I want to write a program that will write a file in 2 steps. It is likely that the file may not exist before the program is run. The filename is fixed.

The problem is that OpenOptions.new().write() can fail. In that case, I want to call a custom function trycreate(). The idea is to create the file instead of opening it and return a handle. Since the filename is fixed, trycreate() has no arguments and I cannot set a lifetime of the returned value.

How can I resolve this problem?

use std::io::Write;
use std::fs::OpenOptions;
use std::path::Path;

fn trycreate() -> &OpenOptions {
    let f = OpenOptions::new().write(true).open("foo.txt");
    let mut f = match f {
        Ok(file)  => file,
        Err(_)  => panic!("ERR"),

fn main() {
        let f = OpenOptions::new().write(true).open(b"foo.txt");
        let mut f = match f {
            Ok(file)  => file,
            Err(_)  => trycreate("foo.txt"),
        let buf = b"test1\n";
        let _ret = f.write(buf).unwrap();
        let f = OpenOptions::new().append(true).open("foo.txt");
        let mut f = match f {
            Ok(file)  => file,
            Err(_)  => panic!("append"),
        let buf = b"test2\n";
        let _ret = f.write(buf).unwrap();

Solution 1

The question you asked

TL;DR: No, you cannot return a reference to a variable that is owned by a function. This applies if you created the variable or if you took ownership of the variable as a function argument.


Instead of trying to return a reference, return an owned object. String instead of &str, Vec<T> instead of &[T], T instead of &T, etc.

If you took ownership of the variable via an argument, try taking a (mutable) reference instead and then returning a reference of the same lifetime.

In rare cases, you can use unsafe code to return the owned value and a reference to it. This has a number of delicate requirements you must uphold to ensure you don't cause undefined behavior or memory unsafety.

See also:

Deeper answer

fjh is absolutely correct, but I want to comment a bit more deeply and touch on some of the other errors with your code.

Let's start with a smaller example of returning a reference and look at the errors:

fn try_create<'a>() -> &'a String {

Rust 2015

error[E0597]: borrowed value does not live long enough
 --> src/lib.rs:2:6
2 |     &String::new()
  |      ^^^^^^^^^^^^^ temporary value does not live long enough
3 | }
  | - temporary value only lives until here
note: borrowed value must be valid for the lifetime 'a as defined on the function body at 1:15...
 --> src/lib.rs:1:15
1 | fn try_create<'a>() -> &'a String {
  |               ^^

Rust 2018

error[E0515]: cannot return reference to temporary value
 --> src/lib.rs:2:5
2 |     &String::new()
  |     ^-------------
  |     ||
  |     |temporary value created here
  |     returns a reference to data owned by the current function

Is there any way to return a reference from a function without arguments?

Technically "yes", but for what you want, "no".

A reference points to an existing piece of memory. In a function with no arguments, the only things that could be referenced are global constants (which have the lifetime &'static) and local variables. I'll ignore globals for now.

In a language like C or C++, you could actually take a reference to a local variable and return it. However, as soon as the function returns, there's no guarantee that the memory that you are referencing continues to be what you thought it was. It might stay what you expect for a while, but eventually the memory will get reused for something else. As soon as your code looks at the memory and tries to interpret a username as the amount of money left in the user's bank account, problems will arise!

This is what Rust's lifetimes prevent - you aren't allowed to use a reference beyond how long the referred-to value is valid at its current memory location.

See also:

Your actual problem

Look at the documentation for OpenOptions::open:

fn open<P: AsRef<Path>>(&self, path: P) -> Result<File>

It returns a Result<File>, so I don't know how you'd expect to return an OpenOptions or a reference to one. Your function would work if you rewrote it as:

fn trycreate() -> File {
        .expect("Couldn't open")

This uses Result::expect to panic with a useful error message. Of course, panicking in the guts of your program isn't super useful, so it's recommended to propagate your errors back out:

fn trycreate() -> io::Result<File> {

Option and Result have lots of nice methods to deal with chained error logic. Here, you can use or_else:

let f = OpenOptions::new().write(true).open("foo.txt");
let mut f = f.or_else(|_| trycreate()).expect("failed at creating");

I'd also return the Result from main. All together, including fjh's suggestions:

use std::{
    io::{self, Write},

fn main() -> io::Result<()> {
    let mut f = OpenOptions::new()



Solution 2

Is there any way to return a reference from a function without arguments?

No (except references to static values, but those aren't helpful here).

However, you might want to look at OpenOptions::create. If you change your first line in main to

let  f = OpenOptions::new().write(true).create(true).open(b"foo.txt");

the file will be created if it does not yet exist, which should solve your original problem.

Solution 3

You can not return a reference pointing to a local variable. You have two alternatives, either return the value or use a static variable.

Here is why:

References are pointers to memory locations. Once functions are executed, local variables are popped off the execution stack and resources are de-allocated. After that point, any reference to a local variable will be pointing to some useless data. Since it is de-allocated, it is not in our program's possession any more and OS may have already given it to another process and our data may have been overwritten.

