如何告诉编译器我正在返回的枚举变体总是没有生命周期？

Question

下面的代码将无法编译,因为编译器认为我不应该分配,t1因为它是借用的,但实际上函数always_returns_no_lifetime将始终返回实际上没有生命周期的枚举的变体,所以它可以我修改t1.如何让编译器理解这一点,或者我应该如何重新组织我的代码以避免发生此错误？

#[derive(Clone)]
enum Types<'a> {
    NoLifetime(i32),
    AlsoNoLifetime(i32),
    AlsoAlsoNoLifetime(i32),
    HasLifetime(&'a str)
}

fn always_returns_no_lifetime<'a>(some_type: &'a Types) -> Types<'a> {
    match *some_type {
        Types::HasLifetime(text) => panic!("I only return the type that has no lifetime"),
        _ => some_type.clone()
    }
}


fn main() {
    let mut t1 = Types::NoLifetime(20);

    let copy = always_returns_no_lifetime(&t1);

    t1 = Types::NoLifetime(30);

}

操场

错误是:

error[E0506]: cannot assign to `t1` because it is borrowed
  --> src/main.rs:23:5
   |
21 |     let copy = always_returns_no_lifetime(&t1);
   |                                            -- borrow of `t1` occurs here
22 |     
23 |     t1 = Types::NoLifetime(30);
   |     ^^^^^^^^^^^^^^^^^^^^^^^^^^ assignment to borrowed `t1` occurs here

Answer 1

函数的返回类型是错误的.如果保证返回值没有任何生命周期,那么它应该被标记为这样,并且不会被束缚到任意生命周期:

fn always_returns_no_lifetime(...) -> Types<'static>;

通过此更改,您实际上不再需要任何输入生命周期,因为它们仅用于绑定输入和输出,从而导致以下签名:

fn always_returns_no_lifetime(some_type: &Types) -> Types<'static>;

不幸的是,这意味着它clone现在不在桌面中,因为它克隆了生命周期,因此实现也必须改变:

fn always_returns_no_lifetime(some_type: &Types) -> Types<'static> {
    match *some_type {
        Types::HasLifetime(_)
            => panic!("I only return values that have no lifetime"),
        Types::NoLifetime(i) => Types::NoLifetime(i),
        Types::AlsoNoLifetime(i) => Types::AlsoNoLifetime(i),
        Types::AlsoAlsoNoLifetime(i) => Types::AlsoAlsoNoLifetime(i),
    }
}

可以在以下示例中演示此实现的好处:

fn tie<'a>(text: &'a str) -> Types<'a> {
    if text[0] == 'a' {
        Types::HasLifetime(text)
    } else {
        Types::NoLifetime(0)
    }
}

fn main() {
    let no_lifetime = {
        let string = String::from("Hello, world");
        let has_lifetime = tie(&*string);
        always_returns_no_lifetime(&has_lifetime)
    };

    //  Requires deriving Debug, all structs really should...
    println!("{:?}", no_lifetime);
}

如果您在不需要时保留生命周期,则无法编译此示例,这是一个不必要的限制.