Aren't <U, T extends U> and <T, U super T> the same?

Question

I have a confusion in 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    
    }

Shouldn't both of the above be valid? With the analogy of If U is parent of T , then T is child of U. Then why does 2nd one gives compilation error?

EDIT:: I think , T extends T and T super T both are 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 can't bound a named generic with super. See also this stackoverflow posting.