Unexpand names of type lambda params

Otherwise we would print Foo[([+scala$Option$$A] => Option[+A])^] which
is ugly and not even coherent

Author: smarter

compiler/src/dotty/tools/dotc/printing/PlainPrinter.scala

@@ -174,7 +174,7 @@ class PlainPrinter(_ctx: Context) extends Printer {
       case tp: ExprType =>
         changePrec(GlobalPrec) { "=> " ~ toText(tp.resultType) }
       case tp: TypeLambda =>
-        def paramText(name: Name, bounds: TypeBounds): Text = name.toString ~ toText(bounds)
+        def paramText(name: Name, bounds: TypeBounds): Text = name.unexpandedName.toString ~ toText(bounds)
         changePrec(GlobalPrec) {
           "[" ~ Text((tp.paramNames, tp.paramInfos).zipped.map(paramText), ", ") ~
           "]" ~ lambdaHash(tp) ~ (" => " provided !tp.resultType.isInstanceOf[MethodType]) ~

tests/repl/i2492.check

@@ -4,5 +4,5 @@ scala> val s: Map {type Map$K =String;type Map$V = Int} = null
 -- [E055] Syntax Error: <console>:5:7 ------------------------------------------
 5 |val s: Map {type Map$K =String;type Map$V = Int} = null
   |       ^^^
-  |missing type parameter for [line1$object$$iw$$iw$Map$$K, line1$object$$iw$$iw$Map$$V] => Map[K, V]
+  |       missing type parameter for [K, V] => Map[K, V]
 scala> :quit

tests/repl/names.check

@@ -0,0 +1,5 @@
+scala> case class Foo[M[_]](x: M[Int]) 
+defined class Foo
+scala> Foo(Option(1))
+val res0: Foo[[+A] => Option[+A]] = Foo(Some(1))
+scala> :quit