Why does Option.as_ref() not deref to Opti

问题:

I expect same result for both 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:

expected type `&std::string::String`
found type `&'static str`

回答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);

Eventually, you can also use Option::deref.

See also:

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