如何 return 使用局部声明变量的结构实例

Question

我想知道如何在本地声明一个变量并在返回的值中使用它。以下是导致问题的代码

use std::io;
use std::string::String;
use std::io::Write; // Used for flush implicitly
use topping::Topping;

pub fn read_line(stdin: io::Stdin, prompt: &str) -> String {
    print!("{}", prompt);
    let _ = io::stdout().flush();
    let mut result = String::new();
    let _ = stdin.read_line(&mut result);
    return result;
}

pub fn handle_topping<'a>(stdin: io::Stdin) -> Topping<'a>{
    let name = read_line(stdin, "Topping name: ");
    //let price = read_line(stdin, "Price: ");
    return Topping {name: &name, price: 0.7, vegetarian: false};
}

我有以下结构作为助手

pub struct Topping<'a> {
    pub name: &'a str,
    pub vegetarian: bool,
    pub price: f32,
}

编译器抛出以下错误

error: `name` does not live long enough
  --> src/helpers.rs:17:28
   |
17 |     return Topping {name: &name, price: 0.7, vegetarian: false};
   |                            ^^^^ does not live long enough
18 | }
   | - borrowed value only lives until here
   |
note: borrowed value must be valid for the lifetime 'a as defined on the body at 14:58...
  --> src/helpers.rs:14:59
   |
14 |   pub fn handle_topping<'a>(stdin: io::Stdin) -> Topping<'a>{
   |  ___________________________________________________________^ starting here...
15 | |     let name = read_line(stdin, "Topping name: ");
16 | |     //let price = read_line(stdin, "Price: ");
17 | |     return Topping {name: &name, price: 0.7, vegetarian: false};
18 | | }
   | |_^ ...ending here

我并不是特别想更改结构，更愿意就我不理解的内容征求一些建议。

Answer 1

只需将 Topping.name 从 &str 切换为 String。

您不能 return 引用 read_line 的结果（一个 String），因为 String 将在 [=16] 的末尾被删除=].但是，您可以将 String 的所有权移至结构中，并将 return 移至 Topping {name: String, veg: bool, ...}.

如何 return 使用局部声明变量的结构实例

How to return a instance of a struct that uses a locally declared variable

rust

rust-cargo