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Q&A What allows a string slice (&str) to outlive its scope?

As a relative newcomer to Rust, I'm trying to understand the behaviour of lifetimes, but I am confused by the following code: let s: &str = "first"; let mut r: &str = s; println!("First ...

2 answers  ·  posted 1y ago by deleted user  ·  last activity 1y ago by Moshi‭

#1: Initial revision by (deleted user) · 2022-07-04T20:44:28Z (over 1 year ago)
What allows a string slice (&str) to outlive its scope?
As a relative newcomer to Rust, I'm trying to understand the behaviour of lifetimes, but I am confused by the following code:

    let s: &str = "first";
    let mut r: &str = s;
    println!("First ref is {}", r);
    {
        let inner: &str = "second";
        r = inner;
    }
    println!("Second ref is {}", r);

Since we are taking a long-lived reference `r` to a string slice `inner` which is destroyed at the end of its scope, I was expecting this code to fail to compile with a "variable does not live long enough" error. But to my surprise, it compiles fine, and prints valid output:

    First ref is first
    Second ref is second

It seems that the reference `r` is somehow keeping the `"second"` string slice alive beyond the end of the scope in which it is defined, but I haven't seen anything in the book which mentions this. What am I missing?