Rust - 生命周期 - 了解对 self 的可变引用的生命周期错误

Question

我敢肯定这是重复的，但是我找不到一个与我的问题相匹配的问题，因为我有几个额外的要求，因为我必须遵守某些无法控制的特征。

这是我的代码。对于那种令人费解的示例，我深表歉意，但这是我尝试使用serde库来实现自定义序列化格式的最小化。

// Doesn't really matter what this struct contains, it just needs an owning method
struct SideStruct;
impl SideStruct {
    fn something_side<A: TraitA>(&self, aval: A) {
        println!("something sideways :)");
        aval.something_a(42)
    }
}

trait TraitA {
    fn something_a(&mut self, data: u32); // this would be the meat of my logic
}

// Note that this struct has an explicit lifetime
struct MainStruct<'a> {
    refr: &'a mut u32
}

// Note that I implement for a mutable reference to MainStruct
impl<'a> TraitA for &'a mut MainStruct<'a> {
    fn something_a(&mut self, data: u32) {
        // Completely arbitrary, can safely ignore this function body
        *self.refr += data;
        println!("We're finally doing something: {}", self.refr);
    }
}

// Implementing for MainStruct itself
impl<'a> MainStruct<'a> {
    // Note, I can't change the signature for this function because it implements a trait
    fn something_indirect(&mut self, ss: &SideStruct) {
        // here is where the error occurs!
        ss.something_side(self)
    }
}

fn main() {
    let mut base_val: u32 = 42;
    let ss = SideStruct {};
    let mut main_val = MainStruct { refr: &mut base_val };
    main_val.something_indirect(&ss);
}

这是我遇到的错误：

error[E0495]: cannot infer an appropriate lifetime due to conflicting requirements
  --> src/main.rs:28:27
   |
28 |         ss.something_side(self)
   |                           ^^^^
   |
note: first, the lifetime cannot outlive the anonymous lifetime defined here...
  --> src/main.rs:27:27
   |
27 |     fn something_indirect(&mut self, ss: &SideStruct) {
   |                           ^^^^^^^^^
note: ...so that the expression is assignable
  --> src/main.rs:28:27
   |
28 |         ss.something_side(self)
   |                           ^^^^
   = note: expected `&mut MainStruct<'a>`
              found `&mut MainStruct<'a>`
note: but, the lifetime must be valid for the lifetime `'a` as defined here...
  --> src/main.rs:26:6
   |
26 | impl<'a> MainStruct<'a> {
   |      ^^
note: ...so that the types are compatible
  --> src/main.rs:28:12
   |
28 |         ss.something_side(self)
   |            ^^^^^^^^^^^^^^
   = note: expected `<&mut MainStruct<'a> as TraitA>`
              found `<&mut MainStruct<'_> as TraitA>`

For more information about this error, try `rustc --explain E0495`.

当编译器指出注意：首先，寿命无法超过此处定义的匿名寿命...时，我不知道编译器的含义。这是否意味着某些约束迫使自我不超越方法sometsing_indirect？那没有道理。还有消息使表达式可分配使我感到困惑。 Mainstruct不应分配something_side在调用时不应分配吗？由于我实现了traita，用于mainscrupt的可变引用，因此我不应该以something> mainsstruct的可变引用来调用someways_side通过传递self？无论如何，感谢您的帮助，祝您有美好的一天！

Answer 1

问题是，要使用定义的方法，您必须将mainsstruct借用具有匿名寿命可变的。在您编写的代码中，您不仅在Mainsstruct中借入'a，还借入Mainsstruct本身。这是不必要的，因为借款具有推断的寿命。您可以通过在特质inplant中删除'A来解决此问题

impl<'a> TraitA for &mut MainStruct<'a> {
    /*...*/
}

，这应该做完全相同的事情，但要删除错误。该错误试图告诉您您编写的代码是错误的，因为它使用结构本身中的寿命借用了Mainsstruct。