Is there any way to return a reference to a variable created in a function?

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:

  • Proper way to return a new string in Rust
  • Return local String as a slice (&str)
  • Why can’t I store a value and a reference to that value in the same struct?

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/
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/
1 | fn try_create<'a>() -> &'a String {
  |               ^^

Rust 2018

error[E0515]: cannot return reference to temporary value
 --> src/
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:

  • Is it possible to return either a borrowed or owned type in Rust?
  • Why can I return a reference to a local literal but not a variable?

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()



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