Str.subst: Add examples for nth(*), etc.

doc/Type/Str.pod6

@@ -823,14 +823,23 @@ regex, not the C<subst> method call.
 Here are other examples of usage:
 
     my $str = "Hey foo foo foo";
-    $str.subst(/foo/, "bar", :g); # global substitution - returns Hey bar bar bar
 
-    $str.subst(/\s+/, :g); # global substitution - returns Heyfoofoofoo
-
-    $str.subst(/foo/, "no subst", :x(0)); # targeted substitution. Number of times to substitute. Returns back unmodified.
-    $str.subst(/foo/, "bar", :x(1)); #replace just the first occurrence.
-
-    $str.subst(/foo/, "bar", :nth(3)); # replace nth match alone. Replaces the third foo. Returns Hey foo foo bar
+    say $str.subst(/foo/, "bar", :g);           # OUTPUT: «Hey foo foo foo␤»
+    say $str.subst(/\s+/, :g);                  # OUTPUT: «Hey foo foo foo␤»
+
+    say $str.subst(/foo/, "bar", :x(0));        # OUTPUT: «Hey foo foo foo␤»
+    say $str.subst(/foo/, "bar", :x(1));        # OUTPUT: «Hey bar foo foo␤»
+    # Can not match 4 times, so no substitutions made
+    say $str.subst(/foo/, "bar", :x(4));        # OUTPUT: «Hey foo foo foo␤»
+    say $str.subst(/foo/, "bar", :x(2..4));     # OUTPUT: «Hey bar bar bar␤»
+    # Replace all of them, identical to :g
+    say $str.subst(/foo/, "bar", :x(*));        # OUTPUT: «Hey bar bar bar␤»
+
+    say $str.subst(/foo/, "bar", :nth(3));      # OUTPUT: «Hey foo foo bar␤»
+    # Replace last match
+    say $str.subst(/foo/, "bar", :nth(*));      # OUTPUT: «Hey foo foo bar␤»
+    # Replace next-to-last last match
+    say $str.subst(/foo/, "bar", :nth(*-1));    # OUTPUT: «Hey foo bar foo␤»
 
 The C<:nth> adverb has readable English-looking variants: