I understand how lifetime parameters apply to functions and structs, but what does it mean for a trait to have a lifetime parameter? Is it a shortcut to introduce a lifetime parameter to its methods, or is it something else?
If you have a trait with a lifetime bound, then implementors of the trait can participate in the same lifetime. Concretely, this allows you to store references with that lifetime. It is not a shortcut for specifying lifetimes on member methods, and difficulty and confusing error messages lie that way!
trait Keeper<'a> {
fn save(&mut self, v: &'a u8);
fn restore(&self) -> &'a u8;
}
struct SimpleKeeper<'a> {
val: &'a u8,
}
impl<'a> Keeper<'a> for SimpleKeeper<'a> {
fn save(&mut self, v: &'a u8) {
self.val = v
}
fn restore(&self) -> &'a u8 {
self.val
}
}
Note how both the struct and the trait are parameterized on a lifetime, and that lifetime is the same.
What would the non-trait versions of save() and restore() look like for SimpleKeeper<'a>?
Very similar, actually. The important part is that the struct stores the reference itself, so it needs to have a lifetime parameter for the values inside.
struct SimpleKeeper<'a> {
val: &'a u8,
}
impl<'a> SimpleKeeper<'a> {
fn save(&mut self, v: &'a u8) {
self.val = v
}
fn restore(&self) -> &'a u8 {
self.val
}
}
And would they mean exactly the same thing as the the trait version?
Yep!
Related
I currently have code that looks kind of like this:
struct People {
names: Vec<String>,
ages: Vec<i32>,
}
impl People {
fn iter_people<'a>(&'a self) -> PeopleIterator<'a> {
return PeopleIterator {
names_iterator: Box::new(self.names.iter()),
ages: Box::new(self.ages.iter()),
};
}
}
struct PeopleIterator<'a> {
names_iterator: Box<dyn Iterator<Item = &'a String>>,
ages: Box<dyn Iterator<Item = &'a i32>>,
}
impl<'a> Iterator for PeopleIterator<'a> {
...snip...
}
I am aware that I should model a person as a struct Person and then have a Vec<Person> to model people but this is just a simplification of my actual code.
Anyway, the Rust compiler tells me this:
lifetime may not live long enough
requirement occurs because of the type PeopleIterator<'_>, which makes the generic argument '_ invariant
I have looked at the suggested link for subtyping and variance but I need to read it a few more times to actually understand it.
What stumps me is that I would expect both my iterators self.names.iter() and self.ages.iter() to live as long as self and I have declared that self should live as long as PeopleIterator. However, when I look at the iter() function, it does not make this constraint but instead has an anonymous lifetime '_. I am guessing this is the problem but I am confused and don't know how to fix it :(
The problem is the lifetime of the iterator itself in Box<dyn Iterator<Item = &'a String>> is by default bound to be 'static, but that's not possible for an iterator containing non static references like anything from &'a self. The solution is to specify an explicit lifetime bound:
struct PeopleIterator<'a> {
names_iterator: Box<dyn Iterator<Item = &'a String> + 'a>,
ages: Box<dyn Iterator<Item = &'a i32> + 'a>,
}
Personally I'd just use generics instead of static dispatch avoiding some indirection and the whole problem from the beginning:
impl People {
fn iter_people(&self) -> PeopleIterator<impl Iterator<Item = &String>, impl Iterator<Item = &i32>> {
return PeopleIterator {
names_iterator: self.names.iter(),
ages: self.ages.iter(),
};
}
}
struct PeopleIterator<N, A> {
names_iterator: N,
ages: A,
}
I want to create a wrapper around (nested) slices for easy operations on multidimensional data, owned by a different struct.
