cannot borrow as immutable because it is also borrowed as mutable

问题

I'm using the structs Foo and Bar from a library and I'm getting a compilation error in the client code. I simplified the code to this:

use std::marker::PhantomData;

struct Foo {
    some_str: &'static str,
}

struct Bar<'a> {
    some_str: &'static str,
    marker: PhantomData<&'a Foo>,
}

impl Foo {
    fn read_data(&self) {
        // add code here
    }
    fn create_bar<'a>(&'a mut self) -> Bar<'a> {
        Bar {
            some_str: "test2",
            marker: PhantomData,
        }
    }
}

fn process(_arr: &mut [Bar]) {}

fn main() {
    let mut foo = Foo { some_str: "test" };
    let mut array: [Bar; 1] = [foo.create_bar()];
    process(&mut array);
    foo.read_data();
}

(playground)

Output:

error[E0502]: cannot borrow `foo` as immutable because it is also borrowed as mutable
  --> src/main.rs:30:5
   |
28 |     let mut array: [Bar; 1] = [foo.create_bar()];
   |                                --- mutable borrow occurs here
29 |     process(&mut array);
30 |     foo.read_data();
   |     ^^^ immutable borrow occurs here
31 | }
   | - mutable borrow ends here

The error in the console output is very clear, but I cannot fix the problem.

回答1:

You can limit the lifetime of the array variable by placing it in a new scope with curly braces ({ ... }):

fn main() {
    let mut foo = Foo { some_str: "test" };
    {
        let mut array: [Bar; 1] = [foo.create_bar()];
        process(&mut array);
    }
    foo.read_data();
}

回答2:

You original code will work as-is once non-lexical lifetimes are enabled by default:

#![feature(nll)]

use std::marker::PhantomData;

struct Foo {
    some_str: &'static str,
}

struct Bar<'a> {
    some_str: &'static str,
    marker: PhantomData<&'a Foo>,
}

impl Foo {
    fn read_data(&self) {
        // add code here
    }
    fn create_bar<'a>(&'a mut self) -> Bar<'a> {
        Bar {
            some_str: "test2",
            marker: PhantomData,
        }
    }
}

fn process(_arr: &mut [Bar]) {}

fn main() {
    let mut foo = Foo { some_str: "test" };
    let mut array: [Bar; 1] = [foo.create_bar()];
    process(&mut array);
    foo.read_data();
}

With NLL, the borrow checker becomes more advanced and precise; it can now understand that you aren't using array after the call to process so it is safe to use foo in a new manner.

来源：https://stackoverflow.com/questions/35054404/cannot-borrow-as-immutable-because-it-is-also-borrowed-as-mutable

标签

rust