相关疑难解决方法(0)

我从此Rust代码收到意外错误：

struct Container<'a> {
    x: &'a i32,
}

trait Reply {}
impl Reply for i32 {}

fn json<T>(_val: &T) -> impl Reply {
    3
}

fn f() -> impl Reply {
    let i = 123;
    let a = Container { x: &i };
    json(&a)
}

操场

错误是：

error[E0597]: `i` does not live long enough
  --> src/lib.rs:14:28
   |
12 | fn f() -> impl Reply {
   |           ---------- opaque type requires that `i` is borrowed for `'static`
13 |     let i …

我有一个关联类型的特征:

pub trait Speak {
    type Error;
    fn speak(&self) -> Result<String, Self::Error>;
}

该特征的实现:

#[derive(Default)]
pub struct Dog;

impl Speak for Dog {
    type Error = ();
    fn speak(&self) -> Result<String, Self::Error> {
        Ok("woof".to_string())
    }
}

并且函数返回该实现的实例:

pub fn speaker() -> impl Speak {
    Dog::default()
}

我知道在这个例子中我只能Dog用作返回类型,但在我的实际代码中我必须使用impl Speak(上面的函数实际上是由宏生成的).

据我了解,该impl Trait符号让出这实际上返回具体类型编译器的身影,所以我希望下面的函数正确编译,因为speaker()回报Dog,并且Dog::Error是类型():

fn test() -> Result<String, ()> {
    speaker().speak()
}

操场

相反,我收到以下错误:

error[E0308]: mismatched types
  --> src/lib.rs:21:5
   | …