Question

我正在使用库中的结构Foo和Bar，并且我在客户端代码中收到了编译错误。我将代码简化为：

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）

输出：

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

控制台输出中的错误非常清楚，但我无法解决问题。

Answer 1

您可以通过将array变量置于带有大括号（{ ... }）的新范围来限制其生命周期：

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

Answer 2

默认情况下，一旦启用non-lexical lifetimes，您的原始代码将按原样工作：

#![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();
}

使用NLL，借阅检查器变得更加先进和精确；现在可以了解到，在调用array之后您没有使用process，因此可以安全地以新方式使用foo。

不能借用不可变的因为它也被借用为可变的

2 个答案: