如何匹配字符串对字符串字面量?

我试图弄清楚如何匹配Rust中的字符串。

我最初尝试这样匹配，但我发现Rust不能隐式地从std::string:: string转换为&str。

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

这有一个错误:

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`

然后我尝试构造新的String对象，因为我找不到一个函数来将String转换为&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"),
    }
}

这给了我以下错误3次:

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

如何在Rust中匹配字符串?

当前回答

使用as_str()获取字符串切片

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

如果你从控制台获取输入并希望对其执行匹配，请确保在as_str()后调用trim()来从输入中删除转义字符，即“\n”。就像在

match stringthing.as_str().trim() {...}

2022-07-04 06:23:46

其他回答

你可以这样做:

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

Rust 1.7.0中还有一个as_str方法:

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

2015-09-25 20:56:57

你也可以

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

See:

std:: string:字符串 std::运维::Deref Deref胁迫

2016-01-22 22:28:19

你可以试试:

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

2020-05-03 07:33:45

编者注:这个答案适用于Rust 1.0之前的版本，在Rust 1.0中不能工作

你可以匹配一个字符串切片。

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

2014-08-19 12:44:40

你可以这样把String转换成&str:

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

2021-01-20 18:03:07

如何匹配字符串对字符串字面量?

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