我正在尝试重写publicsuffix板条箱，以期获得更好的性能。我正在使用与参考相同名称的Python库，并且在简化了它的数据结构之后，我想到了：
#[derive(Debug)]
pub struct Node<T> {
pub name: T,
pub children: Vec<Node<T>>,
}
在收到结构加载良好的信号之前，我感到太懒了，无法实现结构加载，这就是为什么我决定对其进行硬编码的原因。当我尝试进行类似error[E0015]: calls in statics are limited to constant functions, struct and enum constructors的操作时，我会不断得到pub static root = Node { name: "x", children: vec![]};，所以我决定将其包装在一个函数中。我最终遇到this file，这需要花很长时间才能完成（大约10分钟）。在问过IRC＃rust-beginners之后，我得到了一个建议，尝试尝试使用对数组切片的引用，并最终得到以下示例代码：
#[derive(Debug)]
pub struct Node<T: 'static> {
pub name: T,
pub children: &'static [Node<T>],
}
pub fn return_root() -> Node<&'static str> {
return Node {
name: "root",
children: &[Node {
name: "uk",
children: &[
Node {
name: "ac",
children: &[][..],
},
Node {
name: "co",
children: &[
Node {
name: "blogspot",
children: &[][..],
},
Node {
name: "no-ip",
children: &[][..],
},
][..],
},
Node {
name: "gov",
children: &[Node {
name: "service",
children: &[][..],
}][..],
},
Node {
name: "ltd",
children: &[][..],
},
Node {
name: "me",
children: &[][..],
},
Node {
name: "net",
children: &[][..],
},
Node {
name: "nhs",
children: &[][..],
},
Node {
name: "org",
children: &[][..],
},
Node {
name: "plc",
children: &[][..],
},
Node {
name: "police",
children: &[][..],
},
Node {
name: "sch",
children: &[Node {
name: "*",
children: &[][..],
}][..],
},
][..],
}][..],
};
}
这无法编译，我不理解错误消息：
error[E0597]: borrowed value does not live long enough
--> src/main.rs:9:20
|
9 | children: &[Node {
| ____________________^
10 | | name: "uk",
11 | | children: &[
12 | | Node {
... |
71 | | ][..],
72 | | }][..],
| |__________^ temporary value does not live long enough
73 | };
| - temporary value only lives until here
|
= note: borrowed value must be valid for the static lifetime...
= note: consider using a `let` binding to increase its lifetime
error[E0597]: borrowed value does not live long enough
--> src/main.rs:11:24
|
11 | children: &[
| ________________________^
12 | | Node {
13 | | name: "ac",
14 | | children: &[][..],
... |
70 | | },
71 | | ][..],
| |_____________^ temporary value does not live long enough
72 | }][..],
73 | };
| - temporary value only lives until here
|
= note: borrowed value must be valid for the static lifetime...
= note: consider using a `let` binding to increase its lifetime
error[E0597]: borrowed value does not live long enough
--> src/main.rs:18:32
|
18 | children: &[
| ________________________________^
19 | | Node {
20 | | name: "blogspot",
21 | | children: &[][..],
... |
26 | | },
27 | | ][..],
| |_____________________^ temporary value does not live long enough
...
73 | };
| - temporary value only lives until here
|
= note: borrowed value must be valid for the static lifetime...
= note: consider using a `let` binding to increase its lifetime
error[E0597]: borrowed value does not live long enough
--> src/main.rs:31:32
|
31 | children: &[Node {
| ________________________________^
32 | | name: "service",
33 | | children: &[][..],
34 | | }][..],
| |______________________^ temporary value does not live long enough
...
73 | };
| - temporary value only lives until here
|
= note: borrowed value must be valid for the static lifetime...
= note: consider using a `let` binding to increase its lifetime
error[E0597]: borrowed value does not live long enough
--> src/main.rs:66:32
|
66 | children: &[Node {
| ________________________________^
67 | | name: "*",
68 | | children: &[][..],
69 | | }][..],
| |______________________^ temporary value does not live long enough
...
73 | };
| - temporary value only lives until here
|
= note: borrowed value must be valid for the static lifetime...
= note: consider using a `let` binding to increase its lifetime
有没有办法解决这个问题？
答案 0 :(得分：1)
创建切片之后，请使用&[][..]再次引用它。这将创建一个临时对象，之后将其删除。这是错误的根源。
从代码中删除所有[..]时，您可以修复它。
#[derive(Debug)]
pub struct Node<T: 'static> {
pub name: T,
pub children: &'static [Node<T>],
}
pub fn return_root() -> Node<&'static str> {
return Node {
name: "root",
children: &[Node {
name: "uk",
children: &[
Node {
name: "ac",
children: &[],
},
Node {
name: "co",
children: &[
Node {
name: "blogspot",
children: &[],
},
Node {
name: "no-ip",
children: &[],
},
],
},
Node {
name: "gov",
children: &[Node {
name: "service",
children: &[],
}],
},
Node {
name: "ltd",
children: &[],
},
Node {
name: "me",
children: &[],
},
Node {
name: "net",
children: &[],
},
Node {
name: "nhs",
children: &[],
},
Node {
name: "org",
children: &[],
},
Node {
name: "plc",
children: &[],
},
Node {
name: "police",
children: &[],
},
Node {
name: "sch",
children: &[Node {
name: "*",
children: &[],
}],
},
],
}],
};
}
不需要[..]，因为它仅从另一个切片创建一个切片，从而产生相同的切片。