How does returning a reference to Path from the function's stack work?

Question

The following snippet is from rust source code for path.rs

impl AsRef<Path> for String {
    fn as_ref(&self) -> &Path {
        Path::new(self)
    }
}

It seems to be returning a reference of a newly created Path object from the stack frame. How does it bypass Rust's borrow checking rules? A stack object should have a lifetime for the current frame only and returning reference for temporary objects should produce an error.

Answer 1

The returned reference does not point to some object that is newly created inside of as_ref(), but rather to the data of the string that was passed in as an argument.

Let's make the lifetimes in the relevant functions explicit. The prototype

fn as_ref(&self) -> &Path

can be desugared to

fn as_ref(&'a self) -> &'a Path

The type of self is &'a String, a reference to a string with lifetime 'a.

Inside the function body, Path::new() is called. According to the documentation, it is defined like this:

pub fn new<S: AsRef<OsStr> + ?Sized>(s: &S) -> &Path

Adding the elided lifetime gives

pub fn new<S: AsRef<OsStr> + ?Sized>(s: &'a S) -> &'a Path

so the newly created Path reference will have the same lifetime as the reference we passed in, which is exactly the required lifetime of the return value.