Common Rust Lifetime Misconceptions

May 19th, 2020 · 34 minute read · #rust · #lifetimes

Intro

I've held all of these misconceptions at some point and I see many beginners struggle with these misconceptions today. Some of my terminology might be non-standard, so here's a table of shorthand phrases I use and what I intend for them to mean.

Phrase Shorthand for T 1) a set containing all possible types or
2) some type within that set owned type some non-reference type, e.g. i32 , String , Vec , etc 1) borrowed type or
2) ref type some reference type regardless of mutability, e.g. &i32 , &mut i32 , etc 1) mut ref or
2) exclusive ref exclusive mutable reference, i.e. &mut T 1) immut ref or
2) shared ref shared immutable reference, i.e. &T

The Misconceptions

In a nutshell: A variable's lifetime is how long the data it points to can be statically verified by the compiler to be valid at its current memory address. I'll now spend the next ~6000 words going into more detail about where people commonly get confused.

1) `T` only contains owned types

This misconception is more about generics than lifetimes but generics and lifetimes are tightly interwined in Rust so it's not possibly to talk about one without also talking about the other. Anyway:

When I first started learning Rust I understood that i32 , &i32 , and &mut i32 are different types. I also understood that some generic type variable T represents a set which contains all possible types. However, despite understanding both of these things separately, I wasn't able to understand them together. In my newbie Rust mind this is how I thought generics worked:

Type Variable T &T &mut T Examples i32 &i32 &mut i32

T contains all owned types. &T contains all immutably borrowed types. &mut T contains all mutably borrowed types. T , &T , and &mut T are disjoint finite sets. Nice, simple, clean, easy, intuitive, and completely totally wrong. This is how generics actually work in Rust:

Type Variable T &T &mut T Examples i32 , &i32 , &mut i32 , &&i32 , &mut &mut i32 , ... &i32 , &&i32 , &&mut i32 , ... &mut i32 , &mut &mut i32 , &mut &i32 , ...

T , &T , and &mut T are all infinite sets, since it's possible to borrow a type ad-infinitum. T is a superset of both &T and &mut T . &T and &mut T are disjoint sets. Here's a couple examples which validate these concepts:

trait Trait {}

impl<T> Trait for T {}

impl<T> Trait for &T {} // compile error

impl<T> Trait for &mut T {} // compile error

The above program doesn't compile as expected:

error[E0119]: conflicting implementations of trait `Trait` for type `&_`:
 --> src/lib.rs:5:1
  |
3 | impl<T> Trait for T {}
  | ------------------- first implementation here
4 |
5 | impl<T> Trait for &T {}
  | ^^^^^^^^^^^^^^^^^^^^ conflicting implementation for `&_`

error[E0119]: conflicting implementations of trait `Trait` for type `&mut _`:
 --> src/lib.rs:7:1
  |
3 | impl<T> Trait for T {}
  | ------------------- first implementation here
...
7 | impl<T> Trait for &mut T {}
  | ^^^^^^^^^^^^^^^^^^^^^^^^ conflicting implementation for `&mut _`

The compiler doesn't allow us to define an implementation of Trait for &T and &mut T since it would conflict with the implementation of Trait for T which already includes all of &T and &mut T . The program below compiles as expected, since &T and &mut T are disjoint:

trait Trait {}

impl<T> Trait for &T {} // compiles

impl<T> Trait for &mut T {} // compiles

Key Takeaways

T is a superset of both &T and &mut T
&T and &mut T are disjoint sets

2) if `T: 'static` then `T` must be valid for the entire program

Misconception Corollaries

T: 'static should be read as " T has a 'static lifetime"
&'static T and T: 'static are the same thing
if T: 'static then T must be immutable
if T: 'static then T can only be created at compile time

Most Rust beginners get introduced to the 'static lifetime for the first time in a code example that looks something like this:

fn main() {
    let str_literal: &'static str = "str literal";
}

They get told that "str literal" is hardcoded into the compiled binary and it's loaded into read-only memory at run-time so it's immutable and valid for the entire program and that's what makes it 'static . These concepts are further reinforced by the rules surrounding defining static variables using the static keyword.

static BYTES: [u8; 3] = [1, 2, 3];
static mut MUT_BYTES: [u8; 3] = [1, 2, 3];

fn main() {
   MUT_BYTES[0] = 99; // compile error, mutating static is unsafe

    unsafe {
        MUT_BYTES[0] = 99;
        assert_eq!(99, MUT_BYTES[0]);
    }
}

Regarding static variables

they can only be created at compile-time
they should be immutable, mutating them is unsafe
they're valid for the entire program

The 'static lifetime was probably named after the default lifetime of static variables, right? So it makes sense that the 'static lifetime has to follow all the same rules, right?

