我如何迭代 Bevy 查询并保留对迭代值的引用以便我以后可以使用它?
How can I iterate on a Bevy Query and keep a reference to the iterated value so that I can use it later?
我在 empty
变量中有借用,我想延长它的寿命。在注释代码块中,我尝试解决它,但参考不再可用。我必须再次循环以找到匹配项才能对其进行操作。
如何循环查询以寻找最佳匹配,然后在知道它是最佳匹配后对其采取行动,而不必再次循环查找?
use bevy::prelude::*;
struct Person;
struct Name(String);
fn main() {
App::build()
.add_default_plugins()
.add_startup_system(startup.system())
.add_system(boot.system())
.run();
}
fn boot(mut query: Query<(&Person, &mut Name)>) {
let mut temp_str = String::new();
let mut empty: Option<&mut Name> = None;
for (_p, mut n_val) in &mut query.iter() {
if n_val.0.to_lowercase() > temp_str.to_lowercase() {
temp_str = n_val.0.clone();
empty = Some(&mut n_val);
}
}
println!("{}", temp_str);
if let Some(n) = empty {
// ...
}
// for (_p, mut n_val) in &mut query.iter() {
// if n_val.0 == temp_str {
// n_val.0 = "a".to_string();
// }
// }
}
fn startup(mut commands: Commands) {
commands
.spawn((Person, Name("Gene".to_string())))
.spawn((Person, Name("Candace".to_string())))
.spawn((Person, Name("Zany".to_string())))
.spawn((Person, Name("Sarah".to_string())))
.spawn((Person, Name("Carl".to_string())))
.spawn((Person, Name("Robert".to_string())));
}
Cargo.toml:
[package]
name = "sample"
version = "0.1.0"
authors = [""]
edition = "2018"
# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
[dependencies]
bevy = "0.1.3"
具体错误:
error[E0716]: temporary value dropped while borrowed
--> src/main.rs:17:33
|
17 | for (_p, mut n_val) in &mut query.iter() {
| ^^^^^^^^^^^-
| | |
| | temporary value is freed at the end of this statement
| creates a temporary which is freed while still in use
...
24 | if let Some(n) = empty {
| ----- borrow later used here
|
= note: consider using a `let` binding to create a longer lived value
error[E0597]: `n_val` does not live long enough
--> src/main.rs:20:26
|
20 | empty = Some(&mut n_val);
| ^^^^^^^^^^ borrowed value does not live long enough
21 | }
22 | }
| - `n_val` dropped here while still borrowed
23 | println!("{}", temp_str);
24 | if let Some(n) = empty {
| ----- borrow later used here
,但这不是您的问题,错误显示 temporary value dropped while borrowed
,因此您必须延长临时文件的使用寿命。
如果您想知道什么是临时的,编译器还会在错误消息中指出(字面意思):query.iter()
。这是一个函数调用,returned 值没有绑定到任何东西,因此编译器为此创建了一个临时值。然后使用对该临时文件的引用进行迭代。当 for
循环结束时,临时文件将被丢弃,并且任何对它的引用生命周期都将到期。
解决方案是将临时变量绑定到局部变量。这样你就可以将对象的生命周期延长到变量的范围内:
let mut iter = query.iter();
for (_p, n_val) in &mut iter {
if n_val.0.to_lowercase() > temp_str.to_lowercase() {
temp_str = n_val.0.clone();
empty = Some(n_val);
}
}
PS:我发现遍历 &mut iter
的模式很奇怪。我希望 iter()
的 return 直接实现 Iterator
或 IntoIterator
,但它看起来像 this is not the case.
我在 empty
变量中有借用,我想延长它的寿命。在注释代码块中,我尝试解决它,但参考不再可用。我必须再次循环以找到匹配项才能对其进行操作。
如何循环查询以寻找最佳匹配,然后在知道它是最佳匹配后对其采取行动,而不必再次循环查找?
use bevy::prelude::*;
struct Person;
struct Name(String);
fn main() {
App::build()
.add_default_plugins()
.add_startup_system(startup.system())
.add_system(boot.system())
.run();
}
fn boot(mut query: Query<(&Person, &mut Name)>) {
let mut temp_str = String::new();
let mut empty: Option<&mut Name> = None;
for (_p, mut n_val) in &mut query.iter() {
if n_val.0.to_lowercase() > temp_str.to_lowercase() {
temp_str = n_val.0.clone();
empty = Some(&mut n_val);
}
}
println!("{}", temp_str);
if let Some(n) = empty {
// ...
}
// for (_p, mut n_val) in &mut query.iter() {
// if n_val.0 == temp_str {
// n_val.0 = "a".to_string();
// }
// }
}
fn startup(mut commands: Commands) {
commands
.spawn((Person, Name("Gene".to_string())))
.spawn((Person, Name("Candace".to_string())))
.spawn((Person, Name("Zany".to_string())))
.spawn((Person, Name("Sarah".to_string())))
.spawn((Person, Name("Carl".to_string())))
.spawn((Person, Name("Robert".to_string())));
}
Cargo.toml:
[package]
name = "sample"
version = "0.1.0"
authors = [""]
edition = "2018"
# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
[dependencies]
bevy = "0.1.3"
具体错误:
error[E0716]: temporary value dropped while borrowed
--> src/main.rs:17:33
|
17 | for (_p, mut n_val) in &mut query.iter() {
| ^^^^^^^^^^^-
| | |
| | temporary value is freed at the end of this statement
| creates a temporary which is freed while still in use
...
24 | if let Some(n) = empty {
| ----- borrow later used here
|
= note: consider using a `let` binding to create a longer lived value
error[E0597]: `n_val` does not live long enough
--> src/main.rs:20:26
|
20 | empty = Some(&mut n_val);
| ^^^^^^^^^^ borrowed value does not live long enough
21 | }
22 | }
| - `n_val` dropped here while still borrowed
23 | println!("{}", temp_str);
24 | if let Some(n) = empty {
| ----- borrow later used here
temporary value dropped while borrowed
,因此您必须延长临时文件的使用寿命。
如果您想知道什么是临时的,编译器还会在错误消息中指出(字面意思):query.iter()
。这是一个函数调用,returned 值没有绑定到任何东西,因此编译器为此创建了一个临时值。然后使用对该临时文件的引用进行迭代。当 for
循环结束时,临时文件将被丢弃,并且任何对它的引用生命周期都将到期。
解决方案是将临时变量绑定到局部变量。这样你就可以将对象的生命周期延长到变量的范围内:
let mut iter = query.iter();
for (_p, n_val) in &mut iter {
if n_val.0.to_lowercase() > temp_str.to_lowercase() {
temp_str = n_val.0.clone();
empty = Some(n_val);
}
}
PS:我发现遍历 &mut iter
的模式很奇怪。我希望 iter()
的 return 直接实现 Iterator
或 IntoIterator
,但它看起来像 this is not the case.