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)
}
}
这似乎有效，但所有方法都需要对self进行可变引用，这是不幸的。我试着给内部的可变性：
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)
}
}
但是，我似乎经常遇到问题。大多数是各种各样的How do I return a reference to something inside a RefCell without breaking encapsulation?
我在这里尝试了很多变种，但是我在理解中缺少一些基本的东西。有没有办法实现我想要的？
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 | | }
| |_____^
我找到了一个解决方案 - 将HashMap提取为原始指针。这反过来意味着我可以在没有恶作剧的情况下到达HashSet，包括返回迭代器。
作为解决方案，我很高兴。不安全的代码很小并且包含在内，如果我理解编译器抱怨没有不安全的原因，它就不会出现在这段代码中，因为在构造之后，HashMap和HashSet都没有被删除或替换。
这是一项很大的努力。
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_as_ptr(&self) -> *mut HashMap<i64, HashSet<i64>> {
self.hmhs.borrow_mut().entry(0).or_insert(HashSet::new());
self.hmhs.as_ptr()
}
fn mut_hs_for_hmhs(&mut self) -> &mut HashSet<i64> {
unsafe { (*self.hs_as_ptr()).get_mut(&0).unwrap() }
}
fn hs_for_hmhs(&self) -> &HashSet<i64> {
unsafe { (*self.hs_as_ptr()).get(&0).unwrap() }
}
fn iter_for_hmhs<'a>(&'a self) -> impl Iterator<Item = &'a i64> + 'a {
self.hs_for_hmhs().iter()
}
fn insert_for_hmhs(&mut self, i: i64) -> bool {
self.mut_hs_for_hmhs().insert(i)
}
}
fn main() {
let mut r = R {
hmhs: HashMap::new(),
};
let mut s = S {
hmhs: RefCell::new(HashMap::new()),
};
r.insert_for_hmhs(10);
s.insert_for_hmhs(20);
println!("r next: {:?}", r.iter_for_hmhs().next());
println!("s next: {:?}", s.iter_for_hmhs().next());
}