string 如何使用字符串成员创建 Rust 结构？
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How to create a Rust struct with string members?
提问by vladimir
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.
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。
回答by BurntSushi5
If the string has to be owned by the struct, then you should use String. Alternatively, you could use an &strwith a static lifetime (i.e., the lifetime of the program). For example:
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&str具有静态生命周期（即程序的生命周期）的 。例如：
struct Foo {
bar: String,
baz: &'static str,
}
fn main() {
let foo = Foo {
bar: "bar".to_string(),
baz: "baz",
};
println!("{}, {}", foo.bar, foo.baz);
}
If the lifetime of the string is unknown, then you can parameterize Foowith a lifetime:
如果字符串的生命周期未知，那么您可以Foo使用生命周期参数化：
struct Foo<'a> {
baz: &'a str,
}
See also:
也可以看看：
If you're not sure whether the string will be owned or not (useful for avoiding allocations), then you can use borrow::Cow:
如果您不确定该字符串是否会被拥有（用于避免分配），那么您可以使用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);
}
Note that the Cowtype is parameterized over a lifetime. The lifetime refers to the lifetime of the borrowedstring (i.e., when it is a Borrowed). If you have a Cow, then you can use borrowand 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 borrowisn't required because of deref coercions. Namely, Cowvalues will dereference to their borrowed form automatically, so &*valwhere valhas type Cow<'a, str>will produce a &str.
请注意，该Cow类型是在整个生命周期内参数化的。生命周期是指借用字符串的生命周期（即，当它是 a 时Borrowed）。如果您有 a Cow，那么您可以使用borrow和获取 a &'a str，您可以使用它进行正常的字符串操作，而无需担心是否分配新字符串。通常，borrow由于 deref 强制，不需要显式调用 of 。即，Cow值将自动取消引用其借用形式，因此&*valwhere valhas typeCow<'a, str>将产生一个&str.
