相关疑难解决方法(0)

我经常Option<String>从计算中获得一个,我想使用这个值或默认的硬编码值.

这对于一个整数来说是微不足道的:

let opt: Option<i32> = Some(3);
let value = opt.unwrap_or(0); // 0 being the default

但是使用a String和a &str,编译器会抱怨类型不匹配:

let opt: Option<String> = Some("some value".to_owned());
let value = opt.unwrap_or("default string");

这里的确切错误是:

error[E0308]: mismatched types
 --> src/main.rs:4:31
  |
4 |     let value = opt.unwrap_or("default string");
  |                               ^^^^^^^^^^^^^^^^
  |                               |
  |                               expected struct `std::string::String`, found reference
  |                               help: try using a conversion method: `"default string".to_string()`
  |
  = note: expected type `std::string::String`
             found type `&'static str`

一种选择是将字符串切片转换为拥有的String,如rustc所示:

let value …

use std::fs::File;
use std::io::Read;

pub struct Foo {
    maybe_file: Option<File>,
}

impl Foo {
    pub fn init(&mut self) {
        self.maybe_file = Some(File::open("/proc/uptime").unwrap());
    }

    pub fn print(&mut self) {
        let mut file = self.maybe_file.unwrap();
        let mut s = String::new();
        file.read_to_string(&mut s).unwrap();
        println!("Uptime: {}", s);
    }
}

fn main() {}

编译这将给我:

error[E0507]: cannot move out of borrowed content
  --> src/main.rs:14:24
   |
14 |         let mut file = self.maybe_file.unwrap();
   |                        ^^^^ cannot move out of borrowed content

为什么会这样？我该怎么做才能解决它？