struct - 如何为包含引用的结构实现 AsRef

Question

如果我有一个包含这样的引用的结构:

struct Struct<'a> {
    reference: &'a str
}

如何为 Struct 实现 AsRef？我试过这个:

impl<'a> AsRef<Struct<'a>> for Struct<'a> {
    fn as_ref(&self) -> &Struct {
        self
    }
}

但未能满足编译要求:

cannot infer an appropriate lifetime for lifetime parameter in generic type due to conflicting requirements

Answer 1

与 fn as_ref(&self) -> &Struct编译器必须推断返回类型中的(隐式)泛型生命周期，但未能这样做。编译器期待一个 Struct<'a>但签名 promise 一个免费参数。这就是为什么你得到

      expected fn(&Struct<'a>) -> &Struct<'a>
         found fn(&Struct<'a>) -> &Struct<'_>  // '_ is some anonymous lifetime,
                                               // which needs to come from somewhere

解决方法是修改签名返回Struct<'a>而不是 Struct .更短更清晰:

impl<'a> AsRef<Struct<'a>> for Struct<'a> {
    fn as_ref(&self) -> &Self {  // Self is Struct<'a>, the type for which we impl AsRef
        self
    }
}