Returning a string from a method in Rust

Question

Editor's note: The syntax in this question predates Rust 1.0 and the 1.0-updated syntax generates different errors, but the overall concepts are still the same in Rust 1.0.

I have a struct T with a name field, I'd like to return that string from the name function. I don't want to copy the whole string, just the pointer:

struct T {
     name: ~str,
}

impl Node for T {
     fn name(&self) -> &str { self.name }
     // this doesn't work either (lifetime error)
     // fn name(&self) -> &str { let s: &str = self.name; s }
}

trait Node {
     fn name(&self) -> &str;
}

Why is this wrong? What is the best way to return an &str from a function?

Shepmaster · Accepted Answer · 2017-12-18 14:30:15Z

5

You have to take a reference to self.name and ensure that the lifetimes of &self and &str are the same:

struct T {
    name: String,
}

impl Node for T {
    fn name<'a>(&'a self) -> &'a str {
        &self.name
    }
}

trait Node {
    fn name(&self) -> &str;
}

You can also use lifetime elision, as shown in the trait definition. In this case, it automatically ties together the lifetimes of self and the returned &str.

Take a look at these book chapters:

edited Dec 18, 2017 at 14:30

Shepmaster

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answered Apr 11, 2014 at 21:28

Arjan

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3 Comments

cloudhead Over a year ago

So I got it working with fn name<'a>(&'a self) -> &'a str { let s: &'a str = self.name; s } is there any way for it work without the let binding?

Arjan Over a year ago

I had in my memory that ~ to & conversioon worked, maybe was changed (or was always just for fuction parameters?), you have to just call as_slice

Vladimir Matveev Over a year ago

@cloudhead, for strings or owned vectors (~[T]) you have to call as_slice(). For other boxed types &*value should suffice. This will probably change when dynamically sized types land.

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