Rust 为什么在 trait 中返回 `Self` 有效，但返回 `Option` 需要 `Sized`

这个trait 定义可以很好地编译:

trait Works {
    fn foo() -> Self;
}

然而，这会导致以下错误:

trait Errors {
    fn foo() -> Option<Self>;
}

error[E0277]: the size for values of type `Self` cannot be known at compilation time
 --> src/lib.rs:6:5
  |
6 |     fn foo() -> Option<Self>;
  |     ^^^^^^^^^^^^^^^^^^^^^^^^^ doesn't have a size known at compile-time
  |
  = help: the trait `std::marker::Sized` is not implemented for `Self`
  = note: to learn more, visit <https://doc.rust-lang.org/book/second-edition/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>
  = help: consider adding a `where Self: std::marker::Sized` bound
  = note: required by `std::option::Option`

有了: Sized个supertrait绑定，它就可以工作了.

我知道，Self种类型的trait 并不是自动地就一定是Sized种.我知道Option<Self>不能被返回(通过堆栈)，除非它被调整大小(反过来，需要调整Self的大小).然而，Self作为返回类型也是一样的，对吗？它也不能存储在堆栈上，除非大小合适.

Why doesn't the first trait definition already trigger that error?

(100 is related, but it doesn't answer my exact question – unless I didn't understand it.)