【发布时间】:2019-02-10 15:50:19
【问题描述】:
我有这样的数据结构:
struct R {
hmhs: HashMap<i64, HashSet<i64>>,
}
impl R {
fn hs_for_hmhs(&mut self) -> &mut HashSet<i64> {
if let None = self.hmhs.get(&0) {
self.hmhs.insert(0, HashSet::new());
}
self.hmhs.get_mut(&0).unwrap()
}
fn iter_for_hmhs<'a>(&'a mut self) -> impl Iterator<Item = &'a i64> {
self.hs_for_hmhs().iter()
}
fn insert_for_hmhs(&mut self, i: i64) -> bool {
self.hs_for_hmhs().insert(i)
}
}
这似乎可行,但所有方法都需要一个可变的
参考self,这是不幸的。我试着给
内部可变性一展身手:
struct S {
hmhs: RefCell<HashMap<i64, HashSet<i64>>>,
}
impl S {
fn hs_for_hmhs(&self) -> &HashSet<i64> {
if let None = self.hmhs.borrow().get(&0) {
self.hmhs.borrow_mut().insert(0, HashSet::new());
}
self.hmhs.borrow_mut().get_mut(&0).unwrap()
}
fn iter_for_hmhs(&mut self) -> impl Iterator<Item = &i64> {
self.hs_for_hmhs().iter()
}
fn insert_for_hmhs(&mut self, i: i64) -> bool {
self.hs_for_hmhs().insert(i)
}
}
但是,我似乎经常遇到问题。大多是How do I return a reference to something inside a RefCell without breaking encapsulation?的各种@
我在这里尝试了很多变体,但我遗漏了一些东西 在我的理解中是基本的。有没有办法实现我的目标 想要吗?
use std::cell::RefCell;
use std::collections::{HashMap, HashSet};
struct R {
hmhs: HashMap<i64, HashSet<i64>>,
}
impl R {
fn hs_for_hmhs(&mut self) -> &mut HashSet<i64> {
if let None = self.hmhs.get(&0) {
self.hmhs.insert(0, HashSet::new());
}
self.hmhs.get_mut(&0).unwrap()
}
fn iter_for_hmhs<'a>(&'a mut self) -> impl Iterator<Item = &'a i64> {
self.hs_for_hmhs().iter()
}
fn insert_for_hmhs(&mut self, i: i64) -> bool {
self.hs_for_hmhs().insert(i)
}
}
struct S {
hmhs: RefCell<HashMap<i64, HashSet<i64>>>,
}
impl S {
fn hs_for_hmhs(&self) -> &mut HashSet<i64> {
if let None = self.hmhs.borrow().get(&0) {
self.hmhs.borrow_mut().insert(0, HashSet::new());
}
self.hmhs.borrow_mut().get_mut(&0).unwrap()
}
fn iter_for_hmhs(&self) -> impl Iterator<Item = &i64> {
self.hs_for_hmhs().iter()
}
fn insert_for_hmhs(&self, i: i64) -> bool {
self.hs_for_hmhs().insert(i)
}
}
fn main() {}
编译器消息:
error[E0597]: borrowed value does not live long enough
--> src/main.rs:36:9
|
36 | self.hmhs.borrow_mut().get_mut(&0).unwrap()
| ^^^^^^^^^^^^^^^^^^^^^^ temporary value does not live long enough
37 | }
| - temporary value only lives until here
|
note: borrowed value must be valid for the anonymous lifetime #1 defined on the method body at 31:5...
--> src/main.rs:31:5
|
31 | / fn hs_for_hmhs(&self) -> &mut HashSet<i64> {
32 | | if let None = self.hmhs.borrow().get(&0) {
33 | | self.hmhs.borrow_mut().insert(0, HashSet::new());
34 | | }
35 | |
36 | | self.hmhs.borrow_mut().get_mut(&0).unwrap()
37 | | }
| |_____^
【问题讨论】:
-
为什么需要可变借用?所有方法都返回对其内容的不可变引用?
-
哪个?
HashMap和第二个borrow_mut调用。我不知道。如果我不这样做,它会抱怨更多。 -
啊,我没注意到
insert_for_hmhs调用insert。 -
哦,
HashSet-- 是的,这部分将实现结构的公共接口。 HashMap 确实是一个实现细节。 -
这是一个棘手的问题,我想知道您是否应该尝试不同的方法。例如,this 将让你成功——一个抽象的智能指针指向
HashSet。但是很难将迭代器返回给它。
标签: rust borrow-checker mutability