I thought it would not compile, because a takes read ownership, and push needs read + write access. But I was wrong!
It does compile! It’s due to a Rust feature called Non-Lexical Lifetimes (added in 2018). So along with the other rules, the borrow checker also asks: “Is there any path where the borrow is used after this point?” If the answer is no, the borrow ends there!
let mut v = vec![1, 2, 3];
let a = &v[0]; // immutable borrow starts
println!("{a}"); // last use of `a` - borrow ENDS here with NLL
v.push(4); // RW mutable borrow is fine, no conflict
I hope this helps others learning Rust (its a beautiful language)