我的结构在parking_lot::RwLock中有一个向量,一个成员函数必须从该向量返回一个有保护的元素:
use parking_lot::*;
struct S {
v: RwLock<Vec<String>>,
}
impl S {
fn f(&self, i: usize) -> MappedRwLockReadGuard<'_, Option<&String>> {
RwLockReadGuard::map(self.v.read(), |unlocked| &unlocked.get(i))
}
}(这是问题的简化核心,而不是真正的代码)
代码通过类型检查,但我得到了f的终身错误。是否有一种方法来修改代码,使它是健全的,并通过借用检查?
错误是:
error[E0495]: cannot infer an appropriate lifetime for lifetime parameter in function call due to conflicting requirements
--> src/lib.rs:9:66
|
9 | RwLockReadGuard::map(self.v.read(), |unlocked| &unlocked.get(i))
| ^^^
|
note: first, the lifetime cannot outlive the anonymous lifetime #2 defined on the body at 9:45...
--> src/lib.rs:9:45
|
9 | RwLockReadGuard::map(self.v.read(), |unlocked| &unlocked.get(i))
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^
note: ...so that reference does not outlive borrowed content
--> src/lib.rs:9:57
|
9 | RwLockReadGuard::map(self.v.read(), |unlocked| &unlocked.get(i))
| ^^^^^^^^
note: but, the lifetime must be valid for the anonymous lifetime #1 defined on the method body at 8:5...
--> src/lib.rs:8:5
|
8 | / fn f(&self, i: usize) -> MappedRwLockReadGuard<'_, Option<&String>> {
9 | | RwLockReadGuard::map(self.v.read(), |unlocked| &unlocked.get(i))
10 | | }
| |_____^
note: ...so that the expression is assignable
--> src/lib.rs:9:9
|
9 | RwLockReadGuard::map(self.v.read(), |unlocked| &unlocked.get(i))
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
= note: expected `lock_api::rwlock::MappedRwLockReadGuard<'_, _, std::option::Option<&std::string::String>>`
found `lock_api::rwlock::MappedRwLockReadGuard<'_, _, std::option::Option<&std::string::String>>`发布于 2020-02-28 18:11:10
解决方案是使用RwLockReadGuard::try_map()并将返回类型从MappedRwLockReadGuard<'_, Option<&_>>更改为Option<MappedRwLockReadGuard<'a, _>>
impl S {
fn f(&self, i: usize) -> Option<MappedRwLockReadGuard<'_, String>> {
RwLockReadGuard::try_map(self.v.read(), |unlocked| unlocked.get(i)).ok()
}
}https://stackoverflow.com/questions/60452298
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