Well yes, but a type with a 'static lifetime is different from a type bounded by a 'static lifetime. The latter can be dynamically allocated at run-time, can be safely and freely mutated, can be dropped, and can live for arbitrary durations.

It's important at this point to distinguish &'static T from T: 'static .

&'static T is an immutable reference to some T that can be safely held indefinitely long, including up until the end of the program. This is only possible if T itself is immutable and does not move after the reference was created . T does not need to be created at compile-time. It's possible to generate random dynamically allocated data at run-time and return 'static references to it at the cost of leaking memory, e.g.

use rand;

// generate random 'static str refs at run-time
fn rand_str_generator() -> &'static str {
    let rand_string = rand::random::<u64>().to_string();
    Box::leak(rand_string.into_boxed_str())
}

T: 'static is some T that can be safely held indefinitely long, including up until the end of the program. T: 'static includes all &'static T however it also includes all owned types, like String , Vec , etc. The owner of some data is guaranted that data will never get invalidated as long as the owner holds onto it, therefore the owner can safely hold onto the data indefinitely long, including up until the end of the program. T: 'static should be read as " T is bounded by a 'static lifetime" not " T has a 'static lifetime" . A program to help illustrate these concepts:

use rand;

fn drop_static<T: 'static>(t: T) {
    std::mem::drop(t);
}

fn main() {
    let mut strings: Vec<String> = Vec::new();
    for _ in 0..10 {
        if rand::random() {
            // all the strings are randomly generated
            // and dynamically allocated at run-time
            let string = rand::random::<u64>().to_string();
            strings.push(string);
        }
    }

    // strings are owned types so they're bounded by 'static
    for mut string in strings {
        // all the strings are mutable
        string.push_str("a mutation");
        // all the strings are droppable
        drop_static(string); // compiles
    }

    // all the strings have been invalidated before the end of the program
    println!("i am the end of the program");
}

Key Takeaways

T: 'static should be read as " T is bounded by a 'static lifetime"
if T: 'static then T can be a borrowed type with a 'static lifetime or an owned type
since T: 'static includes owned types that means T
- can be dynamically allocated at run-time
- does not have to be valid for the entire program
- can be safely and freely mutated
- can be dynamically dropped at run-time
- can have lifetimes of different durations

3) `&'a T` and `T: 'a` are the same thing

This misconception is a generalized version of the one above.

&'a T requires and implies T: 'a since a reference to T of lifetime 'a cannot be valid for 'a if T itself is not valid for 'a . For example, the Rust compiler will never allow the construction of the type &'static Ref<'a, T> because if Ref is only valid for 'a we can't make a 'static reference to it.

T: 'a includes all &'a T but the reverse is not true.

// only takes ref types bounded by 'a
fn t_ref<'a, T: 'a>(t: &'a T) {}

// takes any types bounded by 'a
fn t_bound<'a, T: 'a>(t: T) {}

// owned type which contains a reference
struct Ref<'a, T: 'a>(&'a T);

fn main() {
    let string = String::from("string");

    t_bound(&string); // compiles
    t_bound(Ref(&string)); // compiles
    t_bound(&Ref(&string)); // compiles

    t_ref(&string); // compiles
    t_ref(Ref(&string)); // compile error, expected ref, found struct
    t_ref(&Ref(&string)); // compiles

    // string var is bounded by 'static which is bounded by 'a
    t_bound(string); // compiles
}

Key Takeaways

T: 'a is more general and more flexible than &'a T
T: 'a accepts owned types, owned types which contain references, and references
&'a T only accepts references
if T: 'static then T: 'a since 'static >= 'a for all 'a

4) my code isn't generic and doesn't have lifetimes

Misconception Corollaries

it's possible to avoid using generics and lifetimes

This comforting misconception is kept alive thanks to Rust's lifetime elision rules, which allow you to omit lifetime annotations in functions because the Rust borrow checker will infer them following these rules:

every input ref to a function gets a distinct lifetime
if there's exactly one input lifetime it gets applied to all output refs
if there's multiple input lifetimes but one of them is &self or &mut self then the lifetime of self is applied to all output refs
otherwise output lifetimes have to be made explicit

That's a lot to take in so lets look at some examples:

