如何使用字符串成员创建 Rust 结构? [英] How to create a Rust struct with string members?

问题描述

我希望成员归结构所有.对不起，这个简单的问题，但我找不到一个例子.我正在寻找结构和实例化示例的正确声明.

I want the members to be owned by the struct. Sorry for the simple question, but I wasn't able to find an example. I'm looking for the correct declaration of a struct and instantiation examples.

推荐答案

如果字符串必须归结构所有，那么您应该使用 String.或者，您可以使用具有静态生命周期(即程序的生命周期)的 &str.例如:

If the string has to be owned by the struct, then you should use String. Alternatively, you could use an &str with a static lifetime (i.e., the lifetime of the program). For example:

struct Foo {
    bar: String,
    baz: &'static str,
}

fn main() {
    let foo = Foo {
        bar: "bar".to_string(),
        baz: "baz",
    };
    println!("{}, {}", foo.bar, foo.baz);
}

如果字符串的生命周期未知，那么你可以用生命周期参数化<​​code>Foo:

If the lifetime of the string is unknown, then you can parameterize Foo with a lifetime:

struct Foo<'a> {
    baz: &'a str,
}

另见:

• Rust 的 `String` 和 `str` 有什么区别?

如果您不确定字符串是否会被拥有(用于避免分配)，那么您可以使用 borrow::Cow:

If you're not sure whether the string will be owned or not (useful for avoiding allocations), then you can use borrow::Cow:

use std::borrow::Cow;

struct Foo<'a> {
    baz: Cow<'a, str>,
}

fn main() {
    let foo1 = Foo {
        baz: Cow::Borrowed("baz"),
    };
    let foo2 = Foo {
        baz: Cow::Owned("baz".to_string()),
    };
    println!("{}, {}", foo1.baz, foo2.baz);
}

请注意，Cow 类型是在整个生命周期内参数化的.生命周期指的是 borrowed 字符串的生命周期(即，当它是 Borrowed 时).如果你有一个 Cow，那么你可以使用 borrow 并得到一个 &'a str，你可以用它进行正常的字符串操作，而无需担心是否分配一个新字符串.通常，由于取消引用强制，不需要显式调用 borrow.即，Cow 值将自动取消引用它们的借用形式，因此 &*val 其中 val 的类型为 Cow<'a，str> 将产生一个 &str.

Note that the Cow type is parameterized over a lifetime. The lifetime refers to the lifetime of the borrowed string (i.e., when it is a Borrowed). If you have a Cow, then you can use borrow and get a &'a str, with which you can do normal string operations without worrying about whether to allocate a new string or not. Typically, explicit calling of borrow isn't required because of deref coercions. Namely, Cow values will dereference to their borrowed form automatically, so &*val where val has type Cow<'a, str> will produce a &str.

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