Not <U, T extends U> and <T, U super T> the same?

Question

I have confusion in the following two method declarations:

private <U, T extends U> T funWorks(T child, U parent) { // No compilation errors } private <T, U super T> T funNotWorks(T child, U parent) { // compilation errors } 

Should both of the above be valid? By analogy with If U is the parent of T, then T is a child of U. Then why does the second give a compilation error?

EDIT :: I think T extends T and T super T both valid. right?

Answer 1

• Type parameters (your example) can only use extends ( JLS # 4.4 ):

 TypeParameter: TypeVariable TypeBoundopt TypeBound: extends TypeVariable extends ClassOrInterfaceType AdditionalBoundListopt AdditionalBoundList: AdditionalBound AdditionalBoundList AdditionalBound AdditionalBound: & InterfaceType 

• Wildcards can use either extends or super ( JLS # 4.5.1 ):

 TypeArguments: < TypeArgumentList > TypeArgumentList: TypeArgument TypeArgumentList , TypeArgument TypeArgument: ReferenceType Wildcard Wildcard: ? WildcardBoundsopt WildcardBounds: extends ReferenceType super ReferenceType 

Answer 2

You cannot associate named generic with super. See also this stackoverflow posting.