这里是否有一种分配生存期的方法,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。
游乐场:
[我认为,不像您当前的特征范围要求那样将"abc"
保持为&'static str
是不可能的。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
}
...并以|e| e.key
而不是|e| &e.key
的形式传递闭包,然后代码可以正常编译。
函数实现没有改变,该谓词仍返回一个引用,但现在该引用存在于'static
。 |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。
删除两个'static
特征的F
界限,并将闭包更改为|e| e.key.clone()
,但我想这不是您要查找的内容,因为您将不再返回参考。