I want to define a class that implements the generic Comparable interface. While in my class I also defined a generic type element T
. In order to implement the interface, I delegate the comparison to T
. Here is my code:
public class Item<T extends Comparable<T>> implements Comparable<Item> { private int s; private T t; public T getT() { return t; } @Override public int compareTo(Item o) { return getT().compareTo(o.getT()); } }
When I try to compile it, I get the following error information:
Item.java:11: error: method compareTo in interface Comparable<T#2> cannot be applied to given types; return getT().compareTo(o.getT()); ^ required: T#1 found: Comparable reason: actual argument Comparable cannot be converted to T#1 by method invocation conversion where T#1,T#2 are type-variables: T#1 extends Comparable<T#1> declared in class Item T#2 extends Object declared in interface Comparable 1 error
Can anybody tell me why and how to fix it?
Item
(without any type argument) is a raw type, so:
We could pass any kind of Item
to Item.compareTo
. For example, this would compile:
new Item<String>().compareTo(new Item<Integer>())
The method o.getT()
returns Comparable
instead of T
, which causes the compilation error.
In the example under the 1st point, after passing Item<Integer>
to Item.compareTo
, we would then erroneously pass an Integer
to String.compareTo
. The compilation error prevents us from writing the code which does that.
I think you just need to remove the raw types:
public class Item<T extends Comparable<T>> implements Comparable<Item<T>> { ... @Override public int compareTo(Item<T> o) { return getT().compareTo(o.getT()); } }
You're using raw types in your class definition (Item<T>
is generic, but you're omitting the type parameter <T>
), change it to:
class Item<T extends Comparable<T>> implements Comparable<Item<T>>
(Note the last <T>
)
The compareTo
method will then have to be changed as well:
public int compareTo(Item<T> o) { // again, use generics return getT().compareTo(o.getT()); }
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