更改闭包内的向量会导致"借用移动值"错误

Question

我正在尝试更改闭包内向量的元素：

pub struct Foo<'a, T> {
cb: Box<dyn FnMut(Vec<T>) + 'a>,
}
impl<'a, T> Foo<'a, T> {
pub fn new<F: FnMut(Vec<T>) + 'a>(cb: F) -> Self {
Self { cb: Box::new(cb) }
}
pub fn fun(&mut self, v: T) {
let vector = vec![v];
(self.cb)(vector);
}
}
fn main() {
let mut a = Vec::new();
let mut foo = Foo::new(move |v| {
for i in v {
a.push(i);
}
});
foo.fun(1);
println!("{:?}", a);
}

操场

我收到错误：

error[E0382]: borrow of moved value: `a`
--> src/main.rs:24:22
|
17 |     let mut a = Vec::new();
|         ----- move occurs because `a` has type `std::vec::Vec<i32>`, which does not implement the `Copy` trait
18 |     let mut foo = Foo::new(move |v| {
|                            -------- value moved into closure here
19 |         for i in v {
20 |             a.push(i);
|             - variable moved due to use in closure
...
24 |     println!("{:?}", a);
|                      ^ value borrowed here after move

我知道 Rust 无法在闭包中复制a的值，因为Vec没有实现 traitCopy，所以它必须移动它，并且a移动为可变性使其以后无法使用println!。

我是否正确存储了封盖？这里的使用寿命'a正确吗？我应该将向量包装成Box或Cell之类的东西吗？

Answer 1

这是解决方案(游乐场(：

pub struct Foo<'a, T> {
cb: Box<dyn FnMut(Vec<T>) + 'a>,
}
impl<'a, T> Foo<'a, T> {
pub fn new<F: FnMut(Vec<T>) + 'a>(cb: F) -> Self {
Self {
cb: Box::new(cb),
}
}
pub fn fun(&mut self, v: T) {
let vector = vec![v];
(self.cb)(vector);
}
}
fn main() {
let mut a = Vec::new();
// new scope to make sure that `foo` isn't alive when `a` is borrowed later
{
// no `move` to prevent moving `a` into the closure
let mut foo = Foo::new(|v| {
a = v.clone();
});
foo.fun(1);
}
println!("{:?}", a);
}