Return local String as a slice (&str)

ghz 7months ago ⋅ 87 views

There are several questions that seem to be about the same problem I'm having. For example see here and here. Basically I'm trying to build a String in a local function, but then return it as a &str. Slicing isn't working because the lifetime is too short. I can't use str directly in the function because I need to build it dynamically. However, I'd also prefer not to return a String since the nature of the object this is going into is static once it's built. Is there a way to have my cake and eat it too?

Here's a minimal non-compiling reproduction:

fn return_str<'a>() -> &'a str {
    let mut string = "".to_string();

    for i in 0..10 {
        string.push_str("ACTG");
    }

    &string[..]
}

Answers

The problem you're encountering stems from Rust's strict lifetime rules, which prevent you from returning a reference to a value that will be dropped when the function exits. In your case, you're trying to return a &str that points to a String created within the function, but that String is deallocated when the function returns, making the reference invalid.

To achieve your goal, you generally have two main options:

  1. Return a String instead of a &str: This is the most straightforward and idiomatic way to handle dynamically built strings in Rust. Although you mentioned you'd prefer not to return a String, it's the safest and most flexible option.

  2. Use a smart pointer or container to manage lifetimes: If you really need to return a reference, you can use a smart pointer like Box<str> or Arc<str> to transfer ownership and manage the lifetime.

Here's how you can modify your function for both approaches:

1. Return a String

This is the simplest solution. You can convert the String to a &str when needed by the caller:

fn return_string() -> String {
    let mut string = String::new();

    for _ in 0..10 {
        string.push_str("ACTG");
    }

    string
}

fn main() {
    let s = return_string();
    println!("{}", s);
    // If needed as &str
    let s_ref: &str = &s;
    println!("{}", s_ref);
}

2. Use Box<str>

This approach involves converting the String to a Box<str> which allows you to manage the memory allocation while still providing a &str reference:

fn return_boxed_str() -> Box<str> {
    let mut string = String::new();

    for _ in 0..10 {
        string.push_str("ACTG");
    }

    string.into_boxed_str()
}

fn main() {
    let boxed_str = return_boxed_str();
    let s_ref: &str = &boxed_str;
    println!("{}", s_ref);
}

3. Use Arc<str>

If you need shared ownership with reference counting, you can use Arc<str>:

use std::sync::Arc;

fn return_arc_str() -> Arc<str> {
    let mut string = String::new();

    for _ in 0..10 {
        string.push_str("ACTG");
    }

    Arc::from(string)
}

fn main() {
    let arc_str = return_arc_str();
    let s_ref: &str = &arc_str;
    println!("{}", s_ref);
}

Explanation of Lifetime Rules

The core of the issue is Rust's ownership and borrowing rules. In your original function:

fn return_str<'a>() -> &'a str {
    let mut string = String::new();

    for i in 0..10 {
        string.push_str("ACTG");
    }

    &string[..]
}

The String string is created and destroyed within the function's scope. When you try to return a reference to its data, Rust prevents you because the reference would point to deallocated memory.

Using a smart pointer or returning the String itself ensures the data remains valid for as long as needed, adhering to Rust's safety guarantees.

Conclusion

While it might seem convenient to return a &str, Rust's design encourages returning String for dynamically constructed strings to ensure memory safety. If you need to work with &str, smart pointers like Box<str> or Arc<str> provide a robust alternative by managing the lifetime and ownership of the string data.

#rust