Fix regression with runnableExamples in generic expr

The examples should not enter the generic analysis at all.
The regression was introduced in 4cf704bb as a fix for #8694.

Fixes #9130
Fixes #8694

tests/generics/t8694.nim

@@ -0,0 +1,31 @@
+discard """
+  output: '''
+true
+true
+true
+'''
+"""
+
+when true:
+  # Error: undeclared identifier: '|'
+  proc bar[T](t:T): bool =
+    runnableExamples:
+      type Foo = int | float
+    true
+  echo bar(0)
+
+when true:
+  # ok
+  proc bar(t:int): bool =
+    runnableExamples:
+      type Foo = int | float
+    true
+  echo bar(0)
+
+when true:
+  # Error: undeclared identifier: '|'
+  proc bar(t:typedesc): bool =
+    runnableExamples:
+      type Foo = int | float
+    true
+  echo bar(int)

tests/generics/t9130.nim

@@ -0,0 +1,33 @@
+when true:
+  # stack overflow
+  template baz1*(iter: untyped): untyped =
+    runnableExamples:
+      import sugar
+      proc fun(a: proc(x:int): int) = discard
+      baz1(fun(x:int => x))
+    discard
+
+  proc foo1[A](ts: A) =
+    baz1(ts)
+
+when true:
+  # ok
+  template baz2*(iter: untyped): untyped =
+    runnableExamples:
+      import sugar
+      proc fun(a: proc(x:int): int) = discard
+      baz2(fun(x:int => x))
+    discard
+
+  proc foo2(ts: int) =
+    baz2(ts)
+
+when true:
+  # stack overflow
+  template baz3*(iter: untyped): untyped =
+    runnableExamples:
+      baz3(fun(x:int => x))
+    discard
+
+  proc foo3[A](ts: A) =
+    baz3(ts)