CODEWITHSUNDEEP
javagenericscollectionstype-parameter
Priyank Doshi
Priyank Doshi

Reputation: 13151

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

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 ?

Upvotes: 8

Views: 422

Answers (2)

assylias
assylias

Reputation: 328598

  • 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

Upvotes: 7

Tim Lamballais
Tim Lamballais

Reputation: 1054

You can't bound a named generic with super. See also this stackoverflow posting.

Upvotes: 2

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