如何返回一個使用本地聲明變量的結構的實例

Question

我想弄清楚如何在本地聲明一個變量並在正在返回的值中使用它。以下是造成問題的原因

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}; 
} 

的代碼，我有以下結構作爲輔助

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

編譯器會引發以下錯誤

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 

我不是特別想要改變結構，更願意得到一些關於我不瞭解的東西的建議。

Answer 1

只需將Topping.name從&str切換爲String即可。

您不能返回對read_line（a String）結果的引用，因爲String將在handle_topping的末尾丟失。但是，您可以將String的所有權移動到結構中並返回Topping {name: String, veg: bool, ...}。