Temporarily move out of borrowed content

Question

I'm tring to replace a value in a mutable borrow; moving part of it into the new value:

enum Foo<T> {
    Bar(T),
    Baz(T),
}

impl<T> Foo<T> {
    fn switch(&mut self) {
        *self = match self {
            &mut Foo::Bar(val) => Foo::Baz(val),
            &mut Foo::Baz(val) => Foo::Bar(val),
        }
    }
}

The code above doesn't work, and understandibly so, moving the value out of self breaks the integrity of it. But since that value is dropped immediately afterwards, I (if not the compiler) could guarantee it's safety.

Is there some way to achieve this? I feel like this is a job for unsafe code, but I'm not sure how that would work.

Answer 1

The code above doesn't work, and understandibly so, moving the value out of self breaks the integrity of it.

This is not exactly what happens here. For example, same thing with self would work nicely:

impl<T> Foo<T> {
    fn switch(self) {
        self = match self {
            Foo::Bar(val) => Foo::Baz(val),
            Foo::Baz(val) => Foo::Bar(val),
        }
    }
}

Rust is absolutely fine with partial and total moves. The problem here is that you do not own the value you're trying to move - you only have a mutable borrowed reference. You cannot move out of any reference, including mutable ones.

This is in fact one of the frequently requested features - a special kind of reference which would allow moving out of it. It would allow several kinds of useful patterns. You can find more here and here.

In the meantime for some cases you can use std::mem::replace and std::mem::swap. These functions allow you to "take" a value out of mutable reference, provided you give something in exchange.