// elided
fn print(s: &str);

// expanded
fn print<'a>(s: &'a str);

// elided
fn trim(s: &str) -> &str;

// expanded
fn trim<'a>(s: &'a str) -> &'a str;

// illegal, can't determine output lifetime, no inputs
fn get_str() -> &str;

// explicit options include
fn get_str<'a>() -> &'a str; // pointlessly generic, since 'a must equal 'static
fn get_str() -> &'static str; // better, more explicit

// illegal, can't determine output lifetime, mutiple inputs
fn overlap(s: &str, t: &str) -> &str;

// explicit (but still partially elided) options include
fn overlap<'a>(s: &'a str, t: &str) -> &'a str; // output can't outlive s
fn overlap<'a>(s: &str, t: &'a str) -> &'a str; // output can't outlive t
fn overlap<'a>(s: &'a str, t: &'a str) -> &'a str; // output can't outlive s & t
fn overlap(s: &str, t: &str) -> &'static str; // output can outlive s & t
fn overlap<'a>(s: &str, t: &str) -> &'a str; // no relationship between input & output lifetimes

// expanded
fn overlap<'a, 'b>(s: &'a str, t: &'b str) -> &'a str;
fn overlap<'a, 'b>(s: &'a str, t: &'b str) -> &'b str;
fn overlap<'a>(s: &'a str, t: &'a str) -> &'a str;
fn overlap<'a, 'b>(s: &'a str, t: &'b str) -> &'static str;
fn overlap<'a, 'b, 'c>(s: &'a str, t: &'b str) -> &'c str;

// elided
fn compare(&self, s: &str) -> &str;

// expanded
fn compare<'a, 'b>(&'a self, &'b str) -> &'a str;

If you've ever written

a struct method
a function which takes references
a function which returns references
a generic function
a trait object (more on this later)
a closure (more on this later)

then your code has generic elided lifetime annotations all over it.

Key Takeaways

almost all Rust code is generic code and there's elided lifetime annotations everywhere

5) if it compiles then my lifetime annotations are correct

Misconception Corollaries

Rust's lifetime elision rules for functions are always right
Rust's borrow checker is always right, technically and semantically
Rust knows more about the semantics of my program than I do

It's possible for a Rust program to be technically compilable but still semantically wrong. Take this for example:

struct ByteIter<'a> {
    remainder: &'a [u8]
}

impl<'a> ByteIter<'a> {
    fn next(&mut self) -> Option<&u8> {
        if self.remainder.is_empty() {
            None
        } else {
            let byte = &self.remainder[0];
            self.remainder = &self.remainder[1..];
            Some(byte)
        }
    }
}

fn main() {
    let mut bytes = ByteIter { remainder: b"1" };
    assert_eq!(Some(&b'1'), bytes.next());
    assert_eq!(None, bytes.next());
}

ByteIter is an iterator that iterates over a slice of bytes. We're skipping the Iterator trait implementation for conciseness. It seems to work fine, but what if we want to check a couple bytes at a time?

fn main() {
    let mut bytes = ByteIter { remainder: b"1123" };
    let byte_1 = bytes.next();
    let byte_2 = bytes.next();
    if byte_1 == byte_2 {
        // do something
    }
}

Uh oh! Compile error:

error[E0499]: cannot borrow `bytes` as mutable more than once at a time
  --> src/main.rs:20:18
   |
19 |     let byte_1 = bytes.next();
   |                  ----- first mutable borrow occurs here
20 |     let byte_2 = bytes.next();
   |                  ^^^^^ second mutable borrow occurs here
21 |     if byte_1 == byte_2 {
   |        ------ first borrow later used here

I guess we can copy each byte. Copying is okay when we're working with bytes but if we turned ByteIter into a generic slice iterator that can iterate over any &'a [T] then we might want to use it in the future with types that may be very expensive or impossible copy / clone. Oh well, I guess there's nothing we can do about that, the code compiles so the lifetime annotations must be right, right?

