How to satisfy Rust borrow checker with this simple code?

Question

I'm getting a Rust compile error from the borrow checker, and don't know how to fix it.
The code below is simple, and no problem with similar code in C++.

fn main() {
    let mut nums = vec![1, 2, 3];
    if let Some(x) = nums.last() {
        nums.push(*x);
    }
}

Here is the error：

message: 'cannot borrow `nums` as mutable because it is also borrowed as immutable (4, 9)'

Answer 1

When you call .last() you borrow nums as immutable, as mutating it would invalidate the reference x that you hold. You then call .push, which borrows nums as mutable.

The problem is that you now have an immutable and a mutable borrow of the same value at the same time, which is against rusts memory safety guarantees (multiple readers or one single writer guarantee that you will never have invalid memory anywhere).

fn main() {
    let mut nums = vec![1, 2, 3];
    if let Some(x) = nums.last() { // Immutable borrow starts here
        nums.push(*x);             // Mutable borrow starts here
    }                              // Immutable and mutable borrows end here
}

The solution would be to lower the scope of the immutable borrow by immediately dropping the reference of its result, as per @DanielSanchez's example:

let mut nums = vec![1, 2, 3];
if let Some(&x) = nums.last() { // Immutable borrow starts and ends here
    nums.push(x);               // Mutable borrow starts here
}                               // Mutable borrow ends here

Answer 2

You can dereference the reference in the guard clause:

fn main() {
    let mut nums = vec![1, 2, 3];
    if let Some(&x) = nums.last() {
        nums.push(x);
    }
}

Rust has a powerful pattern matching feature, you can unpack almost everything if you know its type. Check the Rust pattern matching documentation.