循环中的可变借位 [英] Mutable borrow in a loop

问题描述

我有以下代码:

struct Baz {
    x: usize,
    y: usize,
}

struct Bar {
    baz: Baz,
}

impl Bar {
    fn get_baz_mut(&mut self) -> &mut Baz {
        &mut self.baz
    }
}

struct Foo {
    bar: Bar,
}

impl Foo {
    fn foo(&mut self) -> Option<&mut Baz> {
        for i in 0..4 {
            let baz = self.bar.get_baz_mut();
            if baz.x == 0 {
                return Some(baz);
            }
        }
        None
    }
}

铁锈操场

它无法编译为:

error[E0499]: cannot borrow `self.bar` as mutable more than once at a time
  --> src/main.rs:23:23
   |
23 |             let baz = self.bar.get_baz_mut();
   |                       ^^^^^^^^ mutable borrow starts here in previous iteration of loop
...
29 |     }
   |     - mutable borrow ends here

但是，如果我从Foo::foo返回Some(baz.x)(并将返回类型更改为Option<usize>)，则代码将编译.这使我相信问题不在于循环，即使编译器似乎表明了这一点.更具体地说，我相信局部可变引用baz在循环的下一次迭代时将超出范围，从而导致这不是问题.上面的代码的生命周期问题是什么?

However, if I return Some(baz.x) from Foo::foo (and change the return type to Option<usize>), the code compiles. This makes me believe the problem is not with the loop even though the compiler seems to indicate so. More specifically, I believe the local mutable reference baz would go out of scope at the next iteration of the loop, causing this to be a non-problem. What is the lifetime problem with the above code?

以下问题类似:

• 可变的借入循环
• 将self和引用的生命周期链接到方法

• Mutable borrow in loop
• Linking the lifetimes of self and a reference in method
• Cannot borrow as mutable more than once at a time in one code - but can in another very similar

但是，它们处理显式声明的生存期(特别是这些显式生存期是答案的一部分).我的代码忽略了这些生命周期，因此删除它们是一个非解决方案.

However, they deal with explicitly declared lifetimes (and specifically these explicit lifetimes are part of the answer). My code omits these lifetimes so removing them is a non-solution.

推荐答案

它不起作用，因为返回借用的值会将借用扩展到函数的末尾.

It does not work because returning a borrowed value extends the borrow to the end of the function.

有关一些有用的详细信息，请参见此处.

See here for some useful details.

这可用于非词汇生存期和 1.27 夜间版本:

#![feature(nll)]

struct Baz {
    x: usize,
    y: usize,
}

// ...

非词条生命周期RFC解释了生命周期的实际工作:

The non-lexical lifetimes RFC explains the actual working of lifetimes:

但是，当您具有跨多个语句的引用时，就会出现问题.在那种情况下，编译器要求生命周期是包含两个语句的最里面的表达式(通常是一个块)，并且通常比实际需要或期望的要大得多

Problems arise however when you have a reference that spans multiple statements. In that case, the compiler requires the lifetime to be the innermost expression (which is often a block) that encloses both statements, and that is typically much bigger than is really necessary or desired

rustc每晚1.28

@pnkfelix指出从每晚1.28开始不再编译以上代码.

As pointed out by @pnkfelix, the non-lexical lifetimes implementation starting from nightly 1.28 no longer compiles the above code.

但是，有(重新)-启用更强大的NLL分析.

这篇关于循环中的可变借位的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！