返回几乎相同特征的特征的通用方法

Question

我想创建这个特征：

trait Functor<A> {
    fn map<B, F>(self, f: F) -> Functor<B>
    where
        F: Fn(A) -> B;
}

我有此错误：

warning: trait objects without an explicit `dyn` are deprecated
 --> src/lib.rs:2:33
  |
2 |     fn map<B, F>(self, f: F) -> Functor<B>
  |                                 ^^^^^^^^^^ help: use `dyn`: `dyn Functor<B>`
  |
  = note: `#[warn(bare_trait_objects)]` on by default

error: associated item referring to unboxed trait object for its own trait
 --> src/lib.rs:2:33
  |
1 | trait Functor<A> {
  |       ------- in this trait
2 |     fn map<B, F>(self, f: F) -> Functor<B>
  |                                 ^^^^^^^^^^
  |
help: you might have meant to use `Self` to refer to the implementing type
  |
2 |     fn map<B, F>(self, f: F) -> Self
  |                                 ^^^^

error[E0038]: the trait `Functor` cannot be made into an object
 --> src/lib.rs:2:33
  |
1 | trait Functor<A> {
  |       ------- this trait cannot be made into an object...
2 |     fn map<B, F>(self, f: F) -> Functor<B>
  |        ---                      ^^^^^^^^^^ the trait `Functor` cannot be made into an object
  |        |
  |        ...because method `map` has generic type parameters
  |
  = help: consider moving `map` to another trait

似乎没有办法实现这种事情。

我宁愿返回类似Functor<B>之类的内容，也不希望从map方法中返回Self<B>，因为我希望实现返回自身（使用通用B而不是A），但我不知道该怎么写。

起初，我尝试使用关联类型而不是通用类型，因为我认为它更合适，但效果并不更好。