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Build ID: I20070621-1340 Steps To Reproduce: 1. Open 'New Java Class' dialog 2. Set some Name 3. Set java.lang.String as a Superclass and click 'Finish' More information:
Suppose you have a(String str); and a(SomeInterface s); when you write a(new InterfaceImpl()); select new InterfaceImpl() and use quick assist 'create class InterfaceImpl()' - it will show a create class dialog with superclass field set to java.lang.String, which is wrong, because it should be interface SomeInterface, as String is final, so it is a call to a(SomeInterface s);
*** Bug 405796 has been marked as a duplicate of this bug. ***
This bug hasn't had any activity in quite some time. Maybe the problem got resolved, was a duplicate of something else, or became less pressing for some reason - or maybe it's still relevant but just hasn't been looked at yet. If you have further information on the current state of the bug, please add it. The information can be, for example, that the problem still occurs, that you still want the feature, that more information is needed, or that the bug is (for whatever reason) no longer relevant. -- The automated Eclipse Genie.
Still an issue in 4.12 M1.
Still occurs in 4.20 M1
(In reply to Marcin Kowalski from comment #0) > Build ID: I20070621-1340 > > Steps To Reproduce: > 1. Open 'New Java Class' dialog > 2. Set some Name > 3. Set java.lang.String as a Superclass and click 'Finish' > > > More information: This first scenario has been fixed. Using 2023-03.