为什么选择<String> .as_ref() 不解除对 Option&lt;&amp;str&gt; 的引用？

Question

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");

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`

Answer 1

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

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

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

Instead, you can add in String::as_str :

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

Or the shorter:

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

As of Rust 1.40, you can also use Option::as_deref .

maybe_string.as_deref().unwrap_or("none");

See also:

• Converting from Option<String> to Option<&str>