Why can't I store a value and a reference to that value in the same struct?

Question

I have a value and I want to store that value and a reference to something inside that value in my own type:

struct Thing {
    count: u32,
}

struct Combined<'a>(Thing, &'a u32);

fn make_combined<'a>() -> Combined<'a> {
    let thing = Thing { count: 42 };

    Combined(thing, &thing.count)
}

Sometimes, I have a value and I want to store that value and a reference to that value in the same structure:

struct Combined<'a>(Thing, &'a Thing);

fn make_combined<'a>() -> Combined<'a> {
    let thing = Thing::new();

    Combined(thing, &thing)
}

Sometimes, I'm not even taking a reference of the value and I get the same error:

struct Combined<'a>(Parent, Child<'a>);

fn make_combined<'a>() -> Combined<'a> {
    let parent = Parent::new();
    let child = parent.child();

    Combined(parent, child)
}

In each of these cases, I get an error that one of the values "does not live long enough". What does this error mean?

Answer 1

Let's look at a simple implementation of this:

struct Parent {
    count: u32,
}

struct Child<'a> {
    parent: &'a Parent,
}

struct Combined<'a> {
    parent: Parent,
    child: Child<'a>,
}

impl<'a> Combined<'a> {
    fn new() -> Self {
        let parent = Parent { count: 42 };
        let child = Child { parent: &parent };

        Combined { parent, child }
    }
}

fn main() {}

This will fail with the error:

error[E0515]: cannot return value referencing local variable `parent`
  --> src/main.rs:19:9
   |
17 |         let child = Child { parent: &parent };
   |                                     ------- `parent` is borrowed here
18 | 
19 |         Combined { parent, child }
   |         ^^^^^^^^^^^^^^^^^^^^^^^^^^ returns a value referencing data owned by the current function

error[E0505]: cannot move out of `parent` because it is borrowed
  --> src/main.rs:19:20
   |
14 | impl<'a> Combined<'a> {
   |      -- lifetime `'a` defined here
...
17 |         let child = Child { parent: &parent };
   |                                     ------- borrow of `parent` occurs here
18 | 
19 |         Combined { parent, child }
   |         -----------^^^^^^---------
   |         |          |
   |         |          move out of `parent` occurs here
   |         returning this value requires that `parent` is borrowed for `'a`

To completely understand this error, you have to think about how the values are represented in memory and what happens when you move those values. Let's annotate Combined::new with some hypothetical memory addresses that show where values are located:

let parent = Parent { count: 42 };
// `parent` lives at address 0x1000 and takes up 4 bytes
// The value of `parent` is 42 
let child = Child { parent: &parent };
// `child` lives at address 0x1010 and takes up 4 bytes
// The value of `child` is 0x1000
         
Combined { parent, child }
// The return value lives at address 0x2000 and takes up 8 bytes
// `parent` is moved to 0x2000
// `child` is ... ?

What should happen to child? If the value was just moved like parent was, then it would refer to memory that no longer is guaranteed to have a valid value in it. Any other piece of code is allowed to store values at memory address 0x1000. Accessing that memory assuming it was an integer could lead to crashes and/or security bugs, and is one of the main categories of errors that Rust prevents.

This is exactly the problem that lifetimes prevent. A lifetime is a bit of metadata that allows you and the compiler to know how long a value will be valid at its current memory location. That's an important distinction, as it's a common mistake Rust newcomers make. Rust lifetimes are not the time period between when an object is created and when it is destroyed!

As an analogy, think of it this way: During a person's life, they will reside in many different locations, each with a distinct address. A Rust lifetime is concerned with the address you currently reside at, not about whenever you will die in the future (although dying also changes your address). Every time you move it's relevant because your address is no longer valid.

It's also important to note that lifetimes do not change your code; your code controls the lifetimes, your lifetimes don't control the code. The pithy saying is "lifetimes are descriptive, not prescriptive".

Let's annotate Combined::new with some line numbers which we will use to highlight lifetimes:

{                                          // 0
    let parent = Parent { count: 42 };     // 1
    let child = Child { parent: &parent }; // 2
                                           // 3
    Combined { parent, child }             // 4
}                                          // 5

The concrete lifetime of parent is from 1 to 4, inclusive (which I'll represent as [1,4]). The concrete lifetime of child is [2,4], and the concrete lifetime of the return value is [4,5]. It's possible to have concrete lifetimes that start at zero - that would represent the lifetime of a parameter to a function or something that existed outside of the block.

Note that the lifetime of child itself is [2,4], but that it refers to a value with a lifetime of [1,4]. This is fine as long as the referring value becomes invalid before the referred-to value does. The problem occurs when we try to return child from the block. This would "over-extend" the lifetime beyond its natural length.

This new knowledge should explain the first two examples. The third one requires looking at the implementation of Parent::child. Chances are, it will look something like this:

impl Parent {
    fn child(&self) -> Child { /* ... */ }
}

This uses lifetime elision to avoid writing explicit generic lifetime parameters. It is equivalent to:

impl Parent {
    fn child<'a>(&'a self) -> Child<'a> { /* ... */ }
}

In both cases, the method says that a Child structure will be returned that has been parameterized with the concrete lifetime of self. Said another way, the Child instance contains a reference to the Parent that created it, and thus cannot live longer than that Parent instance.

