rust - 从闭包调用可变方法时，无法推断autoref的生存期

Question

这是一个再现错误的游乐场链接:https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=86ec4f11f407f5d04a8653cc904f991b
我有一个特性FooTraitMut，可以访问BarStruct内部的特定范围的数据，并且我想对此特性进行概括，以便它可以在锁步中访问多个BarStruct上的相同范围。所以我有一个MutChannels特性，它像一个类型级别的函数一样来生成访问者需要的引用元组，例如(T, U) --> (&mut T, &mut U)。
我实际上还没有到达使用Channels2的地步，因为我无法使用更简单的Channels1案例来工作。
在操场上，对不变特性FooTraitRef进行了相同的操作，该特性可以按预期工作。但是由于autoref生存期问题，可变项已损坏。我认为self的生命周期正在发生某种隐式转换，因为我可以内联indexer函数，并且可以正常工作。
任何帮助将不胜感激。
有问题的代码:

struct BarStruct<T> {
    data: [T; 1],
}

pub struct Channels1<T>(T);
pub struct Channels2<T, U>(T, U);

fn indexer(mut f: impl FnMut(usize)) {
    f(0)
}

trait FooMutTrait {
    type Data: for<'a> MutChannels<'a>;

    fn foo<'a, F>(&'a mut self, f: F)
    where
        F: FnMut(<Self::Data as MutChannels<'a>>::Mut);
}

trait MutChannels<'a> {
    type Mut;
}

impl<'a, T: 'a> MutChannels<'a> for Channels1<T> {
    type Mut = &'a mut T;
}
impl<'a, T: 'a, U: 'a> MutChannels<'a> for Channels2<T, U> {
    type Mut = (&'a mut T, &'a mut U);
}

impl<T> BarStruct<T> {
    fn get_data_mut<'a>(&'a mut self, i: usize) -> &'a mut T {
        &mut self.data[i]
    }
}

impl<T> FooMutTrait for BarStruct<T>
where
    T: 'static,
{
    type Data = Channels1<T>;

    #[inline]
    fn foo<'a, F>(&'a mut self, mut f: F)
    where
        F: FnMut(<Self::Data as MutChannels<'a>>::Mut),
    {
        indexer(|i| f(self.get_data_mut(i)))
        
        // This works.
        // f(self.get_data_mut(0))
    }
}

错误:

error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements
  --> src/lib.rs:85:28
   |
85 |         indexer(|i| f(self.get_data_mut(i)))
   |                            ^^^^^^^^^^^^
   |
note: first, the lifetime cannot outlive the lifetime `'_` as defined on the body at 85:17...
  --> src/lib.rs:85:17
   |
85 |         indexer(|i| f(self.get_data_mut(i)))
   |                 ^^^^^^^^^^^^^^^^^^^^^^^^^^^
note: ...so that closure can access `self`
  --> src/lib.rs:85:23
   |
85 |         indexer(|i| f(self.get_data_mut(i)))
   |                       ^^^^
note: but, the lifetime must be valid for the lifetime `'a` as defined on the method body at 81:12...
  --> src/lib.rs:81:12
   |
81 |     fn foo<'a, F>(&'a mut self, mut f: F)
   |            ^^
note: ...so that reference does not outlive borrowed content
  --> src/lib.rs:85:23
   |
85 |         indexer(|i| f(self.get_data_mut(i)))
   |                       ^^^^^^^^^^^^^^^^^^^^

Answer 1

此错误可以通过以下示例重现:

fn indexer(mut f: impl FnMut()) {}

fn foo<'a, F>(a: &'a mut String, mut f: F)
where
    F: FnMut(&'a mut str),
{
    indexer(|| f(a.as_mut_str()));
}

error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements
 --> src/lib.rs:7:20
  |
7 |     indexer(|| f(a.as_mut_str()));
  |                    ^^^^^^^^^^
  |
note: first, the lifetime cannot outlive the lifetime `'_` as defined on the body at 7:13...
 --> src/lib.rs:7:13
  |
7 |     indexer(|| f(a.as_mut_str()));
  |             ^^^^^^^^^^^^^^^^^^^^
note: ...so that closure can access `a`
 --> src/lib.rs:7:18
  |
7 |     indexer(|| f(a.as_mut_str()));
  |                  ^
note: but, the lifetime must be valid for the lifetime `'a` as defined on the function body at 3:8...
 --> src/lib.rs:3:8
  |
3 | fn foo<'a, F>(a: &'a mut String, mut f: F)
  |        ^^
note: ...so that reference does not outlive borrowed content
 --> src/lib.rs:7:18
  |
7 |     indexer(|| f(a.as_mut_str()));
  |                  ^^^^^^^^^^^^^^

发生的情况是，键入 F以期望有 'a引用，但这不是闭包可以提供的。闭包将 &'a mut T转换为生命周期较短的 &'_ mut T。根据我的理解， FnMut保留外部生存期是不合理的，因为该函数可能会泄漏超出其范围的引用，并且可能违反Rust的引用保证。这个问题不会带来不可变的借贷，因为借贷的限制较少，而且关闭也不会缩短借贷的期限。
可以通过允许 F在任何生命周期内工作来解决此问题:

fn indexer(mut f: impl FnMut()) {}

fn foo<'a, F>(a: &'a mut String, mut f: F)
where
    F: FnMut(&mut str), // <--------
{
    indexer(|| f(a.as_mut_str()));
}

或者使用 FnOnce，因为它的功能受到更多限制，并且不需要缩短生存期:

fn indexer(f: impl FnOnce()) {} // <--------

fn foo<'a, F>(a: &'a mut String, mut f: F)
where
    F: FnOnce(&'a mut str), // <--------
{
    indexer(move || f(a.as_mut_str())); // added move so that the reference isn't reborrowed
}

FnOnce更改对您的情况而言是微不足道的。但是，放宽 F使其在整个生命周期中都能正常工作会遇到有关 MutChannels<'_>::Mut与 &'_ mut T不同的错误，我不确定该如何处理。