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How do I make format! return a &str from a conditional expression?

I happened upon this problem where format! creates a temporary value in a pattern that is not anchored to anything, as far as I understand it.

let x = 42;
let category = match x {
    0...9 => "Between 0 and 9",
    number @ 10 => format!("It's a {}!", number).as_str(),
    _ if x < 0 => "Negative",
    _ => "Something else",
};

println!("{}", category);

In this code, the type of category is a &str, which is satisfied by returning a literal like "Between 0 and 9". If I want to format the matched value to a slice using as_str(), then I get an error:

error[E0716]: temporary value dropped while borrowed
 --> src/main.rs:5:24
  |
3 |     let category = match x {
  |         -------- borrow later stored here
4 |         0...9 => "Between 0 and 9",
5 |         number @ 10 => format!("It's a {}!", number).as_str(),
  |                        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^        - temporary value is freed at the end of this statement
  |                        |
  |                        creates a temporary which is freed while still in use
  |
  = note: consider using a `let` binding to create a longer lived value
  = note: this error originates in a macro outside of the current crate (in Nightly builds, run with -Z external-macro-backtrace for more info)

I have done some reading, and found people with similar problems, but I couldn't seem to find any solution.

A simple workaround would be to have category be a String instead of a &str, but I don't like the idea of having to put .to_string() on the end of every literal in the pattern, as it's not as clean.

Is there a way to solve the problem, or do I just need to work around it?

like image 954
Bjarke Pedersen Avatar asked Jan 16 '19 18:01

Bjarke Pedersen


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2 Answers

This is 90% a duplicate of Return local String as a slice (&str), see that for multiple other solutions.

There's one extra possibility since this is all in one function: You can declare a variable for the String and only set it when you need to allocate. The compiler (obliquely) suggests this:

consider using a let binding to create a longer lived value

fn main() {
    let x = 42;
    let tmp;

    let category = match x {
        0...9 => "Between 0 and 9",
        number @ 10 => {
            tmp = format!("It's a {}!", number);
            &tmp
        }
        _ if x < 0 => "Negative",
        _ => "Something else",
    };

    println!("{}", category);
}

This is mostly the same as using a Cow, just handled by the compiler instead of a specific type.

See also:

  • How can I conditionally provide a default reference without performing unnecessary computation when it isn't used?
  • Are polymorphic variables allowed?
like image 129
Shepmaster Avatar answered Oct 26 '22 23:10

Shepmaster


format! can't return &str because it will always allocate String. What is possible to do is to return a &str from a String, which is what you did in your code.

As the compiler hinted, the created String is immediately dropped after its creation because it went out of the current scope and one way around could be an external variable that is not bounded to the match scope. E.g.:

use std::fmt::Write;

fn main() {
    let mut buffer = String::with_capacity(20);
    buffer.push_str("It's a ");

    let x = 10;
    let category = match x {
        0...9 => "Between 0 and 9",
        number @ 10 => {
            write!(&mut buffer, "{}", number).unwrap();
            buffer.as_str()
        }
        _ if x < 0 => "Negative",
        _ => "Something else",
    };

    println!("{}", category);
}

If you want an [no_std] environment or don't want to do any dynamic allocation, you can take a look at this limited code snippet:

use core::str;

fn each_digit<F>(mut number: u32, mut f: F)
where
    F: FnMut(u8),
{
    while number > 0 {
        f((number % 10) as u8);
        number /= 10;
    }
}

fn main() {
    const BUFFER_LEN: usize = 20;
    let mut buffer = [0u8; BUFFER_LEN];

    let x = 12344329;
    let category = match x {
        0...9 => "Between 0 and 9",
        number @ 123443219 => {
            let mut idx = BUFFER_LEN;
            each_digit(number, |digit| {
                let ascii = digit + 48;
                idx -= 1;
                buffer[idx] = ascii;
            });
            str::from_utf8(&buffer[idx..BUFFER_LEN]).unwrap()
        },
        _ => "Something else",
    };

    assert_eq!("123443219", category);
}
like image 23
Caio Avatar answered Oct 26 '22 23:10

Caio