当我尝试创建自定义迭代器时,我遇到了一些问题。不可变版本工作得很好,但当复制相同的函数来创建可变版本时,出现了终生错误。这是我的问题的一个简化版本:
struct Test {
map: HashMap<u32, String>
}
impl Test {
pub fn iter(&self, start: u32, end: u32) -> impl Iterator<Item = &String> {
(start..=end).filter_map(move |i| {
self.map.get(&i)
})
}
pub fn iter_mut(&mut self, start: u32, end: u32) -> impl Iterator<Item = &mut String> {
(start..=end).filter_map(move |i| {
self.map.get_mut(&i)
})
}
}iter函数运行良好,但iter_mut函数无法编译,出现以下错误:
error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements
--> src/main.rs:21:22
|
21 | self.map.get_mut(&i)
| ^^^^^^^
|
note: first, the lifetime cannot outlive the lifetime `'_` as defined on the body at 20:34...
--> src/main.rs:20:34
|
20 | (start..=end).filter_map(|i| {
| ^^^
note: ...so that closure can access `self`
--> src/main.rs:21:13
|
21 | self.map.get_mut(&i)
| ^^^^^^^^
note: but, the lifetime must be valid for the anonymous lifetime defined on the method body at 19:21...
--> src/main.rs:19:21
|
19 | pub fn iter_mut(&mut self, start: u32, end: u32) -> impl Iterator<Item = &mut String> {
| ^^^^^^^^^
note: ...so that the types are compatible
--> src/main.rs:19:57
|
19 | pub fn iter_mut(&mut self, start: u32, end: u32) -> impl Iterator<Item = &mut String> {
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
= note: expected `std::option::Option<&mut std::string::String>`
found `std::option::Option<&mut std::string::String>`发布于 2021-07-27 18:26:34
正如托德所说,iter_mut上的编译错误可能是由于创建了许多对同一HashMap的可变引用而发生的,但我不确定这一点。你可以这样做:
struct Test {
map: HashMap<u32, String>
}
impl Test {
pub fn iter(&self, start: u32, end: u32) -> impl Iterator<Item=&String> {
self.map
.iter()
.filter_map(move |k| {
if (start..=end).contains(k.0) {
Some(k.1)
} else {
None
}
})
}
pub fn iter_mut(&mut self, start: u32, end: u32) -> impl Iterator<Item=&mut String> {
self.map
.iter_mut()
.filter_map(move |k| {
if (start..=end).contains(k.0) {
Some(k.1)
} else {
None
}
})
}
}https://stackoverflow.com/questions/68542666
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