borrow-checker lifetime reference rust

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


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?


  • 2

    For the latter example, a definition of Parent and Child could help…

    Aug 31, 2015 at 6:32

  • 2

    @MatthieuM. I debated that, but decided against it based on the two linked questions. Neither of those questions looked at the definition of the struct or the method in question, so I thought it would be best to mimic that to that people can more easily match this question to their own situation. Note that I do show the method signature in the answer.

    Aug 31, 2015 at 14:25


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/
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/
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
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

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.

Examples of problems solved with Rental:

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:

struct WhatAboutThis<'a> {
    name: String,
    nickname: Option<&'a str>,

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

struct WhatAboutThis<'a> {
    name: String,
    nickname: Option<&'a str>,

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

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

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

See also:

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:


  • 1

    Is something like this ( considered idiomatic? Ie, to expose the data via methods instead of the raw data.

    Jan 4, 2016 at 22:05

  • 2

    @PeterHall sure, it just means that Combined owns the Child which owns the Parent. That may or may not make sense depending on the actual types that you have. Returning references to your own internal data is pretty typical.

    Jan 4, 2016 at 22:42

  • 2

    What is the solution to the heap problem?

    Nov 14, 2016 at 11:25

  • @derekdreery perhaps you could expand on your comment? Why is the entire paragraph talking about the owning_ref crate insufficient?

    Nov 14, 2016 at 13:57

  • 2

    @FynnBecker it’s still impossible to store a reference and a value to that reference. Pin is mostly a way to know the safety of a struct containing a self referential pointer. The ability to use a raw pointer for the same purpose has existed since Rust 1.0.

    Mar 4, 2019 at 17:52


A slightly different issue which causes very similar compiler messages is object lifetime dependency, rather than storing an explicit reference. An example of that is the ssh2 library. When developing something bigger than a test project, it is tempting to try to put the Session and Channel obtained from that session alongside each other into a struct, hiding the implementation details from the user. However, note that the Channel definition has the 'sess lifetime in its type annotation, while Session doesn’t.

This causes similar compiler errors related to lifetimes.

One way to solve it in a very simple way is to declare the Session outside in the caller, and then for annotate the reference within the struct with a lifetime, similar to the answer in this Rust User’s Forum post talking about the same issue while encapsulating SFTP. This will not look elegant and may not always apply – because now you have two entities to deal with, rather than one that you wanted!

Turns out the rental crate or the owning_ref crate from the other answer are the solutions for this issue too. Let’s consider the owning_ref, which has the special object for this exact purpose:
OwningHandle. To avoid the underlying object moving, we allocate it on the heap using a Box, which gives us the following possible solution:

use ssh2::{Channel, Error, Session};
use std::net::TcpStream;

use owning_ref::OwningHandle;

struct DeviceSSHConnection {
    tcp: TcpStream,
    channel: OwningHandle<Box<Session>, Box<Channel<'static>>>,

impl DeviceSSHConnection {
    fn new(targ: &str, c_user: &str, c_pass: &str) -> Self {
        use std::net::TcpStream;
        let mut session = Session::new().unwrap();
        let mut tcp = TcpStream::connect(targ).unwrap();

        session.userauth_password(c_user, c_pass).unwrap();

        let mut sess = Box::new(session);
        let mut oref = OwningHandle::new_with_fn(
            unsafe { |x| Box::new((*x).channel_session().unwrap()) },
        let ret = DeviceSSHConnection {
            tcp: tcp,
            channel: oref,

The result of this code is that we can not use the Session anymore, but it is stored alongside with the Channel which we will be using. Because the OwningHandle object dereferences to Box, which dereferences to Channel, when storing it in a struct, we name it as such. NOTE: This is just my understanding. I have a suspicion this may not be correct, since it appears to be quite close to discussion of OwningHandle unsafety.

One curious detail here is that the Session logically has a similar relationship with TcpStream as Channel has to Session, yet its ownership is not taken and there are no type annotations around doing so. Instead, it is up to the user to take care of this, as the documentation of handshake method says:

This session does not take ownership of the socket provided, it is
recommended to ensure that the socket persists the lifetime of this
session to ensure that communication is correctly performed.

It is also highly recommended that the stream provided is not used
concurrently elsewhere for the duration of this session as it may
interfere with the protocol.

So with the TcpStream usage, is completely up to the programmer to ensure the correctness of the code. With the OwningHandle, the attention to where the “dangerous magic” happens is drawn using the unsafe {} block.

A further and a more high-level discussion of this issue is in this Rust User’s Forum thread – which includes a different example and its solution using the rental crate, which does not contain unsafe blocks.


    I’ve found the Arc (read-only) or Arc<Mutex> (read-write with locking) patterns to be sometimes quite useful tradeoff between performance and code complexity (mostly caused by lifetime-annotation).


    use std::sync::Arc;
    struct Parent {
        child: Arc<Child>,
    struct Child {
        value: u32,
    struct Combined(Parent, Arc<Child>);
    fn main() {
        let parent = Parent { child: Arc::new(Child { value: 42 }) };
        let child = parent.child.clone();
        let combined = Combined(parent, child.clone());
        assert_eq!(combined.0.child.value, 42);
        assert_eq!(child.value, 42);
        // combined.0.child.value = 50; // fails, Arc is not DerefMut

    Arc + Mutex:

    use std::sync::{Arc, Mutex};
    struct Child {
        value: u32,
    struct Parent {
        child: Arc<Mutex<Child>>,
    struct Combined(Parent, Arc<Mutex<Child>>);
    fn main() {
        let parent = Parent { child: Arc::new(Mutex::new(Child {value: 42 }))};
        let child = parent.child.clone();
        let combined = Combined(parent, child.clone());
        assert_eq!(combined.0.child.lock().unwrap().value, 42);
        assert_eq!(child.lock().unwrap().value, 42);
        child.lock().unwrap().value = 50;
        assert_eq!(combined.0.child.lock().unwrap().value, 50);

    See also RwLock (When or why should I use a Mutex over an RwLock?)