For the following example, x is created when the function runs and dropped off when the function completes executing. It is local to the function and lives on this particular function's stack. Function's stack holds local variables.

When run is pop off the execution stack, any reference to x, &x, will be pointing to some garbage data. That is what people call a dangling pointer. The Rust compiler does not allow to use dangling pointers since it is not safe.

fn run() -> &u32 {
    let x: u32 = 42;

    return &x;
} // x is dropped here

fn main() {
    let x = run();

So, that is why we can not return a reference to a local variable. We have two options: either return the value or use a static variable.

Returning the value is the best option here. By returning the value, you will be passing the result of your calculation to the caller, in Rust's terms x will be owned by the caller. In our case it is main. So, no problem there.

Since a static variable lives as long as the process runs, its references will be pointing to the same memory location both inside and outside the function. No problem there either.

Note: @navigaid advises using a box, but it does not make sense because you are moving readily available data to heap by boxing it and then returning it. It does not solve the problem, you are still returning the local variable to the caller but using a pointer when accessing it. It adds an unnecessary indirection due to de-referencing hence incurring additional cost. Basically you will be using & for the sake of using it, nothing more.

Solution 4

This is an elaboration on snnsnn's answer, which briefly explained the problem without being too specific.

Rust doesn't allow return a reference to a variable created in a function. Is there a workaround? Yes, simply put that variable in a Box then return it. Example:

fn run() -> Box<u32> {
    let x: u32 = 42;
    return Box::new(x);

fn main() {
    println!("{}", run());

code in rust playground

As a rule of thumb, to avoid similar problems in Rust, return an owned object (Box, Vec, String, ...) instead of reference to a variable:

  • Box<T> instead of &T
  • Vec<T> instead of &[T]
  • String instead of &str

For other types, refer to The Periodic Table of Rust Types to figure out which owned object to use.

Of course, in this example you can simply return the value (T instead of &T or Box<T>)

fn run() -> u32 {
    let x: u32 = 42;
    return x;

Solution 5

Yes! But you have to find a way to extend the lifetime. One way to do is to provide mutable reference to a dummy/default value (&mut T) to the function and then fill/replace the value in the function and then return a reference to that value (&T). This way you can specify the lifetime so that the returned reference get the lifetime of a value outside the function.


//&mut T -> &T
fn example2<'a>(life: &'a mut Vec<i32>) -> &'a Vec<i32> {
    *life = vec![1, 2, 3, 4];

fn test_example2() {
    //Could also use Vec::new()
    let mut life = Vec::default();
    let res = example2(&mut life);
    println!("{:?}", res)

fn test2_example2() {
    let life = &mut Vec::default();
    let res = example2(life);
    println!("{:?}", res)

//shows real use case
fn get_check_test_slices2<'a>(
    lifetime: &'a mut Vec<usize>,
    limit: usize,
) -> impl Iterator<Item = (&'a [usize], &'a [usize])> + 'a {
    *lifetime = primes1_iter_bitvec(limit).collect::<Vec<_>>();

Edit: How is that different from just giving &mut T to the function? (asked by Chayim Friedman (see old solution below)): It's basically the same... I lost against the borrow checker previously, and that is why I didn't just use &mut T. But after renewed battles I have finally managed to just use &mut T. Thank you for the insightful question.

Old solution: My solution works by creating a default value before calling the function, which the function later replaces/fills and returns a reference to.

/// Used to return references to values created in a function.
/// fn example<'a>(lt:& 'a mut LifeExtender<Vec<i32>>) -> &'a Vec<i32> {
///     lt.set(vec![1,2,3,4]);
///     lt.get()
/// }
pub struct LifeExtender<T> {
    value: T,

impl<T> Default for LifeExtender<T>
    T: Default,
    /// using T default.
    pub fn default() -> Self {
        Self {
            value: T::default(),

impl<T> LifeExtender<T> {
    /// If T doesn't have default.
    pub fn new(value: T) -> Self {
        Self { value }
    /// set value to be returned by reference
    pub fn set(&mut self, new_value: T) {
        self.value = new_value;
    /// Get a reference with lifetime self.
    pub fn get<'a>(&'a self) -> &'a T {
    /// Get a mut reference with lifetime self.
    pub fn get_mut<'a>(&'a mut self) -> &'a mut T {
        &mut self.value

fn example<'a>(life: &'a mut LifeExtender<Vec<i32>>) -> &'a Vec<i32> {
    let local_value = vec![1, 2, 3, 4];

//prints: [1,2,3,4]
pub fn test_example() {
    let mut life = LifeExtender::default();
    let res = example(&mut life);
    println!("{:?}", res);

//Real example code snippet, where I used this solution: 
fn get_check_slices2<'a>(
    lifetime: &'a mut LifeExtender<Vec<usize>>,
    limit: usize,
) -> impl Iterator<Item = (&'a [usize], &'a [usize])> + 'a {