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我有一些平面缓冲区 IDL：

table Filter {
  text: string (required);
}

生成以下.rs：

...
pub struct Filter<'a> {
  pub _tab: flatbuffers::Table<'a>,
}
...

我正在尝试为其添加一些特征 impl：

pub trait TFilter {
    fn as_any(&self) -> &dyn Any;
    fn as_any_mut(&mut self) -> &mut dyn Any;
}
...
impl TFilter for crate::schema_generated::Filter<'_> {
    fn as_any(&self) -> &dyn Any { self }
    fn as_any_mut(&mut self) -> &mut dyn Any { self }
}

我得到以下信息：

cannot infer an appropriate lifetime due to conflicting requirements

note: ...so that the type `schema_generated::Filter<'_>` will meet its required lifetime bounds
note: but, the lifetime must be valid for the static lifetime...
note: ...so that the expression is assignable

如果我添加生命周期：

// flatbuffers-based impl
impl<'a> TFilter for crate::schema_generated::Filter<'a> {
    fn as_any(&self) -> &'a dyn Any { self }
    fn as_any_mut(&mut self) -> &'a mut dyn Any { self }
}

我得到以下信息：

the type `schema_generated::Filter<'a>` does not fulfill the required lifetime

note: type must satisfy the static lifetimerustc(E0477)

我不确定我明白这是什么意思。如果我理解正确，我需要标记as_any生活时间不超过crate::schema_generated::Filter，对吗？

有什么线索吗？需求从何'static而来？

PS。我找到了以下内容，但我不确定它是否相关。

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0 回答 0