This also lets us recognize that something is really wrong with our creation function:

fn make_combined<'a>() -> Combined<'a> { /* ... */ }

Although you are more likely to see this written in a different form:

impl<'a> Combined<'a> {
    fn new() -> Combined<'a> { /* ... */ }
}

In both cases, there is no lifetime parameter being provided via an argument. This means that the lifetime that Combined will be parameterized with isn't constrained by anything - it can be whatever the caller wants it to be. This is nonsensical, because the caller could specify the 'static lifetime and there's no way to meet that condition.

How do I fix it?

The easiest and most recommended solution is to not attempt to put these items in the same structure together. By doing this, your structure nesting will mimic the lifetimes of your code. Place types that own data into a structure together and then provide methods that allow you to get references or objects containing references as needed.

There is a special case where the lifetime tracking is overzealous: when you have something placed on the heap. This occurs when you use a Box<T>, for example. In this case, the structure that is moved contains a pointer into the heap. The pointed-at value will remain stable, but the address of the pointer itself will move. In practice, this doesn't matter, as you always follow the pointer.

Some crates provide ways of representing this case, but they require that the base address never move. This rules out mutating vectors, which may cause a reallocation and a move of the heap-allocated values.

• rental (no longer maintained or suppported)
• owning_ref
• ouroboros

Examples of problems solved with Rental:

• Is there an owned version of String::chars?
• Returning a RWLockReadGuard independently from a method
• How can I return an iterator over a locked struct member in Rust?
• How to return a reference to a sub-value of a value that is under a mutex?
• How do I store a result using Serde Zero-copy deserialization of a Futures-enabled Hyper Chunk?
• How to store a reference without having to deal with lifetimes?

In other cases, you may wish to move to some type of reference-counting, such as by using Rc or Arc.

More information

After moving parent into the struct, why is the compiler not able to get a new reference to parent and assign it to child in the struct?

While it is theoretically possible to do this, doing so would introduce a large amount of complexity and overhead. Every time that the object is moved, the compiler would need to insert code to "fix up" the reference. This would mean that copying a struct is no longer a very cheap operation that just moves some bits around. It could even mean that code like this is expensive, depending on how good a hypothetical optimizer would be:

let a = Object::new();
let b = a;
let c = b;

Instead of forcing this to happen for every move, the programmer gets to choose when this will happen by creating methods that will take the appropriate references only when you call them.

A type with a reference to itself

There's one specific case where you can create a type with a reference to itself. You need to use something like Option to make it in two steps though:

#[derive(Debug)]
struct WhatAboutThis<'a> {
    name: String,
    nickname: Option<&'a str>,
}

fn main() {
    let mut tricky = WhatAboutThis {
        name: "Annabelle".to_string(),
        nickname: None,
    };
    tricky.nickname = Some(&tricky.name[..4]);

    println!("{:?}", tricky);
}

This does work, in some sense, but the created value is highly restricted - it can never be moved. Notably, this means it cannot be returned from a function or passed by-value to anything. A constructor function shows the same problem with the lifetimes as above:

fn creator<'a>() -> WhatAboutThis<'a> { /* ... */ }

If you try to do this same code with a method, you'll need the alluring but ultimately useless &'a self. When that's involved, this code is even more restricted and you will get borrow-checker errors after the first method call:

#[derive(Debug)]
struct WhatAboutThis<'a> {
    name: String,
    nickname: Option<&'a str>,
}

impl<'a> WhatAboutThis<'a> {
    fn tie_the_knot(&'a mut self) {
       self.nickname = Some(&self.name[..4]); 
    }
}

fn main() {
    let mut tricky = WhatAboutThis {
        name: "Annabelle".to_string(),
        nickname: None,
    };
    tricky.tie_the_knot();

    // cannot borrow `tricky` as immutable because it is also borrowed as mutable
    // println!("{:?}", tricky);
}

See also:

• Cannot borrow as mutable more than once at a time in one code - but can in another very similar

What about Pin?

Pin, stabilized in Rust 1.33, has this in the module documentation:

A prime example of such a scenario would be building self-referential structs, since moving an object with pointers to itself will invalidate them, which could cause undefined behavior.

It's important to note that "self-referential" doesn't necessarily mean using a reference. Indeed, the example of a self-referential struct specifically says (emphasis mine):

We cannot inform the compiler about that with a normal reference, since this pattern cannot be described with the usual borrowing rules. Instead we use a raw pointer, though one which is known to not be null, since we know it's pointing at the string.

The ability to use a raw pointer for this behavior has existed since Rust 1.0. Indeed, owning-ref and rental use raw pointers under the hood.

The only thing that Pin adds to the table is a common way to state that a given value is guaranteed to not move.

See also:

• How to use the Pin struct with self-referential structures?