我有一个简单的结构及其两个实例,如下所示:
#[derive(Debug)]
struct User {
first: String,
last: String,
age: u32,
}
let u1 = User {
first: String::from("John"),
last: String::from("Doe"),
age: 22,
};
let u2 = User {
first: String::from("Mary"),
..u1
};
println!("user: {:#?}", u1);
错误消息:
error[E0382]: borrow of moved value: `u1`
--> src/main.rs:20:29
|
15 | let u2 = User {
| ______________-
16 | | first: String::from("Mary"),
17 | | ..u1
18 | | };
| |_____- value moved here
19 |
20 | println!("user: {:#?}", u1);
| ^^ value borrowed here after partial move
|
= note: move occurs because `u1.last` has type `std::string::String`, which does not implement the `Copy` trait
我试图将其修改为..&u1
,希望它能通过借入检查,以便我可以将基本结构(u1)扩展到u2,但无济于事,想知道我是否想在这里做些什么?
我理解这是因为u1.last
是String
,因此需要传递引用,但是我不确定如何在此情况下使其正常工作。
您的User
类型包含类型String
,该类型拥有它拥有的字符串数据(并且不表示Copy
),这就是为什么两个用户不能在内存中指向相同的名称。] >
您可能想要的解决方案:
#[derive(Debug, Clone)] struct User { first: String, last: String, age: u32, } fn main() { let u1 = User { first: String::from("John"), last: String::from("Doe"), age: 22, }; let u2 = User { first: String::from("Mary"), ..u1.clone() // Copy the strings into the new user // (it also copies first, which is then thrown away? Maybe the compiler optimizes this away) }; println!("user: {:#?}", u1); }
但是如果您真的想让两个用户指向内存中的相同名称(肯定不会),有两个选择:
您可以将String
更改为&'static str
。但是,这意味着在编译时必须指定它。 (您不能在运行时输入用户名并将其存储在用户中)
#[derive(Debug)]
struct User {
first: &'static str,
last: &'static str,
age: u32,
}
fn main() {
let u1 = User {
first: "John",
last: "Doe",
age: 22,
};
let u2 = User {
first: "Mary",
..u1
};
println!("user: {:#?}", u1);
}
Rc
来跟踪对内存的引用。这样,您不必担心一生以及谁拥有什么。 (尽管花费很少的运行时间)use std::rc::Rc;
#[derive(Debug, Clone)]
struct User {
first: Rc<String>,
last: Rc<String>,
age: u32,
}
fn main() {
let u1 = User {
first: Rc::new(String::from("John")),
last: Rc::new(String::from("Doe")),
age: 22,
};
let u2 = User {
first: Rc::new(String::from("Mary")),
..u1.clone() // Clone the references, not the actual string. For strings with just a couple characters, the time difference is completely negligible)
};
println!("user: {:#?}", u1);
}
Rc<Mutex<String>>
。