How can I swap in a new value for a field in a mutable reference to a structure?

If your type implements Default, you can use std::mem::take:

#[derive(Default)]
struct SomeType;

fn foo(a: &mut A) {
    let mut my_local_var = std::mem::take(&mut a.field);
}

If your field happens to be an Option, there's a specific method you can use — Option::take:

struct A {
    field: Option<SomeType>,
}

fn foo(a: &mut A) {
    let old = a.field.take();
    // a.field is now None, old is whatever a.field used to be
}

The implementation of Option::take uses mem::take, just like the more generic answer above shows, but it is wrapped up nicely for you:

pub fn take(&mut self) -> Option<T> {
    mem::take(self)
}

See also:

• Temporarily move out of borrowed content
• Change enum variant while moving the field to the new variant

Use std::mem::swap().

fn foo(a: &mut A) {
    let mut my_local_var = SomeType::new();
    mem::swap(&mut a.field, &mut my_local_var);
}

Or std::mem::replace().

fn foo(a: &mut A) {
    let mut my_local_var = mem::replace(&mut a.field, SomeType::new());
}