Nope, the current lifetime annotations are actually the source of the bug! It's particularly hard to spot because the buggy lifetime annotations are elided. Lets expand the elided lifetimes to get a clearer look at the problem:

struct ByteIter<'a> {
    remainder: &'a [u8]
}

impl<'a> ByteIter<'a> {
    fn next<'b>(&'b mut self) -> Option<&'b u8> {
        if self.remainder.is_empty() {
            None
        } else {
            let byte = &self.remainder[0];
            self.remainder = &self.remainder[1..];
            Some(byte)
        }
    }
}

That didn't help at all. I'm still confused. Here's a hot tip that only Rust pros know: give your lifetime annotations descriptive names. Lets try again:

struct ByteIter<'remainder> {
    remainder: &'remainder [u8]
}

impl<'remainder> ByteIter<'remainder> {
    fn next<'mut_self>(&'mut_self mut self) -> Option<&'mut_self u8> {
        if self.remainder.is_empty() {
            None
        } else {
            let byte = &self.remainder[0];
            self.remainder = &self.remainder[1..];
            Some(byte)
        }
    }
}

Each returned byte is annotated with 'mut_self but the bytes are clearly coming from 'remainder ! Lets fix it.

struct ByteIter<'remainder> {
    remainder: &'remainder [u8]
}

impl<'remainder> ByteIter<'remainder> {
    fn next(&mut self) -> Option<&'remainder u8> {
        if self.remainder.is_empty() {
            None
        } else {
            let byte = &self.remainder[0];
            self.remainder = &self.remainder[1..];
            Some(byte)
        }
    }
}

fn main() {
    let mut bytes = ByteIter { remainder: b"1123" };
    let byte_1 = bytes.next();
    let byte_2 = bytes.next();
    std::mem::drop(bytes); // we can even drop the iterator now!
    if byte_1 == byte_2 { // compiles
        // do something
    }
}

Now that we look back on the previous version of our program it was obviously wrong, so why did Rust compile it? The answer is simple: it was memory safe.

The Rust borrow checker only cares about the lifetime annotations in a program to the extent it can use them to statically verify the memory safety of the program. Rust will happily compile programs even if the lifetime annotations have semantic errors, and the consequence of this is that the program becomes unnecessarily restrictive.

Here's a quick example that's the opposite of the previous example: Rust's lifetime elision rules happen to be semantically correct in this instance but we unintentionally write a very restrictive method with our own unnecessary explicit lifetime annotations.

#[derive(Debug)]
struct NumRef<'a>(&'a i32);

impl<'a> NumRef<'a> {
    // my struct is generic over 'a so that means I need to annotate
    // my self parameters with 'a too, right? (answer: no, not right)
    fn some_method(&'a mut self) {}
}

fn main() {
    let mut num_ref = NumRef(&5);
    num_ref.some_method(); // mutably borrows num_ref for the rest of its lifetime
    num_ref.some_method(); // compile error
    println!("{:?}", num_ref); // also compile error
}

If we have some struct generic over 'a we almost never want to write a method with a &'a mut self receiver. What we're communicating to Rust is "this method will mutably borrow the struct for the entirety of the struct's lifetime". In practice this means Rust's borrow checker will only allow at most one call to some_method before the struct becomes permanently mutably borrowed and thus unusable. The use-cases for this are extremely rare but the code above is very easy for confused beginners to write and it compiles. The fix is to not add unnecessary explicit lifetime annotations and let Rust's lifetime elision rules handle it:

#[derive(Debug)]
struct NumRef<'a>(&'a i32);

impl<'a> NumRef<'a> {
    // no more 'a on mut self
    fn some_method(&mut self) {}

    // above line desugars to
    fn some_method_desugared<'b>(&'b mut self){}
}

fn main() {
    let mut num_ref = NumRef(&5);
    num_ref.some_method();
    num_ref.some_method(); // compiles
    println!("{:?}", num_ref); // compiles
}

Key Takeaways

Rust's lifetime elision rules for functions are not always right for every situation
Rust does not know more about the semantics of your program than you do
give your lifetime annotations descriptive names
try to be mindful of where you place explicit lifetime annotations and why

6) boxed trait objects don't have lifetimes

Earlier we discussed Rust's lifetime elision rules for functions . Rust also has lifetime elision rules for trait objects, which are:

if a trait object is used as a type argument to a generic type then its life bound is inferred from the containing type
- if there's a unique bound from the containing then that's used
- if there's more than one bound from the containing type then an explicit bound must be specified
if the above doesn't apply then
- if the trait is defined with a single lifetime bound then that bound is used
- if 'static is used for any lifetime bound then 'static is used
- if the trait has no lifetime bounds then its lifetime is inferred in expressions and is 'static outside of expressions

All of that sounds super complicated but can be simply summarized as "a trait object's lifetime bound is inferred from context." After looking at a handful of examples we'll see the lifetime bound inferences are pretty intuitive so we don't have to memorize the formal rules:

use std::cell::Ref;

trait Trait {}

// elided
type T1 = Box<dyn Trait>;
// expanded, Box<T> has no lifetime bound on T, so inferred as 'static
type T2 = Box<dyn Trait + 'static>;

