如何在Rust中匹配字符串与字符串文字？

Question

我正在试图弄清楚如何匹配StringRust.

我最初尝试过像这样的匹配,但我发现Rust不能暗中强制转换std::string::String为&str.

fn main() {
    let stringthing = String::from("c");
    match stringthing {
        "a" => println!("0"),
        "b" => println!("1"),
        "c" => println!("2"),
    }
}

这有错误:

error[E0308]: mismatched types
 --> src/main.rs:4:9
  |
4 |         "a" => println!("0"),
  |         ^^^ expected struct `std::string::String`, found reference
  |
  = note: expected type `std::string::String`
             found type `&'static str`

然后我尝试构造新String对象,因为我找不到将a String转换为a的函数&str.

fn main() {
    let stringthing = String::from("c");
    match stringthing {
        String::from("a") => println!("0"),
        String::from("b") => println!("1"),
        String::from("c") => println!("2"),
    }
}

这给了我3次以下错误:

error[E0164]: `String::from` does not name a tuple variant or a tuple struct
 --> src/main.rs:4:9
  |
4 |         String::from("a") => return 0,
  |         ^^^^^^^^^^^^^^^^^ not a tuple variant or struct

如何String在Rust中实际匹配s？

Answer 1

as_slice不推荐使用,您现在应该使用特征std::convert::AsRef:

match stringthing.as_ref() {
    "a" => println!("0"),
    "b" => println!("1"),
    "c" => println!("2"),
    _ => println!("something else!"),
}

请注意,您还必须显式处理catch-all案例.

Answer 2

你可以这样做:

match &stringthing[..] {
    "a" => println!("0"),
    "b" => println!("1"),
    "c" => println!("2"),
    _ => println!("something else!"),
}

as_strRust 1.7.0 还有一个方法:

match stringthing.as_str() {
    "a" => println!("0"),
    "b" => println!("1"),
    "c" => println!("2"),
    _ => println!("something else!"),
}

Answer 3

你也可以这样做

match &stringthing as &str {
    "a" => println!("0"),
    "b" => println!("1"),
    "c" => println!("2"),
    _ => println!("something else!"),
}

看到:

• std::string::String
• std::ops::Deref
• Deref胁迫

Answer 4

编者注:这个答案适用于1.0之前的Rust版本,并且在Rust 1.0中不起作用

您可以匹配字符串切片.

match stringthing.as_slice() {
    "a" => println!("0"),
    "b" => println!("1"),
    "c" => println!("2"),
    _ => println!("something else!"),
}