The most basic version of the mutable version of my slice wrapper might look like this:
struct MySliceMut<'a> {
data: Vec<&'a mut [f32]>,
}
impl<'a, 'b> MySliceMut<'a> {
fn get(&'b mut self) -> &'a mut [&'b mut [f32]] {
self.data.as_mut_slice()
}
}
Now if I want to implement a trait, for instance AddAssign, Rust does not seem to infer the lifetime of &mut self from the implementing type. The compiler complains that &mut self might outlive 'a:
impl<'a> AddAssign<MySlice<'a>> for MySliceMut<'a> { // lifetime 'a
fn add_assign(&mut self, rhs: MySlice<'a>) { // lifetime '1
let a = self.get(); // lifetime may not live long enough, '1 must outlive 'a
let b = rhs.get();
// do inplace addition here
}
}
Full Code - Rust Playground
I tried to figure out the issue with the lifetimes, but can't find it. Would the trait impl require any additional annotations?
struct MySlice<'a> {
data: Vec<&'a [f32]>,
}
impl<'a, 'b> MySlice<'a> {
fn get(&'b self) -> &'a [&'b [f32]] {
self.data.as_slice()
}
}
Problem with your code is that fn get(&'b self) returns variable with wrong lifetime. Associated lifetime 'a of MySlice<'a> is lifetime of inner slice. Associated lifetime 'b of fn get(...) is lifetime of the self. So I guess the function probably should return &'b [&'a [f32]] instead.
-- Edited --
Make sure to change fn get(...) of MySliceMut either.
I am having trouble figuring out what lifetime parameter will work for this, so my current workarounds include transmutes or raw pointers. I have a structure holding a function pointer with a generic as a parameter:
struct CB<Data> {
cb: fn(Data) -> usize
}
I would like to store an instance of that, parameterized by some type containing a reference, in some other structure that implements a trait with one method, and use that trait method to call the function pointer in CB.
struct Holder<'a> {
c: CB<Option<&'a usize>>
}
trait Exec {
fn exec(&self, v: &usize) -> usize;
}
impl<'a> Holder<'a> {
fn exec_aux(&self, v: &'a usize) -> usize {
(self.c.cb)(Some(v))
}
}
impl<'a> Exec for Holder<'a> {
fn exec(&self, v: &usize) -> usize
{
self.exec_aux(v)
}
}
This gives me a lifetime error for the 'Exec' impl of Holder:
error[E0495]: cannot infer an appropriate lifetime for lifetime parameter `'a` due to conflicting requirements
Simply calling exec_aux works fine as long as I don't define that Exec impl:
fn main() {
let h = Holder { c: CB{cb:cbf}};
let v = 12;
println!("{}", h.exec_aux(&v));
}
Also, making CB not generic also makes this work:
struct CB {
cb: fn(Option<&usize>) -> usize
}
The parameter in my actual code is not a usize but something big that I would rather not copy.
The lifetimes in your Exec trait are implicitly this:
trait Exec {
fn exec<'s, 'a>(&'s self, v: &'a usize) -> usize;
}
In other words, types that implement Exec need to accept any lifetimes 's and 'a. However, your Holder::exec_aux method expects a specific lifetime 'a that's tied to the lifetime parameter of the Holder type.
To make this work, you need to add 'a as a lifetime parameter to the Exec trait instead, so that you can implement the trait specifically for that lifetime:
trait Exec<'a> {
// ^^^^ vv
fn exec(&self, v: &'a usize) -> usize;
}
impl<'a> Exec<'a> for Holder<'a> {
// ^^^^ vv
fn exec(&self, v: &'a usize) -> usize
{
self.exec_aux(v)
}
}
The problem here is that the Exec trait is too generic to be used in this way by Holder. First, consider the definition:
trait Exec {
fn exec(&self, v: &usize) -> usize;
}
This definition will cause the compiler to automatically assign two anonymous lifetimes for &self and &v in exec. It's basically the same as
fn exec<'a, 'b>(&'a self, v: &'b usize) -> usize;
Note that there is no restriction on who needs to outlive whom, the references just need to be alive for the duration of the method call.
