问题
I am trying to create an mutable iterator for a vector of type: Vec<Vec<(K, V)>>
The iterator code:
pub struct IterMut<'a, K: 'a, V: 'a> {
iter: &'a mut Vec<Vec<(K, V)>>,
ix: usize,
inner_ix: usize,
}
impl<'a, K, V> Iterator for IterMut<'a, K, V> {
type Item = (&'a K, &'a mut V);
#[inline]
fn next(&mut self) -> Option<(&'a K, &'a mut V)> {
while self.iter.len() < self.ix {
while self.iter[self.ix].len() < self.inner_ix {
self.inner_ix += 1;
let (ref k, ref mut v) = self.iter[self.ix][self.inner_ix];
return Some((&k, &mut v));
}
self.ix += 1;
}
return None;
}
}
The error I get is:
error[E0495]: cannot infer an appropriate lifetime for lifetime parameter in function call due to conflicting requirements
--> src/main.rs:16:42
|
16 | let (ref k, ref mut v) = self.iter[self.ix][self.inner_ix];
| ^^^^^^^^^^^^^^^^^^
|
help: consider using an explicit lifetime parameter as shown: fn next(&'a mut self) -> Option<(&'a K, &'a mut V)>
--> src/main.rs:11:5
|
11 | fn next(&mut self) -> Option<(&'a K, &'a mut V)> {
| ^
Apparently I have lifetime problems, but I don't know how to tell the compiler that this should work.
Is this how you should implement the mutable iterator or is there a better way?
回答1:
When debugging cryptic error messages, I've found it easier to try and isolate the issue as much as possible.
The first step is to break the expression into its essential constituents, let's start by splitting the indexing steps:
fn next(&mut self) -> Option<(&'a K, &'a mut V)> {
while self.iter.len() < self.ix {
while self.iter[self.ix].len() < self.inner_ix {
self.inner_ix += 1;
let outer: &'a mut Vec<_> = self.iter;
let inner: &'a mut Vec<_> = &mut outer[self.ix];
let (ref k, ref mut v) = inner[self.inner_ix];
return Some((&k, &mut v));
}
self.ix += 1;
}
return None;
}
The Index trait assumes that the lifetime of its output is linked to that of its receiver, so to get a 'a lifetime we need the receiver to have a &'a lifetime, and it propagates upward, leading to the above code.
However there's an issue here: let outer: &'a mut Vec<_> = self.iter; will not compile because mutable references are not Copy.
So, how does one get a mutable reference from a mutable reference (which must be possible since IndexMut gets a mutable reference)?
One uses re-borrowing: let outer: &'a mut Vec<_> = &mut *self.iter;.
And, oh:
error[E0495]: cannot infer an appropriate lifetime for borrow expression due to conflicting requirements --> <anon>:16:45 | 16 | let outer: &'a mut Vec<_> = &mut *self.iter; | ^^^^^^^^^^^^^^^ |
The reborrowed reference is not valid for 'a, it's valid only for the (unnamed) lifetime of self!
Why Rust? Why?
Because doing otherwise would be unsafe.
&mut T is guaranteed NOT to be aliasing, however your method could create aliasing references (if you forgot to advance the index):
#[inline]
fn next(&mut self) -> Option<(&'a K, &'a mut V)> {
let (ref k, ref mut v) = self.iter[self.ix][self.inner_ix];
return Some((&k, &mut v));
}
And even if you don't, there's not guarantee that you don't have a rewind method that would allow "stepping back".
TL;DR: You were about to step on a landmine, you were steered toward Stack Overflow instead ;)
Alright, but how do you implement the iterator!.
Well, using iterators, of course. As Shepmaster (briefly) answers, there is the equivalent in the standard library already in the guise of FlatMap. The trick is to use existing iterators for the nitty-gritty details!
Something like:
use std::slice::IterMut;
pub struct MyIterMut<'a, K: 'a, V: 'a> {
outer: IterMut<'a, Vec<(K, V)>>,
inner: IterMut<'a, (K, V)>,
}
Then you consume from inner as long as it provides items, and when empty you refill it from outer.
impl<'a, K, V> MyIterMut<'a, K, V> {
fn new(v: &'a mut Vec<Vec<(K, V)>>) -> MyIterMut<'a, K, V> {
let mut outer = v.iter_mut();
let inner = outer.next()
.map(|v| v.iter_mut())
.unwrap_or_else(|| (&mut []).iter_mut());
MyIterMut { outer: outer, inner: inner }
}
}
impl<'a, K, V> Iterator for MyIterMut<'a, K, V> {
type Item = (&'a K, &'a mut V);
#[inline]
fn next(&mut self) -> Option<(&'a K, &'a mut V)> {
loop {
match self.inner.next() {
Some(r) => return Some((&r.0, &mut r.1)),
None => (),
}
match self.outer.next() {
Some(v) => self.inner = v.iter_mut(),
None => return None,
}
}
}
}
A quick test case:
fn main() {
let mut v = vec![
vec![(1, "1"), (2, "2")],
vec![],
vec![(3, "3")]
];
let iter = MyIterMut::new(&mut v);
let c: Vec<_> = iter.collect();
println!("{:?}", c);
}
Prints:
[(1, "1"), (2, "2"), (3, "3")]
as expected, so it's not completely broken, but I wish I did not have to rely on the &[] is 'static trick (ie, that std::slice::IterMut implemented Default).
回答2:
You've provided no reason that you are reimplementing the standard Iterator::flat_map, so I'd just use that and another map to remove the mutability you don't need:
fn main() {
let mut a: Vec<Vec<(u8, u8)>> = Default::default();
let c = a.iter_mut()
.flat_map(|x| x.iter_mut())
.map(|&mut (ref a, ref mut b)| (a, b))
.count();
println!("{}", c);
}
Once you have that, you can just return the iterator in one of the many ways.
#[derive(Debug, Default)]
struct Thing<K, V>(Vec<Vec<(K, V)>>);
impl<K, V> Thing<K, V> {
fn iter_mut<'a>(&'a mut self) -> Box<Iterator<Item = (&'a K, &'a mut V)> + 'a> {
Box::new(self.0
.iter_mut()
.flat_map(|x| x.iter_mut())
.map(|&mut (ref a, ref mut b)| (a, b)))
}
}
fn main() {
let mut a = Thing::<u8, u8>::default();
let c = a.iter_mut().count();
println!("{}", c);
}
来源：https://stackoverflow.com/questions/42446844/mutable-iterator-for-vecveck-v