rust - 使用谓词返回引用时存在生命周期冲突(使用谓词实现 “split at mut”)

Question

这里是否可以分配生命周期，这样即使对于谓词返回引用也可以实现这样的功能？

fn group_by_into_slices_mut<'a, T, F, K>(data: &'a mut [T], key: F, res: &mut Vec<&'a mut [T]>)
where
    K: PartialEq,
    F: Fn(&T) -> K + 'static,
{
    let mut data = data;
    while !data.is_empty() {
        let j = find_j(&data, &key);
        let (s1, s2) = data.split_at_mut(j);
        res.push(s1);
        data = s2;
    }
}

fn find_j<'a, T, F, K>(data: &'a [T], key: &F) -> usize
where
    K: PartialEq,
    F: Fn(&T) -> K + 'static,
{
    let current_key = key(&data[0]);
    for i in 1..data.len() {
        if current_key != key(&data[i]) {
            return i;
        }
    }
    data.len()
}


struct Struct {
    key: String,
}

fn main() {
    let v = vec![Struct { key: "abc".to_string() }];
    let res = vec![];
    group_by_into_slices_mut(&mut v, |e| &e.key, &mut res);
}

  --> src/main.rs:37:42
   |
37 |     group_by_into_slices_mut(&mut v, |e| &e.key, &mut res);
   |                                          ^^^^^^
   |
note: first, the lifetime cannot outlive the anonymous lifetime #2 defined on the body at 37:38...
  --> src/main.rs:37:38
   |
37 |     group_by_into_slices_mut(&mut v, |e| &e.key, &mut res);
   |                                      ^^^^^^^^^^
note: ...so that reference does not outlive borrowed content
  --> src/main.rs:37:42
   |
37 |     group_by_into_slices_mut(&mut v, |e| &e.key, &mut res);
   |                                          ^^^^^^
note: but, the lifetime must be valid for the expression at 37:5...
  --> src/main.rs:37:5
   |
37 |     group_by_into_slices_mut(&mut v, |e| &e.key, &mut res);
   |     ^^^^^^^^^^^^^^^^^^^^^^^^
note: ...so type `for<'a, 'r> fn(&'a mut [Struct], [closure@src/main.rs:37:38: 37:48], &'r mut std::vec::Vec<&'a mut [Struct]>) {group_by_into_slices_mut::<Struct, [closure@src/main.rs:37:38: 37:48], &std::string::String>}` of expression is valid during the expression
  --> src/main.rs:37:5
   |
37 |     group_by_into_slices_mut(&mut v, |e| &e.key, &mut res);
   |     ^^^^^^^^^^^^^^^^^^^^^^^^

我不确定为什么这行不通。我试图明确地添加一些生命，包括更高等级的特质界限，但是没有运气。

操场:

https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=1663e12bf67838dd0d1440f759a72b4e

Answer 1

解决方案1:
我认为如果不像现在的特征范围那样将"abc"保留为&'static str，就不可能实现。如果可以将Struct对象更改为此:

struct Struct<'a> {
    key: &'a str
}

fn main() {
    let _ = Struct { key: "abc" }; //never convert "abc" to an owned type
}

...并将您的闭包作为 |e| e.key而不是 |e| &e.key传递，然后您的代码可以正常编译。
函数实现不会改变，谓词仍会返回引用，但是现在 'static的引用已存在。 |e| &e.key闭包将不起作用，因为仅允许闭包主体存在返回的引用。
解决方案2:
删除两个 'static特征上的 F界限，并将闭包更改为 |e| e.key.clone()，但是我想这不是您要查找的内容，因为您将不再返回引用。