Now consider the definition
impl<'a> Holder<'a> {
fn exec_aux(&self, v: &'a usize) -> usize {
// ... doesn't matter
}
}
Since we know that &self is a &Holder<'a> (this is what the impl refers to), we need to have at least a &'a Holder<'a> here, because &'_ self can't have a lifetime shorter than 'a in Holder<'a>. So this is saying that the two parameters have the same lifetime: &'a self, &'a usize.
Where it all goes wrong is when you try to combine the two. The trait forces you into the following signature, which (again) has two distinct implicit lifetimes. But the actual Holder which you then try to call a method on forces you to have the same lifetimes for &self and &v.
fn exec(&self, v: &usize) -> usize {
// Holder<'a> needs `v` to be `'a` when calling exec_aux
// But the trait doesn't say so.
self.exec_aux(v)
}
One solution is to redefine the trait as
trait Exec<'a> {
fn exec(&'a self, v: &'a usize) -> usize;
}
and then implement it as
impl<'a> Exec<'a> for Holder<'a> {
fn exec(&'a self, v: &'a usize) -> usize {
self.exec_aux(v)
}
}
I try to write my own RefCell-like mutable memory location but without runtime borrow checking (no overhead). I adopted the code architecture from RefCell (and Ref, and RefMut). I can call .borrow() without problems but if I call .borrow_mut() then the rust compiler says cannot borrow as mutable. I don't see the problem, my .borrow_mut() impl looks fine?
code that fails:
let real_refcell= Rc::from(RefCell::from(MyStruct::new()));
let nooverhead_refcell = Rc::from(NORefCell::from(MyStruct::new()));
// works
let refmut_refcell = real_refcell.borrow_mut();
// cannot borrow as mutable
let refmut_norefcell = nooverhead_refcell.borrow_mut();
norc.rs (No Overhead RefCell)
use crate::norc_ref::{NORefMut, NORef};
use std::cell::UnsafeCell;
use std::borrow::Borrow;
#[derive(Debug)]
pub struct NORefCell<T: ?Sized> {
value: UnsafeCell<T>
}
impl<T> NORefCell<T> {
pub fn from(t: T) -> NORefCell<T> {
NORefCell {
value: UnsafeCell::from(t)
}
}
pub fn borrow(&self) -> NORef<'_, T> {
NORef {
value: unsafe { &*self.value.get() }
}
}
pub fn borrow_mut(&mut self) -> NORefMut<'_, T> {
NORefMut {
value: unsafe { &mut *self.value.get() }
}
}
}
norc_ref.rs (data structure returned by NORefCell.borrow[_mut]()
use std::ops::{Deref, DerefMut};
#[derive(Debug)]
pub struct NORef<'b, T: ?Sized + 'b> {
pub value: &'b T,
}
impl<T: ?Sized> Deref for NORef<'_, T> {
type Target = T;
#[inline]
fn deref(&self) -> &T {
self.value
}
}
/// No Overhead Ref Cell: Mutable Reference
#[derive(Debug)]
pub struct NORefMut<'b, T: ?Sized + 'b> {
pub value: &'b mut T,
}
impl<T: ?Sized> Deref for NORefMut<'_, T> {
type Target = T;
#[inline]
fn deref(&self) -> &T {
self.value
}
}
impl<T: ?Sized> DerefMut for NORefMut<'_, T> {
#[inline]
fn deref_mut(&mut self) -> &mut T {
self.value
}
}
NORefCell::borrow_mut() takes &mut self, which requires a DerefMut on the Rc in which it is wrapped. This won't work because Rc does not give mutable references just by asking nicely (you need it to check if the reference count is exactly one, otherwise there would be multiple mutable borrows).
borrow_mut has to take &self instead of &mut self.
As mentioned in my comment: What you are basically doing is providing a safe-looking abstraction around an UnsafeCell. This is incredibly dangerous. Notice the docs regarding UnsafeCell:
The compiler makes optimizations based on the knowledge that &T is not mutably aliased or mutated, and that &mut T is unique. UnsafeCell is the only core language feature to work around the restriction that &T may not be mutated.