// elided
impl dyn Trait {}
// expanded
impl dyn Trait + 'static {}

// elided
type T3<'a> = &'a dyn Trait;
// expanded, &'a T requires T: 'a, so inferred as 'a
type T4<'a> = &'a (dyn Trait + 'a);

// elided
type T5<'a> = Ref<'a, dyn Trait>;
// expanded, Ref<'a, T> requires T: 'a, so inferred as 'a
type T6<'a> = Ref<'a, dyn Trait + 'a>;

trait GenericTrait<'a>: 'a {}

// elided
type T7<'a> = Box<dyn GenericTrait<'a>>;
// expanded
type T8<'a> = Box<dyn GenericTrait<'a> + 'a>;

// elided
impl<'a> dyn GenericTrait<'a> {}
// expanded
impl<'a> dyn GenericTrait<'a> + 'a {}

Concrete types which implement traits can have references and thus they also have lifetime bounds, and so their corresponding trait objects have lifetime bounds. Also you can implement traits directly for references which obviously have lifetime bounds:

trait Trait {}

struct Struct {}
struct Ref<'a, T>(&'a T);

impl Trait for Struct {}
impl Trait for &Struct {} // impl Trait directly on a ref type
impl<'a, T> Trait for Ref<'a, T> {} // impl Trait on a type containing refs

Anyway, this is worth going over because it often confuses beginners when they refactor a function from using trait objects to generics or vice versa. Take this program for example:

use std::fmt::Display;

fn dynamic_thread_print(t: Box<dyn Display + Send>) {
    std::thread::spawn(move || {
        println!("{}", t);
    }).join();
}

fn static_thread_print<T: Display + Send>(t: T) {
    std::thread::spawn(move || {
        println!("{}", t);
    }).join();
}

It throws these compile errors:

error[E0310]: the parameter type `T` may not live long enough
  --> src/lib.rs:10:5
   |
9  | fn static_thread_print<T: Display + Send>(t: T) {
   |                        -- help: consider adding an explicit lifetime bound...: `T: 'static +`
10 |     std::thread::spawn(move || {
   |     ^^^^^^^^^^^^^^^^^^
   |
note: ...so that the type `[closure@src/lib.rs:10:24: 12:6 t:T]` will meet its required lifetime bounds
  --> src/lib.rs:10:5
   |
10 |     std::thread::spawn(move || {
   |     ^^^^^^^^^^^^^^^^^^

Okay great, the compiler tells us how to fix the issue so lets fix the issue.

use std::fmt::Display;

fn dynamic_thread_print(t: Box<dyn Display + Send>) {
    std::thread::spawn(move || {
        println!("{}", t);
    }).join();
}

fn static_thread_print<T: Display + Send + 'static>(t: T) {
    std::thread::spawn(move || {
        println!("{}", t);
    }).join();
}

It compiles now but these two functions look awkward next to each other, why does the second function require a 'static bound on T where the first function doesn't? That's a trick question. Using the lifetime elision rules Rust automatically infers a 'static bound in the first function so both actually have 'static bounds. This is what the Rust compiler sees:

use std::fmt::Display;

fn dynamic_thread_print(t: Box<dyn Display + Send + 'static>) {
    std::thread::spawn(move || {
        println!("{}", t);
    }).join();
}

fn static_thread_print<T: Display + Send + 'static>(t: T) {
    std::thread::spawn(move || {
        println!("{}", t);
    }).join();
}

Key Takeaways

all trait objects have some inferred default lifetime bounds

7) compiler error messages will tell me how to fix my program

Misconception Corollaries

Rust's lifetime elision rules for trait objects are always right
Rust knows more about the semantics of my program than I do

This misconception is the previous 2 misconceptions combined into one example:

use std::fmt::Display;

fn box_displayable<T: Display>(t: T) -> Box<dyn Display> {
    Box::new(t)
}

Throws this error:

error[E0310]: the parameter type `T` may not live long enough
 --> src/lib.rs:4:5
  |
3 | fn box_displayable<T: Display>(t: T) -> Box<dyn Display> {
  |                    -- help: consider adding an explicit lifetime bound...: `T: 'static +`
4 |     Box::new(t)
  |     ^^^^^^^^^^^
  |
note: ...so that the type `T` will meet its required lifetime bounds
 --> src/lib.rs:4:5
  |
4 |     Box::new(t)
  |     ^^^^^^^^^^^

