generics and keyword final

Question

Whu Do it is a non-valid construction

class A <T extends  String & Comparable<T>>{}

out:

java: java.lang.Comparable cannot be inherited with different arguments: <T> and <java.lang.String>

but it is valid

class A <T extends  Number & Comparable<T>>{}

I noiced that it is related with String is final but Number - not.

But T String is valid at first case I think. Why not?

You didn't get the first error message from the first piece of code. Did it really say `java.lang.Integer`? — user207421, Mar 11 '14 at 05:36
possible duplicate of [BaseFoo cannot be inherited with different arguments: > and >](http://stackoverflow.com/questions/19436415/basefoo-cannot-be-inherited-with-different-arguments-t-x-bart-and-t-x-foo) — Makoto, Mar 11 '14 at 05:37

score 6 · Accepted Answer · answered Mar 11 '14 at 05:37

6

The difference is, String class already implements Comparable<String>, while Number class doesn't. So, with that bound, T would be implementing both Comparable<String> and Comparable<T>, which is not allowed.

A class cannot extend from or implement different parameterized instantiation of a generic type.

answered Mar 11 '14 at 05:37

Rohit Jain

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just for information. more in details here: http://stackoverflow.com/questions/22281553/extends-from-generic-interfaces – gstackoverflow Mar 11 '14 at 10:54

generics and keyword final

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