rust - 我如何初始化一个数组，以便 Rust 知道它是一个 `String` s 而不是 `str` 的数组？

Question

我对 Rust 比较陌生，正在尝试执行以下操作:

pub fn route(request: &[String]) {
    let commands = ["one thing", "another thing", "something else"];

    for command in commands.iter() {
        if command == request {
            // do something
        } else {
            // throw error
        }
    }
}

当我尝试构建它时，出现编译器错误:

error[E0277]: the trait bound `&str: std::cmp::PartialEq<[std::string::String]>` is not satisfied
 --> src/main.rs:5:20
  |
5 |         if command == request {
  |                    ^^ can't compare `&str` with `[std::string::String]`
  |
  = help: the trait `std::cmp::PartialEq<[std::string::String]>` is not implemented for `&str`
  = note: required because of the requirements on the impl of `std::cmp::PartialEq<&[std::string::String]>` for `&&str`

Answer 1

你应该回去重新阅读The Rust Programming Language ，特别是 chapter on strings . String 和 &str 是两种不同的类型。

您可以在 number of ways 中创建 String , 但我通常使用 String::from:

let commands = [
    String::from("one thing"),
    String::from("another thing"),
    String::from("something else"),
];

但是，这是低效的，因为您每次都在分配内存。最好换一种方式，从 &String 到 &str。此外，这并不能解决您的问题，因为您正试图将单个值与集合进行比较。我们可以同时解决这两个问题:

let commands = ["one thing", "another thing", "something else"];

for command in commands.iter() {
    if request.iter().any(|r| r == command) {
        // do something
    } else {
        // throw error
    }
}

另见:

• What are the differences between Rust's `String` and `str`?
• How to create a String directly?