Bugzilla – Full Text Bug Listing
| Summary: | [1.5] parameterized inner type is illegal without parameterized outer type | ||
|---|---|---|---|
| Product: | [Eclipse Project] JDT | Reporter: | Markus Keller <markus.kell.r> |
| Component: | Core | Assignee: | Philipe Mulet <philippe_mulet> |
| Status: | VERIFIED FIXED | QA Contact: | |
| Severity: | minor | ||
| Priority: | P3 | CC: | martinae |
| Version: | 3.1 | ||
| Target Milestone: | 3.1 M7 | ||
| Hardware: | PC | ||
| OS: | Windows XP | ||
| Whiteboard: | |||
Tuned ParameterizedSingleTypeReference to better recognize invalid member type case (when generic). Also increased checking for non-static member type reference in static context. Added GenericTypeTest#test613-616+618. Fixed, will be released along with changes for bug 84496 Released Verified in I20050510-0010 |
I20050209-0800 A parameterized inner type is illegal without an explicit parameterized outer type: --------------------------------------------- package xy; import xy.Map.Entry; class Map<M> { class Entry<E> { } } class User { void a(Entry<String> e) { } // Entry<String> is illegal (eclipse accepts) void c(Map.Entry<String> e) { } // illegal (correctly flagged) void b(Entry e) { } // OK void d(Map<Integer>.Entry<String> e) { } // OK } --------------------------------------------- javac's opinion about "Entry<String>": improperly formed type, type parameters given on a raw type