Replace export wildcards _ with * in doc page

docs/docs/reference/other-new-features/export.md

@@ -23,7 +23,7 @@ class Copier:
    private val scanUnit = new Scanner
 
    export scanUnit.scan
-   export printUnit.{status => _, _}
+   export printUnit.{status => _, *}
 
    def status: List[String] = printUnit.status ++ scanUnit.status
 ```
@@ -58,7 +58,7 @@ of one of the following forms:
  - An _omitting selector_ `x => _` prevents `x` from being aliased by a subsequent
    wildcard selector.
  - A _given selector_ `given x` has an optional type bound `x`. It creates aliases for all eligible given instances that conform to either `x`, or `Any` if `x` is omitted, except for members that are named by a previous simple, renaming, or omitting selector.
- - A _wildcard selector_ `_` creates aliases for all eligible members of `path` except for given instances,
+ - A _wildcard selector_ `*` creates aliases for all eligible members of `path` except for given instances,
    synthetic members generated by the compiler and those members that are named by a previous simple, renaming, or omitting selector.
 
 A member is _eligible_ if all of the following holds:
@@ -144,15 +144,15 @@ Consider the following example:
 ```scala
 class B { val c: Int }
 object a { val b = new B }
-export a._
-export b._
+export a.*
+export b.*
 ```
 
-Is the `export b._` clause legal? If yes, what does it export? Is it equivalent to `export a.b._`? What about if we swap the last two clauses?
+Is the `export b.*` clause legal? If yes, what does it export? Is it equivalent to `export a.b.*`? What about if we swap the last two clauses?
 
 ```
-export b._
-export a._
+export b.*
+export a.*
 ```
 
 To avoid tricky questions like these, we fix the elaboration order of exports as follows.