How come > working as >?

Question

suppose we have

class Fruit implements Comparable<Fruit>
{
    @Override
    public int compareTo(Fruit o) { return 0; }
}

class Apple extends Fruit{}

class Main
{

    public static void main (String args[])
    {
        function(new Apple()); // this shouldn't work because Apple is implementing Comparable<Fruit>, not Comparable<Apple>
    }

    public static<T extends Comparable<T>> void function(T t){}
}

Code is working without any issue.

My question is why <T extends Comparable<T>> working like <T extends Comparable<? super T>>. whats the difference ?

Thanks.

[Edited] - A passage from book Image

Possible duplicate of [Difference between super T> and extends T> in Java](http://stackoverflow.com/questions/4343202/difference-between-super-t-and-extends-t-in-java) — henrybbosa, Nov 19 '16 at 08:45

henrybbosa · Answer 1 · 2016-11-19T09:20:02.190

0

In this case it is beacuse class Fruit is a superclass of itself

The difference is here

<T extends Comparable<T>> accepts Comparable with arguement of type T <T extends Comparable<? super T>> accepts Comparable with arguements of type T and its super classes. Here Since in java every class is a superclass of itself, Fruit is also a superclass of Fruit .Thats why you see like they are working the same way But if you used class Apple as T you will see the difference.

edited Nov 19 '16 at 09:20

answered Nov 19 '16 at 08:51

henrybbosa

1,083
10
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if this is the case then this should also work but it isn't. call : `function(new ArrayList());` --- protoype : `private static> void function(List t){}` it requires `>` as it should be. – belnxkkk Nov 19 '16 at 08:58
@Ibrahim In your case , the comment // this shouldn't work because Apple is implementing Comparable, not Comparable is not right. in this case Apple in implementing Comparable Because you are passing type Apple to function – henrybbosa Nov 19 '16 at 09:03
Okay but we know that Apple is a subclass of Fruit So whether it implements comparable even Apple will fit there beacause apple IS A FRUIT – henrybbosa Nov 19 '16 at 09:21
1

no in generics List is not a subclass of List. – belnxkkk Nov 19 '16 at 09:26
`f=a; // Error : Type mismatch: cannot convert from List to List` – belnxkkk Nov 19 '16 at 09:28
Thanks..Thats where super comes in – henrybbosa Nov 19 '16 at 09:29
yea exactly, but why java going against book. don't know. – belnxkkk Nov 19 '16 at 09:35
Yeah i think its java version issue. as well now – henrybbosa Nov 19 '16 at 09:38

davidxxx · Accepted Answer · 2016-11-19T09:51:13.270

0

You compile with Java 8 I think. Because, before Java 8, you should not pass the compilation and have this error :

Bound mismatch: The generic method function(T) of type Main is not applicable for the arguments (Apple). The inferred type Apple is not a valid substitute for the bounded parameter >

I think what you read in your book refers to generic use before Java 8.
But i didn't know that Java 8 had less constraint on this kind of case.
Is it a side-effect of the large use of inference in Java 8 that Java 8 calls the "improved inference" ?

edited Nov 19 '16 at 09:51

answered Nov 19 '16 at 09:36

davidxxx

104,693
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yes i was using java8. Just Compiled with jdk 1.6 and got compilatation error, while java8 compiles smoothly. – belnxkkk Nov 19 '16 at 09:53
It kinda seems to be java 8 bug because if we change function defination to – belnxkkk Nov 19 '16 at 09:54
`private static> void function(List t){}` – belnxkkk Nov 19 '16 at 09:55
and then call it like – belnxkkk Nov 19 '16 at 09:55
`function(new ArrayList());` – belnxkkk Nov 19 '16 at 09:55
Java 8 also gives compilation error, which seems right. Thanks for clearing. @davidxxx – belnxkkk Nov 19 '16 at 09:57
Just for your information, book is written for java 8. Core Java Volume 1 - Fundamentals 10th edition by Cay S. Horstmann – belnxkkk Nov 19 '16 at 10:03

How come > working as >?

2 Answers2