如何创建包含字符串成员的 Rust 结构？

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);
}

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

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

See also:

• What are the differences between Rust's `String` and `str`?

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);
}

Note that the Cow type is parameterized over a lifetime. The lifetime refers to the lifetime of the Borrowed1 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 Borrowed0.