of the "middle" indices are uninitialized
+            // anyway.
+            let elem = self.ring.get_unchecked_mut(tail);
+
+            // round-trip through a raw pointer to unbound the lifetime from
+            // ourselves
+            Some(&mut *(elem as *mut _))
+        }
+    }
+}
+```
+
+A very subtle but interesting detail in this design is that it *relies on
+privacy to be sound*. Borrowck works on some very simple rules. One of those rules
+is that if we have a live &mut Foo and Foo contains an &mut Bar, then that &mut
+Bar is *also* live. Since IterMut is always live when `next` can be called, if
+`ring` were public then we could mutate `ring` while outstanding mutable borrows
+to it exist!

src/doc/tarpl/phantom-data.md

@@ -0,0 +1,87 @@
+% PhantomData
+
+When working with unsafe code, we can often end up in a situation where
+types or lifetimes are logically associated with a struct, but not actually
+part of a field. This most commonly occurs with lifetimes. For instance, the
+`Iter` for `&'a [T]` is (approximately) defined as follows:
+
+```rust,ignore
+struct Iter<'a, T: 'a> {
+    ptr: *const T,
+    end: *const T,
+}
+```
+
+However because `'a` is unused within the struct's body, it's *unbounded*.
+Because of the troubles this has historically caused, unbounded lifetimes and
+types are *illegal* in struct definitions. Therefore we must somehow refer
+to these types in the body. Correctly doing this is necessary to have
+correct variance and drop checking.
+
+We do this using *PhantomData*, which is a special marker type. PhantomData
+consumes no space, but simulates a field of the given type for the purpose of
+static analysis. This was deemed to be less error-prone than explicitly telling
+the type-system the kind of variance that you want, while also providing other
+useful such as the information needed by drop check.
+
+Iter logically contains a bunch of `&'a T`s, so this is exactly what we tell
+the PhantomData to simulate:
+
+```
+use std::marker;
+
+struct Iter<'a, T: 'a> {
+    ptr: *const T,
+    end: *const T,
+    _marker: marker::PhantomData<&'a T>,
+}
+```
+
+and that's it. The lifetime will be bounded, and your iterator will be variant
+over `'a` and `T`. Everything Just Works.
+
+Another important example is Vec, which is (approximately) defined as follows:
+
+```
+struct Vec<T> {
+    data: *const T, // *const for variance!
+    len: usize,
+    cap: usize,
+}
+```
+
+Unlike the previous example it *appears* that everything is exactly as we
+want. Every generic argument to Vec shows up in the at least one field.
+Good to go!
+
+Nope.
+
+The drop checker will generously determine that Vec<T> does not own any values
+of type T. This will in turn make it conclude that it does *not* need to worry
+about Vec dropping any T's in its destructor for determining drop check
+soundness. This will in turn allow people to create unsoundness using
+Vec's destructor.
+
+In order to tell dropck that we *do* own values of type T, and therefore may
+drop some T's when *we* drop, we must add an extra PhantomData saying exactly
+that:
+
+```
+use std::marker;
+
+struct Vec<T> {
+    data: *const T, // *const for covariance!
+    len: usize,
+    cap: usize,
+    _marker: marker::PhantomData<T>,
+}
+```
+
+Raw pointers that own an allocation is such a pervasive pattern that the
+standard library made a utility for itself called `Unique<T>` which:
+
+* wraps a `*const T` for variance
+* includes a `PhantomData<T>`,
+* auto-derives Send/Sync as if T was contained
+* marks the pointer as NonZero for the null-pointer optimization
+