Can you define a generic bound that both has lower and upper bounds?

Question

Is it possible to define a generic bound that:

• implements an interface SomeInterface
• is a superclass of some class MyClass

Something like:

Collection<? extends SomeInterface & super MyClass> c; // doesn't compile

Answer 1

According to the spec , the answer would be no (you can have super or extends , but not both):

 TypeArguments: \n    < TypeArgumentList > \n\nTypeArgumentList:  \n    TypeArgument \n    TypeArgumentList , TypeArgument \n\nTypeArgument: \n    ReferenceType \n    Wildcard \n\nWildcard: \n    ?  WildcardBounds opt \n\nWildcardBounds: \n    extends ReferenceType \n    super ReferenceType  

Answer 2

You cannot use generic type ( T in your case) with bounds when declaring a variable.

It should be either a wildcard ( ? ), or just use the full generic type of the class.

Eg

// Here only extends is allowed
class My< T extends SomeInterface >
{

  // If using T, then no bounds are allowed
  private Collection<T> var1;

  private Collection< ? extends SomeInterface > var2;

  // Cannot have extends and super on the same wildcard declaration
  private Collection< ? super MyClass > var3;

  // You can use T as a bound for wildcard
  private Collection< ? super T > var4;

  private Collection< ? extends T > var5;

}

In some cases, you may tighten the declaration by adding extra generic parameter to a class (or method) and adding a bound on that particular parameter:

class My <
  T extends MyClass< I >,
  I extends SomeInterface 
>
{
}