[英]Rust borrow checker and early returns
Rust-lang Playground link Rust-lang Playground 链接
struct Foo {
val: i32
}
impl Foo {
pub fn maybe_get(&mut self) -> Option<&mut i32> {
Some(&mut self.val)
}
pub fn definitely_get(&mut self) -> &mut i32 {
{ // Add closure to ensure things have a chance to get dropped
if let Some(val) = self.maybe_get() {
// Explicit return to avoid potential scope sharing with an else block or a match arms.
return val;
}
}
// One would think any mutable references would not longer be at play at this point
&mut self.val
}
}
I have some code that's similar but more complicated than what is provided above that I've been fighting with for quite a while.我有一些与上面提供的代码相似但更复杂的代码,这些代码我已经使用了很长时间。 The borrow checker is unhappy with the implementation of definitely_get
and has the following error借用检查器对 absolute_get 的实现definitely_get
,并出现以下错误
error[E0499]: cannot borrow `self.val` as mutable more than once at a time
--> src/main.rs:19:9
|
10 | pub fn definitely_get(&mut self) -> &mut i32 {
| - let's call the lifetime of this reference `'1`
11 | {
12 | if let Some(val) = self.maybe_get() {
| ---------------- first mutable borrow occurs here
13 | return val;
| --- returning this value requires that `*self` is borrowed for `'1`
...
19 | &mut self.val
| ^^^^^^^^^^^^^ second mutable borrow occurs here
It seems unreasonable for there to be no way to implement fallback logic with a mutable reference in Rust so I can't imagine there isn't a way.没有办法在 Rust 中使用可变引用来实现回退逻辑似乎是不合理的,所以我无法想象没有办法。
I've managed to fix this with an unfortunately expensive alternative implementation due to how maybe_get
is implemented in my non-trivial example.由于maybe_get
在我的非平凡示例中是如何实现的,我已经设法用一个不幸的昂贵的替代实现来解决这个问题。
impl Foo {
pub fn has_maybe_val(&self) -> bool {
// Non-trivial lookup...
true
}
pub fn maybe_get(&mut self) -> Option<&mut i32> {
// Same non-trivial lookup...
Some(&mut self.val)
}
pub fn definitely_get(&mut self) -> &mut i32 {
if self.has_maybe_val() {
self.maybe_get().unwrap() // Ouch!
} else {
&mut self.val
}
}
}
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