在Rust中的match语句中解构Option<Box<_>>
我有一个大物体需要装在另一个物体中,但我不一定总是需要它。所以我想使用 if 语句来获取可选的盒装 TempStructure,但我不确定如何同时解构和取消引用。
例子:
pub struct TempStructure {
lots_of_data: [u64; 64],
}
pub struct Structure {
pending_removal: Option<Box<(TempStructure, bool)>>,
}
impl Structure {
pub fn do_somthing(&mut self) {
// How do I destructure the Option and dereference the Box to get TempStructure?
if let Some((temp_structure, some_boolean)) = self.pending_removal.take() {
// Do something with temp_structure and some_boolean
}
}
}
当我这样做 ^^^ 我得到一个expected struct `std::boxed::Box`, found tuple错误。
回答
匹配后取消引用框:
if let Some(inner) = self.pending_removal.take() {
let (temp_structure, some_boolean) = *inner;
// Do something with temp_structure and some_boolean
}
(操场)
如果您认为这有点笨拙,那么您是对的。在夜间,您可以使用不稳定box_patterns功能为此启用更好的语法(尽管这可能永远不会稳定):
if let Some(box (temp_structure, some_boolean)) = self.pending_removal.take() {
// Do something with temp_structure and some_boolean
}
(操场)
回答
您可以通过在.as_deref()之后添加 a 来解决此问题take():
pub struct TempStructure {
lots_of_data: [u64; 64],
}
pub struct Structure {
pending_removal: Option<Box<(TempStructure, bool)>>,
}
impl Structure {
pub fn do_somthing(&mut self) {
// How do I destructure the Option and dereference the Box to get TempStructure?
if let Some((temp_structure, some_boolean)) = self.pending_removal.take().as_deref() {
// Do something with temp_structure and some_boolean
}
}
}
Box<T>取消引用&T,as_deref()取消引用 的值Option,因此它会给您一个&T超出您的Option<Box<T>>.
编辑:另一种选择是取消引用 Box 以将值移出它:
if let Some((temp_structure, some_boolean)) = self.pending_removal.take().map(|boxed| *boxed) {
// assignments just for demonstration purposes that we now get
// owned values rather than references.
let _: TempStructure = temp_structure;
let _: bool = some_boolean;
}
-
I don't think comments are supposed to have multiline code blocks.
https://meta.stackexchange.com/questions/39260/how-do-i-post-code-in-comments
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