Fix challenge 23

src/challenges/challenge-23.md

@@ -54,29 +54,44 @@ C. The program is guaranteed to output 11112.
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 Answer
-C. The program is guaranteed to output: 111222
-
-The Reference describes Rust's method lookup order. The relevant paragraph is:
-
-Obtain [the candidate receiver type] by repeatedly dereferencing the receiver expression's type, adding each type encountered to the list, then finally attempting an unsized coercion at the end, and adding the result type if that is successful. 
-
-Then, for each candidate T, add &T and &mut T to the list immediately after T.
-
-Applying these rules to the given examples, we have:
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-t.f(): We try to find a function f defined on the type T, but there is none. Next, we search the type &T, and find the first implementation of the Or trait, and we are done. Upon invocation, the resolved call prints 1.
-
-wt.f(): We search for a function f defined on &T, which immediately succeeds. Upon invocation, the function prints 1.
-
-wwt.f(): The search order is &&T -> &&&T -> &mut &&T -> &T, and we're done. Upon invocation, the function prints 1.
-
-wwwt.f(): &&&T -> &&&&T. This prints 2.
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-wwwwt.f(): &&&&T. This prints 2.
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-wwwwwt.f(): &&&&&T -> &&&&&&T -> &mut &&&&&T -> &&&&T. This prints 2.
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-The challenge and solution is by David Tolnay.
+C. The program is guaranteed to output: 11112
+
+To understand Rust's function lookup order we will start by analyzing `&&&&T`:
+
+1. Rust will build a deref chain: `&&&&T` -> `&&&T` -> `&&T` -> `&T` -> `T`.
+2. For each step `U` in the chain it will try to find an implemented function in this order:
+    - `U` (direct-match)
+    - `&U` (auto-ref)
+    - `&mut U` (auto-ref-mut)
+
+Let's walk through the program:
+
+`T`:
+- Deref Chain: `T`
+- Step `T`:
+    - `U` = `T` -> no
+    - `&U` = `&T` -> yes (print 1)
+
+`&T`:
+- Deref Chain: `&T` -> `T`
+- Step `&T`:
+    - `U` = `&T` -> yes (print 1)
+
+`&&T`: 
+- Deref Chain: `&&T` -> `&T` -> `T`
+- Step `&T`:
+    - `U` = `&T` -> yes (print 1)
+
+`&&&T`:
+- Deref Chain: `&&&T` -> `&&T` -> `&T` -> `T`
+- Step `&T`:
+    - `U` = `&T` -> yes (print 1)
+
+`&&&&T`:
+- Deref Chain: `&&&&T` -> `&&&T` -> `&&T` -> `&T` -> `T`
+- Step `&&&&T`:
+  - `U` = `&&&&T` -> no
+  - `&U` = `&&&&&T` -> yes (print 2)
+
+The challenge is originally by David Tolnay.
 </details>
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