Question

在下面的代码中，我了解到Storage::insert返回的引用和Consumer<'a>的生存期之间似乎存在生命周期冲突。但是，我遇到了麻烦：

了解为什么会发生这种冲突-即Consumer::storage和Consumer::current_value具有相同的生存期，因此Rust推论的其他生存期与<'a>冲突< / p>
如何最好地指定生命周期以及为何有效

struct Storage {
    value: i64,
}

impl Storage {
    fn new() -> Self {
        Storage { value: 0 }
    }

    fn insert(&mut self, v: i64) -> &i64 {
        self.value = v;
        &self.value
    }
}

struct Consumer<'a> {
    storage: &'a mut Storage,
    current_value: &'a i64,
}

impl<'a> Consumer<'a> {
    fn new(s: &'a mut Storage) -> Self {
        Consumer {
            storage: s,
            current_value: &s.value,
        }
    }

    fn new_value(&mut self, v: i64) {
        self.current_value = self.storage.insert(v);
    }
}

Playground

此代码导致的错误是：

error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements
  --> src/lib.rs:30:43
   |
30 |         self.current_value = self.storage.insert(v);
   |                                           ^^^^^^
   |
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 29:5...
  --> src/lib.rs:29:5
   |
29 | /     fn new_value(&mut self, v: i64) {
30 | |         self.current_value = self.storage.insert(v);
31 | |     }
   | |_____^
note: ...so that reference does not outlive borrowed content
  --> src/lib.rs:30:30
   |
30 |         self.current_value = self.storage.insert(v);
   |                              ^^^^^^^^^^^^
note: but, the lifetime must be valid for the lifetime 'a as defined on the impl at 21:6...
  --> src/lib.rs:21:6
   |
21 | impl<'a> Consumer<'a> {
   |      ^^
note: ...so that reference does not outlive borrowed content
  --> src/lib.rs:30:30
   |
30 |         self.current_value = self.storage.insert(v);
   |                              ^^^^^^^^^^^^^^^^^^^^^^

我已经阅读了以下内容，但不确定是否遵循它们如何应用于我的问题（也许他们这样做了，而我只是阅读不正确）：

具有引用的结构和引用的返回：由于需求冲突，无法为autoref推断适当的生存期

0 个答案: