Generic denifition T super Number

[Ram Reddy]

public class GenericClassDefinition<T super Number> {
public static void main(String...strings)
{

}

}


T extends Number :- Compiles fine
T super Number :- Giving error.

Why compiler is not allowed me to declare 'T' Type as T super Number?
I searched for similar topic here, not able to find
Can anyone provide me link if it is already discussed here?
Otherwise can anyone explain me it?

rami

[Dean Jones]

Hi Ram,

Lower bound for type parameters are not allowed. Replace super by extends it will work.

[Karl Prenton]

try this...

http://www.angelikalanger.com/GenericsFAQ/FAQSections/TypeParameters.html#Why%20is%20there%20no%20lower%20bound%20for%20type%20parameters?

Basically the compiler doesn't allow it because it's pointless (although I thought java had many instances where pointless and optional/flexible syntax go hand in hand); it implies an "unlimited" range!
[ June 12, 2008: Message edited by: Frank Zito ]

[Vishwanath Krishnamurthi]

Hi Ram,

I thought of something, for why a compiler doesn't
allow a <T super someType>... Its kind of vague though..

Considering,

public class Rental<T> {
public static void main(String...strings)
{
 
}
 
}

Here rental is a template... "Foreseeing" that the code would be just the same for a CarRental or a BikeRental, the programmer would've decided to go for a template like this ( so that it saves a lot of typing...)

If he wanted to allow all types of cars to use the same template he would say that by class Rental<T extends Car>... Thats all he might have wanted to do...

i.e while creating a format, a template, he does so only seeing the "type" or the "subtype" to which it can be applied to, so disallowing the superType looks quite logical...