Doubt on References Rules in Rust

[jomicobasi]

Hi guys!

I just wanted to understand one aspect of the Rust's references Rules:
At any given moment, I can only have an unlimited number of immutable references of a variable OR a single mutable reference of a variable, correct?
In the program below, if I take the mut keyword from line 2, the program will of course complain. But, if the mut keyword on line 2 holds, why does Rust does not complain?
Thank you in advance!

fn main() {
  let mut x = 10;

  let _ref1 = &x;
  let _ref2 = &x;
  
  let _ref_mut = & mut x;
  
  *_ref_mut = 9;
  println!("{}", *_ref_mut);
}

(Playground)

Output:

9

Errors:

   Compiling playground v0.0.1 (/playground)
    Finished dev [unoptimized + debuginfo] target(s) in 0.78s
     Running `target/debug/playground`

[2e71828]

Because all of this code is in a single function, Rust can tell that you’re done with _ref1 and _ref2 before you construct _ref_mut. If you change the last line to:

println!("{}", *_ref1);

Then this is no longer true and your code won’t compile.

[jomicobasi]

Ok, I get it now! Thank you!

I've also just found the snippet of the https://doc.rust-lang.org telling the same thing.

[elomatreb]

Historically, this did indeed not work, until we got an improved borrow checker together with the 2018 Edition which was able to reason about "Non-Lexical Lifetimes" (NLLs).