How to end borrow in for-in loop

#1

This compiles:

struct Container {
    list: Vec<String>
}

fn main() {
    let mut c = Container{ list: vec![String::from("asdf")] };
    
    let borrow = &c.list[0];   // Borrow starts here
    let copy = borrow.clone(); // Last use of borrow

    foo(&mut c);               // Can borrow again
    
    println!("{}", copy);
}

fn foo(_container: &mut Container) { }

This does not:

struct Container {
    list: Vec<String>
}

fn main() {
    let mut c = Container{ list: vec![String::from("asdf")] };
    
    for borrow in &c.list {        // Borrow starts here
        let copy = borrow.clone(); // Last use of borrow

        foo(&mut c);               // Cannot borrow again???
    
        println!("{}", copy);
    }
}

fn foo(_container: &mut Container) { }

error[E0502]: cannot borrow `c` as mutable because it is also borrowed as immutable
  --> src/main.rs:11:13
   |
8  |     for borrow in &c.list {        // Borrow starts here
   |                   -------
   |                   |
   |                   immutable borrow occurs here
   |                   immutable borrow later used here
...
11 |         foo(&mut c);               // Cannot borrow again???
   |             ^^^^^^ mutable borrow occurs here

In my mind this does the exactly same thing, doesn't it?

Do I have to manually "end" the borrow? Can I? How?

#2

What is foo called c.list.clear()? Then the vector would be suddenly invalidated during the iteration. In this first case you're not iterating, so there's no harm in any modifications.

1 Like
#3

This is what the for loop desugars to:

struct Container {
    list: Vec<String>
}

fn main() {
    let mut c = Container{ list: vec![String::from("asdf")] };
    
    let mut iter = (&c.list).into_iter();
    while let Some(borrow) = iter.next() {
        let copy = borrow.clone();

        foo(&mut c); // c is still borrowed by iter
    
        println!("{}", copy);
    }
}

fn foo(_container: &mut Container) { }
1 Like