Return local String as a slice (&str)

ghz 21days ago ⋅ 10 views

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[..]
}

Answer

You're encountering a lifetime issue here because string is a local variable inside the function return_str. When you return a reference to string (i.e., &string[..]), you are trying to return a reference to data that will no longer exist once the function ends. Specifically, string is dropped when the function scope ends, which means that the reference you return is pointing to invalid memory.

Problem Breakdown:

In the function return_str, you are creating a String, modifying it, and then trying to return a slice (&str) that points to the contents of the String. However, the String itself is local to the function, so once the function scope ends, the String is dropped, leaving the returned reference (&str) pointing to invalid memory. The lifetime of the reference needs to be valid for as long as the caller needs it, but in your case, the lifetime of string is limited to the function scope.

Solution:

The key point is that you cannot return a reference to data that is owned locally within the function. To fix this, you need to either:

  1. Return a String directly, because a String owns its data and lives as long as you need it.
  2. Ensure the data lives longer, for example, by having it be owned outside the function, or using static data if applicable.

Here are the possible solutions:

1. Return String instead of &str (Most common solution)

Since you're building the string dynamically and it's not going to be used anywhere else, the simplest solution is to return the String directly. A String owns its data and you don't have to worry about lifetimes.

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

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

    string // return the String
}

This solution works because the String will live as long as needed by the caller, and Rust's ownership system will handle the memory for you. You can then convert it to a &str at the point where you need it, by borrowing it temporarily.

2. Use a static or 'static lifetime (for certain use cases)

If you are building the string once and want it to live forever (i.e., for the entire duration of the program), you could use a static string. This would be useful for things like pre-built constants or configuration data that doesn't change.

For example, if you want a string to be built at runtime but still return a reference that lives forever (in a 'static context):

static mut MY_STATIC_STRING: Option<String> = None;

fn return_str() -> &'static str {
    unsafe {
        if MY_STATIC_STRING.is_none() {
            let mut string = String::new();
            for i in 0..10 {
                string.push_str("ACTG");
            }
            MY_STATIC_STRING = Some(string);
        }
        
        // Use the static string stored in MY_STATIC_STRING
        // Safety: We assume that MY_STATIC_STRING is initialized at this point.
        &MY_STATIC_STRING.as_ref().unwrap() // return a reference to the string
    }
}

Here, MY_STATIC_STRING is a static variable, and it lives for the duration of the program. The downside is that it's marked unsafe because mutable static variables need to be accessed in a safe way (which may involve synchronization or careful handling).

3. Use a Vec<u8> or similar for dynamic content

If you want to return a &str dynamically but without returning ownership of the entire String, you might consider building the string in a Vec<u8>, which is more efficient if you need to avoid unnecessary allocations.

However, this approach can be more complex than simply using a String, and in many cases, returning a String directly is preferable.

Conclusion:

The best and most idiomatic approach in Rust would be to return a String directly. This avoids lifetime issues and is the most flexible solution since you don't need to worry about manually managing lifetimes. If performance or other constraints require that you return a &str, consider using static storage, but for most cases, returning the String is the most straightforward and idiomatic solution.

#rust