Moving to and from a mutably borrowed structure

Question

Can someone explain why moving values into borrowed structs is valid while moving them out is not? For example:

struct S {
    v: Vec<u8>
}

fn move_v_out(s: &mut S) {
    let old_vecotr = s.v; // If removed, program compiles fine
    s.v = vec![];
}

fn main() {
    let mut v = S {
        v: vec![]
    };
    move_v_out(&mut v);
}

In either case, a member of a mutably-borrowed structure is modified. However one causes a compilation error.

Answer 1

The main difference is that when you move a value from a borrowed struct, you leave it into a "partially moved state", which forbids it to be used as a whole struct afterwards.

And this state is forbidden for borrowed values, as they need to be still valid at the end of the function and the compiler doesn't (yet ?) understand that you are setting the value properly afterwards.

However, if what you want to do is extract the old Vec and replace it by a new one, the standard library contains exactly the function you need: std::mem::replace

fn move_v_out(s: &mut S) {
    let old_vector = std::mem::replace(&mut s.v, vec![]);
    // do something with old_vector
}