Okay, let's fix it how the compiler is telling us to fix it, nevermind the fact that it's automatically inferring a 'static lifetime bound for our boxed trait object without telling us and its recommended fix is based on that unstated fact:

use std::fmt::Display;

fn box_displayable<T: Display + 'static>(t: T) -> Box<dyn Display> {
    Box::new(t)
}

So the program compiles now... but is this what we actually want? Probably, but maybe not. The compiler didn't mention any other fixes but this would have also been appropriate:

use std::fmt::Display;

fn box_displayable<'a, T: Display + 'a>(t: T) -> Box<dyn Display + 'a> {
    Box::new(t)
}

This function accepts all the same arguments as the previous version plus a lot more! Does that make it better? Not necessarily, it depends on the requirements and constraints of our program. This example is a bit abstract so lets take a look at a simpler and more obvious case:

fn return_first(a: &str, b: &str) -> &str {
    a
}

Throws:

error[E0106]: missing lifetime specifier
 --> src/lib.rs:1:38
  |
1 | fn return_first(a: &str, b: &str) -> &str {
  |                    ----     ----     ^ expected named lifetime parameter
  |
  = help: this function's return type contains a borrowed value, but the signature does not say whether it is borrowed from `a` or `b`
help: consider introducing a named lifetime parameter
  |
1 | fn return_first<'a>(a: &'a str, b: &'a str) -> &'a str {
  |                ^^^^    ^^^^^^^     ^^^^^^^     ^^^

The error message recommends annotating both inputs and the output with the same lifetime. If we did this our program would compile but this function would overly-constrain the return type. What we actually want is this:

fn return_first<'a>(a: &'a str, b: &str) -> &'a str {
    a
}

Key Takeaways

Rust's lifetime elision rules for trait objects are not always right for every situation
Rust does not know more about the semantics of your program than you do
Rust compiler error messages suggest fixes which will make your program compile which is not that same as fixes which will make you program compile and best suit the requirements of your program

8) lifetimes can grow and shrink at run-time

Misconception Corollaries

container types can swap references at run-time to change their lifetime
Rust borrow checker does advanced control flow analysis

This does not compile:

struct Has<'lifetime> {
    lifetime: &'lifetime str,
}

fn main() {
    let long = String::from("long");
    let mut has = Has { lifetime: &long };
    assert_eq!(has.lifetime, "long");

    {
        let short = String::from("short");
        // "switch" to short lifetime
        has.lifetime = &short;
        assert_eq!(has.lifetime, "short");

        // "switch back" to long lifetime (but not really)
        has.lifetime = &long;
        assert_eq!(has.lifetime, "long");
        // `short` dropped here
    }

    // compile error, `short` still "borrowed" after drop
    assert_eq!(has.lifetime, "long");
}

It throws:

error[E0597]: `short` does not live long enough
  --> src/main.rs:11:24
   |
11 |         has.lifetime = &short;
   |                        ^^^^^^ borrowed value does not live long enough
...
15 |     }
   |     - `short` dropped here while still borrowed
16 |     assert_eq!(has.lifetime, "long");
   |     --------------------------------- borrow later used here

This also does not compile, throws the exact same error as above:

struct Has<'lifetime> {
    lifetime: &'lifetime str,
}

fn main() {
    let long = String::from("long");
    let mut has = Has { lifetime: &long };
    assert_eq!(has.lifetime, "long");

    // this block will never run
    if false {
        let short = String::from("short");
        // "switch" to short lifetime
        has.lifetime = &short;
        assert_eq!(has.lifetime, "short");

        // "switch back" to long lifetime (but not really)
        has.lifetime = &long;
        assert_eq!(has.lifetime, "long");
        // `short` dropped here
    }

    // still a compile error, `short` still "borrowed" after drop
    assert_eq!(has.lifetime, "long");
}

Lifetimes have to be statically verified at compile-time and the Rust borrow checker only does very basic control flow analysis, so it assumes every block in an if-else statement and every match arm in a match statement can be taken and then chooses the shortest possible lifetime for the variable. Once a variable is bounded by a lifetime it is bounded by that lifetime forever . The lifetime of a variable can only shrink, and all the shrinkage is determined at compile-time.

