我有这样一个数据结构:
这似乎可行,但所有的方法都需要对
然而,我似乎经常遇到问题。主要是一些如何在不破坏封装的情况下返回RefCell内部内容的引用?的变体。
我已经尝试了很多变种,但我在理解上缺少了一些基础知识。有没有办法实现我想要的功能? 完整代码:
编译器信息:
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,但所有的方法都需要对
self
的可变引用,这很不幸。我尝试使用内部可变性: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)
}
}
然而,我似乎经常遇到问题。主要是一些如何在不破坏封装的情况下返回RefCell内部内容的引用?的变体。
我已经尝试了很多变种,但我在理解上缺少了一些基础知识。有没有办法实现我想要的功能? 完整代码:
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() {}
编译器信息:
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 | | }
| |_____^
HashMap
和第二个borrow_mut
调用。我不知道。如果我不这样做,它会投诉得更多。 - Phil Lordinsert_for_hmhs
调用了insert
。 - Peter HallHashSet
。但是很难返回该指针的迭代器。 - Peter Hall