为什么Option< String> .as_ref()不取消引用Option<& str&gt ;?

问题描述

我期望这两个代码示例具有相同的结果:

I expect the same result for both of these code samples:

let maybe_string = Some(String::from("foo"));
let string = if let Some(ref value) = maybe_string { value } else { "none" };

let maybe_string = Some(String::from("foo"));
let string = maybe_string.as_ref().unwrap_or("none");

第二个示例给我一个错误:

The second sample gives me an error:

error[E0308]: mismatched types
 --> src/main.rs:3:50
  |
3 |     let string = maybe_string.as_ref().unwrap_or("none");
  |                                                  ^^^^^^ expected struct `std::string::String`, found str
  |
  = note: expected type `&std::string::String`
             found type `&'static str`

推荐答案

因为 Option::as_ref 已定义:

Because that's how Option::as_ref is defined:

impl<T> Option<T> {
    fn as_ref(&self) -> Option<&T>
}

由于您有Option<String>，所以生成的类型必须为Option<&String>.

Since you have an Option<String>, then the resulting type must be Option<&String>.

相反，您可以添加 String::as_str :

Instead, you can add in String::as_str:

maybe_string.as_ref().map(String::as_str).unwrap_or("none");

或更短:

maybe_string.as_ref().map_or("none", String::as_str);

最终，您还可以使用 Option::deref .

Eventually, you can also use Option::deref.

另请参阅:

• 从Option< String>转换为到选项<& str>

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