Key Takeaways

lifetimes are statically verified at compile-time
lifetimes cannot grow or shrink or change in any way at run-time
Rust borrow checker will always choose the shortest possible lifetime for a variable assuming all code paths can be taken

9) downgrading mut refs to shared refs is safe

Misconception Corollaries

re-borrowing a reference ends its lifetime and starts a new one

You can pass a mut ref to a function expecting a shared ref because Rust will implicitly re-borrow the mut ref as immutable:

fn takes_shared_ref(n: &i32) {}

fn main() {
    let mut a = 10;
    takes_shared_ref(&mut a); // compiles
    takes_shared_ref(&*(&mut a)); // above line desugared
}

Intuitively this makes sense, since there's no harm in re-borrowing a mut ref as immutable, right? Surprisingly no, as the program below does not compile:

fn main() {
    let mut a = 10;
    let b: &i32 = &*(&mut a); // re-borrowed as immutable
    let c: &i32 = &a;
    dbg!(b, c); // compile error
}

Throws this error:

error[E0502]: cannot borrow `a` as immutable because it is also borrowed as mutable
 --> src/main.rs:4:19
  |
3 |     let b: &i32 = &*(&mut a);
  |                     -------- mutable borrow occurs here
4 |     let c: &i32 = &a;
  |                   ^^ immutable borrow occurs here
5 |     dbg!(b, c);
  |          - mutable borrow later used here

A mutable borrow does occur, but it's immediately and unconditionally re-borrowed as immutable and then dropped. Why is Rust treating the immutable re-borrow as if it still has the mut ref's exclusive lifetime? While there's no issue in the particular example above, allowing the ability to downgrade mut refs to shared refs does indeed introduce potential memory safety issues:

use std::sync::Mutex;

struct Struct {
    mutex: Mutex<String>
}

impl Struct {
    // downgrades mut self to shared str
    fn get_string(&mut self) -> &str {
        self.mutex.get_mut().unwrap()
    }
    fn mutate_string(&self) {
        // if Rust allowed downgrading mut refs to shared refs
        // then the following line would invalidate any shared
        // refs returned from the get_string method
        *self.mutex.lock().unwrap() = "surprise!".to_owned();
    }
}

fn main() {
    let mut s = Struct {
        mutex: Mutex::new("string".to_owned())
    };
    let str_ref = s.get_string(); // mut ref downgraded to shared ref
    s.mutate_string(); // str_ref invalidated, now a dangling pointer
    dbg!(str_ref); // compile error as expected
}

The point here is that when you re-borrow a mut ref as a shared ref you don't get that shared ref without a big gotcha: it extends the mut ref's lifetime for the duration of the re-borrow even if the mut ref itself is dropped. Using the re-borrowed shared ref is very difficult because it's immutable but it can't overlap with any other shared refs. The re-borrowed shared ref has all the cons of a mut ref and all the cons of a shared ref and has the pros of neither. I believe re-borrowing a mut ref as a shared ref should be considered a Rust anti-pattern. Being aware of this anti-pattern is important so that you can easily spot it when you see code like this:

// downgrades mut T to shared T
fn some_function<T>(some_arg: &mut T) -> &T;

struct Struct;

impl Struct {
    // downgrades mut self to shared self
    fn some_method(&mut self) -> &self;

    // downgrades mut self to shared T
    fn other_method(&mut self) -> &T;
}

Even if you avoid re-borrows in function and method signatures Rust still does automatic implicit re-borrows so it's easy to bump into this problem without realizing it like so:

use std::collections::HashMap;

type PlayerID = i32;

#[derive(Debug, Default)]
struct Player {
    score: i32,
}

fn start_game(player_a: PlayerID, player_b: PlayerID, server: &mut HashMap<PlayerID, Player>) {
    // get players from server or create & insert new players if they don't yet exist
    let player_a: &Player = server.entry(player_a).or_default();
    let player_b: &Player = server.entry(player_b).or_default();

    // do something with players
    dbg!(player_a, player_b); // compile error
}

The above fails to compile. or_default() returns a &mut Player which we're implicitly re-borrowing as &Player because of our explicit type annotations. To do what we want we have to:

use std::collections::HashMap;

type PlayerID = i32;

#[derive(Debug, Default)]
struct Player {
    score: i32,
}

fn start_game(player_a: PlayerID, player_b: PlayerID, server: &mut HashMap<PlayerID, Player>) {
    // drop the returned mut Player refs since we can't use them together anyway
    server.entry(player_a).or_default();
    server.entry(player_b).or_default();

    // fetch the players again, getting them immutably this time, without any implicit re-borrows
    let player_a = server.get(&player_a);
    let player_b = server.get(&player_b);

    // do something with players
    dbg!(player_a, player_b); // compiles
}

Kinda awkward and clunky but this is the sacrifice we make at the Altar of Memory Safety.

