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Step-by-step, how to understand the describe() function example

(assuming I understand this, which is an if, a small if, but an if nonetheless)

Part A

We can implement methods directly on shared reference types themselves. There is no deref coercion then when called on a shared reference that is of that shared reference type (playground)

In a sense then the method implemented, when called on a shared reference, takes ownership of the shared reference, per se, not the value the shared reference points to.

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1. Consider two traits

   1. From<String>

   2. From<&'static str>

2. The enum Cow<'static, str> Cow<str> implements both traits

   1. Cow has the associated function
      fn from(s: String) → Cow<'static, str>Cow<str>

   2. Cow has another associated function
      fn from(s: &'static str) → Cow<'static, str>Cow<str>

3. Consider another trait

   1. Into<Cow<'static, str>>Into<Cow<str>>

4. A From implementation implies that there is a corresponding Into implementation. There is a blanket implementation of Into in the standard library. My guess is that the following code is the blanket implementation

   Code Block
   language rust
   impl<T,U> Into<U> for T where U: From<T> { fn into(self) -> U { U::from(self) } }

5. Therefore String implements the Into<Cow<'static,str>> Into<Cow<str>> trait

   Code Block
   language rust
   implimpl<'a> Into<Cow<'statica, str>> for String { fn into(self) -> Cow<'statica, str> { ... } }

   Consider the additional trait

6. String automatically implements Into, as does str, given that the From traits are implemented

7. Both String and str implement Into with a method of the following form

   fn into(self) → Cow<a', str>

   Also &str implements the Into<Cow<str>> trait

   Code Block
   impl<'a> Into<Cow<'a, str>> for &'a str { fn into(self) -> Cow<'a, str> { ... } }

8. In the first four arms of the match expression, there are string literals upon which the method into() is called. We know the methods must have a return type of these method calls must return a Cow<'static, str>.

9. A string literal is of type &str, a shared reference to a string slice

10. A string literal has a static lifetime.

11. When applying the method Therefore the result of calling the into() method on a string literal is a Cow<'static, the string literal is dereferenced into a string slice strThen the method into() can be applied on the str. We have already seen that this method existsstr>

12. So, we see now how the first four arms of the match expression work.

13. In the last arm of the match expression, there is a String upon which the method into() is called. This method call must return a Cow<'static, str>

14. We can call the into() method on a String to get a Cow<str> without any lifetime limitations between the String and the Cow<str>

15. So the returned Cow<str> can be annotated with a static lifetime, without requiring that the String had to have one, too.

16. So, we see now how the fifth arm of the match expression works.