You are providing a thin wrapper around this powerful object, with no unsafe on the API-boundary. The "No-overhead-RefCell" is really a "no-trigger-guard-foot-gun". It does work, yet be warned about its dangers.
I'd like to do something along the following lines:
trait GetRef<'a> {
fn get_ref(&self) -> &'a [u8];
}
struct Foo<'a> {
buf: &'a [u8]
}
impl <'a> GetRef<'a> for Foo<'a> {
fn get_ref(&self) -> &'a [u8] {
&self.buf[1..]
}
}
struct Bar {
buf: Vec<u8>
}
// this is the part I'm struggling with:
impl <'a> GetRef<'a> for Bar {
fn get_ref(&'a self) -> &'a [u8] {
&self.buf[1..]
}
The point of the explicit lifetime variable in the GetRef trait is to allow the return value of get_ref() on a Foo object to outlive the Foo itself, tying the return value's lifetime to that of the lifetime of Foo's buffer.
However, I haven't found a way to implement GetRef for Bar in a way that the compiler accepts. I've tried several variations of the above, but can't seem to find one that works. Is there any there any reason that this fundamentally cannot be done? If not, how can I do this?
Tying a trait lifetime variable to &self lifetime
Not possible.
Is there any there any reason that this fundamentally cannot be done?
Yes. An owning vector is something different than a borrowed slice. Your trait GetRef only makes sense for things that already represent a “loan” and don't own the slice. For an owning type like Bar you can't safely return a borrowed slice that outlives Self. That's what the borrow checker prevents to avoid dangling pointers.
What you tried to do is to link the lifetime parameter to the lifetime of Self. But the lifetime of Self is not a property of its type. It just depends on the scope this value was defined in. And that's why your approach cannot work.
Another way of looking at it is: In a trait you have to be explicit about whether Self is borrowed by a method and its result or not. You defined the GetRef trait to return something that is not linked to Self w.r.t. lifetimes. So, no borrowing. So, it's not implementable for types that own the data. You can't create a borrowed slice referring to a Vec's elements without borrowing the Vec.
If not, how can I do this?
Depends on what exactly you mean by “this”. If you want to write a “common denominator” trait that can be implemented for both borrowed and owning slices, you have to do it like this:
trait GetRef {
fn get_ref(&self) -> &[u8];
}
The meaning of this trait is that get_ref borrows Self and returns a kind of “loan” because of the current lifetime elision rules. It's equivalent to the more explicit form
trait GetRef {
fn get_ref<'s>(&self) -> &'s [u8];
}
It can be implemented for both types now:
impl<'a> GetRef for Foo<'a> {
fn get_ref(&self) -> &[u8] { &self.buf[1..] }
}
impl GetRef for Bar {
fn get_ref(&self) -> &[u8] { &self.buf[1..] }
}
You could make different lifetimes for &self and result in your trait like that:
trait GetRef<'a, 'b> {
fn get_ref(&'b self) -> &'a [u8];
}
struct Foo<'a> {
buf: &'a [u8]
}
impl <'a, 'b> GetRef<'a, 'b> for Foo<'a> {
fn get_ref(&'b self) -> &'a [u8] {
&self.buf[1..]
}
}
struct Bar {
buf: Vec<u8>
}
// Bar, however, cannot contain anything that outlives itself
impl<'a> GetRef<'a, 'a> for Bar {
fn get_ref(&'a self) -> &'a [u8] {
&self.buf[1..]
}
}
fn main() {
let a = vec!(1 as u8, 2, 3);
let b = a.clone();
let tmp;
{
let x = Foo{buf: &a};
tmp = x.get_ref();
}
{
let y = Bar{buf: b};
// Bar's buf cannot outlive Bar
// tmp = y.get_ref();
}
}