如何在Bevy查询上进行迭代并保留对迭代值的引用,以便以后使用?

问题描述

我在empty变量中有借项,我想延长其寿命。在注释的代码块中,我尝试解决它,但是该参考不再可用。我必须再次遍历循环才能找到匹配项,以便对其进行操作。

如何遍历查询以查找最匹配项,然后在知道最匹配项时对其执行操作,而无需遍历以再次找到它?

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"

特定错误

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

解决方法

You cannot extend the lifetime of a reference,但这不是您要解决的问题,错误提示为temporary value dropped while borrowed,因此您必须延长临时文件的寿命。

如果您想知道什么是临时的,编译器还会在错误消息query.iter()中指向(从字面上看)。这是一个函数调用,返回的值未绑定任何内容,因此编译器为此创建了一个临时值。然后,使用对该临时目录的引用进行迭代。 for循环结束时,该临时项将被删除,并且对其的任何参考生命期都将终止。

解决方案是将临时绑定到局部变量。这样,您可以将对象的生存期扩展到变量的范围:

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:我发现在&mut iter上进行迭代非常奇怪。我希望返回iter()可以直接实现IteratorIntoIterator,但看起来像this is not the case