4.7 KiB
% 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:
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:
let x = [1, 2, 3];
let a = &mut x[0];
let b = &mut x[1];
println!("{} {}", a, b);
<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:
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:
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:
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!