In Rust you can easily take a reference of any value with the & operator. But something I’ve been having in the back of my head recently is why the following works:
struct FancyString(String);
impl FancyString {
fn get_str(&self) -> &str {
&self.0
}
}
After all we are creating a reference in the function and returning it! For some reason this bothered me. You can only return something that outlives the scope it is created in, so what bothered me? Well, the fact that we are ‘reaching into’ the &self reference and take a reference to a value inside of that to return it! Since self.0 is a String this feels like we ‘taking’ something out of a shared reference (&) and then referencing that! This didn’t seem right to me, since Rust exactly prevents shenanigans like that.
But this code does compile so somewhere my assumption was wrong. After a nice chat in the #rust matrix channel I’ve found where I went wrong:
self.0 is not a ‘value’ expression! There is no value that exists, but this is something called a ‘place’ expression. You can read more about them in the rust reference.
But basically place expression evaluate to a place in memory, and value expressions represent an actual value. Since I am taking a reference to a place in memory, the rust compiler can attach to that reference the same lifetime it took to get to it in the first place (i.e. the lifetime of &self).
And that’s why returning that reference is fine, since:
- We are pointing to memory with a ‘live’ lifetime once we exit (it was an input)
-
We never ‘took’ the
Stringout ofself, and thus never had a value