Key Takeaways

try not to re-borrow mut refs as shared refs, or you're gonna have a bad time
re-borrowing a mut ref doesn't end its lifetime, even if the ref is dropped

10) closures follow the same lifetime elision rules as functions

This is more of a Rust Gotcha than a misconception.

Closures, despite being functions, do not follow the same lifetime elision rules as functions.

fn function(x: &i32) -> &i32 {
    x
}

fn main() {
    let closure = |x: &i32| x;
}

Throws:

error: lifetime may not live long enough
 --> src/main.rs:6:29
  |
6 |     let closure = |x: &i32| x;
  |                       -   - ^ returning this value requires that `'1` must outlive `'2`
  |                       |   |
  |                       |   return type of closure is &'2 i32
  |                       let's call the lifetime of this reference `'1`

After desugaring we get:

// input lifetime gets applied to output
fn function<'a>(x: &'a i32) -> &'a i32 {
    x
}

fn main() {
    // input and output each get their own distinct lifetimes
    let closure = for<'a, 'b> |x: &'a i32| -> &'b i32 { x };
    // note: the above line is not valid syntax, but we need it for illustrative purposes
}

There's no good reason for this discrepancy. Closures were first implemented with different type inference semantics than functions and now we're stuck with it forever because to unify them at this point would be a breaking change. So how can we explicitly annotate a closure's type? Our options include:

fn main() {
    // cast to trait object, becomes unsized, oops, compile error
    let identity: dyn Fn(&i32) -> &i32 = |x: &i32| x;

    // can allocate it on the heap as a workaround but feels clunky
    let identity: Box<dyn Fn(&i32) -> &i32> = Box::new(|x: &i32| x);

    // can skip the allocation and just create a static reference
    let identity: &dyn Fn(&i32) -> &i32 = &|x: &i32| x;

    // previous line desugared :)
    let identity: &'static (dyn for<'a> Fn(&'a i32) -> &'a i32 + 'static) = &|x: &i32| -> &i32 { x };

    // this would be ideal but it's invalid syntax
    let identity: impl Fn(&i32) -> &i32 = |x: &i32| x;

    // this would also be nice but it's also invalid syntax
    let identity = for<'a> |x: &'a i32| -> &'a i32 { x };

    // since "impl trait" works in the function return position
    fn return_identity() -> impl Fn(&i32) -> &i32 {
        |x| x
    }
    let identity = return_identity();

    // more generic version of the previous solution
    fn annotate<T, F>(f: F) -> F where F: Fn(&T) -> &T {
        f
    }
    let identity = annotate(|x: &i32| x);
}

As I'm sure you've already noticed from the examples above, when closure types are used as trait bounds they do follow the usual function lifetime elision rules.

There's no real lesson or insight to be had here, it just is what it is.

Key Takeaways

every language has gotchas

Conclusion

T is a superset of both &T and &mut T
&T and &mut T are disjoint sets
T: 'static should be read as " T is bounded by a 'static lifetime"
if T: 'static then T can be a borrowed type with a 'static lifetime or an owned type
since T: 'static includes owned types that means T
- can be dynamically allocated at run-time
- does not have to be valid for the entire program
- can be safely and freely mutated
- can be dynamically dropped at run-time
- can have lifetimes of different durations
T: 'a is more general and more flexible than &'a T
T: 'a accepts owned types, owned types which contain references, and references
&'a T only accepts references
if T: 'static then T: 'a since 'static >= 'a for all 'a
almost all Rust code is generic code and there's elided lifetime annotations everywhere
Rust's lifetime elision rules are not always right for every situation
Rust does not know more about the semantics of your program than you do
give your lifetime annotations descriptive names
try to be mindful of where you place explicit lifetime annotations and why
all trait objects have some inferred default lifetime bounds
Rust compiler error messages suggest fixes which will make your program compile which is not that same as fixes which will make you program compile and best suit the requirements of your program
lifetimes are statically verified at compile-time
lifetimes cannot grow or shrink or change in any way at run-time
Rust borrow checker will always choose the shortest possible lifetime for a variable assuming all code paths can be taken
try not to re-borrow mut refs as shared refs, or you're gonna have a bad time
re-borrowing a mut ref doesn't end its lifetime, even if the ref is dropped
every language has gotchas

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Common Rust Lifetime Misconceptions