我的目标是使用many_iter_mut()
方法创建一个NewType(Arc<RwLock<Vec<T>>>)
,该方法返回一个迭代器,该迭代器在内部Vec<T>
上持有一个写锁(RwLockWriteGuard
)。
因此,写锁将由ManyIterMut
迭代器持有,直到迭代器被删除。这与每次调用ManyIterMut::next()
都获得写锁不同。(我已经在其他地方工作了)
这对于使调用者能够同时设置多个元素和读取多个元素非常有用,但原子地。因此,从一批写入到下一批写入,读取器将始终看到一致的视图。(我以前的ManyIterMut
实现失败了并发测试。)
问题是由于生存期问题,我无法编译ManyIterMut::next()
方法。编译器似乎不喜欢write_lock
被ManyIterMut
拥有,而不是被迭代的LockedVec
拥有。
这是playground。
生成错误的代码是:
impl<'a, V, T: 'a> LendingIterator for ManyIterMut<'a, V, T>
where
V: StorageVecWrites<T> + ?Sized,
V::LockedData: StorageVecReads<T>,
T: Copy,
{
type Item<'b> = StorageSetter<'b, V, T>
where
Self: 'b;
// This is the fn that won't compile.
//
// we can't change the fn signature, it is defined by lending_iterator::LendingIterator trait.
fn next(&mut self) -> Option<Self::Item<'_>> {
if let Some(i) = Iterator::next(&mut self.indices) {
let value = self.write_lock.get_at(i).unwrap();
Some(StorageSetter {
phantom: Default::default(),
write_lock: &mut self.write_lock, // <--- this seems to be the problem. Because it has &mut self lifetime, not Self::Item<'_> lifetime
index: i,
value,
})
} else {
None
}
}
}
字符串
错误是:
error: lifetime may not live long enough
--> src/main.rs:164:13
|
148 | impl<'a, V, T: 'a> LendingIterator for ManyIterMut<'a, V, T>
| -- lifetime `'a` defined here
...
161 | fn next(&mut self) -> Option<Self::Item<'_>> {
| - let's call the lifetime of this reference `'1`
...
164 | / Some(StorageSetter {
165 | | phantom: Default::default(),
166 | | write_lock: &mut self.write_lock, // <--- this seems to be the problem. Because it has &mut self lifetime, not Self::It...
167 | | index: i,
168 | | value,
169 | | })
| |______________^ method was supposed to return data with lifetime `'a` but it is returning data with lifetime `'1`
|
= note: requirement occurs because of the type `StorageSetter<'_, V, T>`, which makes the generic argument `'_` invariant
= note: the struct `StorageSetter<'c, V, T>` is invariant over the parameter `'c`
= help: see <https://doc.rust-lang.org/nomicon/subtyping.html> for more information about variance
error: could not compile `playground` (bin "playground") due to previous error
型
那么,我错过了什么?这似乎是可能的。或者有一个更简单的方法?
1条答案
按热度按时间0sgqnhkj1#
字符串
这个模式(
&'a mut T<'a>
)(几乎?)总是错误的。At允许使用数据 * 最多 * 一次。更多信息请参见Borrowing something forever。解决方案是引入另一个生命周期。这需要一些修正,主要是在
type Item<'b> = StorageSetter<'b, V, T>
中将其更改为type Item<'b> = StorageSetter<'a, 'b, V, T>
。