Question

我想实现一个实例化一个新元素的函数，将它添加到一个链表然后返回它刚刚创建的元素的引用。这就是我想出的：

use std::collections::LinkedList;

struct Bar {
    ll: LinkedList<Foo>
}

struct Foo {}


impl Bar {
    fn foo_alloc<'a>(&'a mut self) -> &Option<&'a mut Foo> {
        let foo = Foo{};
        self.ll.push_back(foo);
        &self.ll.front_mut()
    }
}

我认为当我将返回的引用的生命周期绑定到Bar实例（通过&'a mut self）时，这应该足够了，但显然不是。

这是错误：

error: borrowed value does not live long enough
  --> src/main.rs:14:10
   |
14 |         &self.ll.front_mut()
   |          ^^^^^^^^^^^^^^^^^^^ does not live long enough
15 |     }
   |     - temporary value only lives until here
   |
note: borrowed value must be valid for the lifetime 'a as defined on the body at 11:59...
  --> src/main.rs:11:60
   |
11 |       fn foo_alloc<'a>(&'a mut self) -> &Option<&'a mut Foo> {
   |  ____________________________________________________________^ starting here...
12 | |         let foo = Foo{};
13 | |         self.ll.push_back(foo);
14 | |         &self.ll.front_mut()
15 | |     }
   | |_____^ ...ending here

Answer 1

问题不是Option内的引用，而是Option对象本身。按值返回，而不是通过引用。

impl Bar {
    fn foo_alloc(&mut self) -> Option<&mut Foo> {
        let foo = Foo{};
        self.ll.push_back(foo);
        self.ll.front_mut()
    }
}

我还删除了生命周期注释，因为默认的生命周期扣除在这里做了正确的事情。

返回对Linkedlist元素的引用

1 个答案: