Summary: | [1.5][compiler] Semantics for Generics not the same as for Sun | ||
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Product: | [Eclipse Project] JDT | Reporter: | Eric Goff <goffster> |
Component: | Core | Assignee: | Philipe Mulet <philippe_mulet> |
Status: | RESOLVED WORKSFORME | QA Contact: | |
Severity: | major | ||
Priority: | P3 | ||
Version: | 3.1 | ||
Target Milestone: | 3.1 M5 | ||
Hardware: | PC | ||
OS: | Linux | ||
Whiteboard: |
Description
Eric Goff
2004-11-27 15:04:39 EST
sorry, here are the errors Severity Description Resource In Folder Location Creation Time 2 Bound mismatch: The type View<?,?> is not a valid substitute for the bounded parameter <B extends PropertiedObject<B>> of the type PropertiedObject<B> View.java base/src/org/liberatis/sphaera/mvc line 45 November 27, 2004 1:21:22 PM Severity Description Resource In Folder Location Creation Time 2 The type View cannot extend or implement PropertiedObject<View<?,?>>. A supertype may not specify any wildcard View.java base/src/org/liberatis/sphaera/mvc line 45 November 27, 2004 1:21:22 PM Just thought I would add what I am trying to accomplish. It mightb e helpful. I have an abstract hierarchical object called PropertiedObject. It allows a generic type for every member in the tree. Every member of the tree extends the generic type (which also extends PropertiedObject) i.e. A tree of "Model"'s A tree of "View<?,?>"'s Both Model and View extend PropertiedObject. When I defined "View": View<T,U> extends PropertiedObject<View<?,?>>, the types for things like children, grandparents, parents, etc. are logically not known. So eclipse's error that View<?,?> must be qualifed is incorrect IMHO. cheers, Eric This was fixed a while ago. Only direct wildcards are complained against. Added GenericTypeTest#test462. |