Problems in casting variables

Question

I am trying to construct an object using the following constructor in Java:

public UnaryNumberContract(Comparable<Number> expectedValue, Operations operator, Invariant inv) {
  super(inv, operator);
  this.expectedValue = expectedValue;
}

I am passing on the following parameters:

Comparable<Number> zero = new Integer(0); 
UnaryNumberContract contract = new UnaryNumberContract(zero, Operations.NOT_EQUALS, (Invariant) inv);

And I keep getting a Type mismatch: cannot convert from Integer to Comparable<Number>

I've checked and Number is a super class of Integer, so I don't quite understand what is goinh wrong here.

Possible duplicate of http://stackoverflow.com/questions/2112343/java-integer-obj-cant-cast-to-comparable — rid, Jun 27 '11 at 18:30
Also possible duplicate of http://stackoverflow.com/questions/480632/why-doesnt-java-lang-number-implement-comparable — Matten, Jun 27 '11 at 18:31

score 3 · Accepted Answer · edited May 23 '17 at 10:08

3

If you replace references to Comparable<Number> with Comparable<? extends Number>, that will solve the issue.

Also, see this When do Java generics require <? extends T> instead of <T> and is there any downside of switching?

edited May 23 '17 at 10:08

Community

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answered Jun 27 '11 at 18:38

Darth Ninja

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I was avoiding generics, but went with it anyway. Thanks. – Tiago Veloso Jun 27 '11 at 18:52

score 2 · Answer 2 · answered Jun 27 '11 at 18:31

2

The problem is that Integer does not implement Comparable<Number>, it implements Comparable<Integer>.

Comparable<Integer> does not inherit from Comparable<Number> and this is why it's not possible to cast it.

What you can do is use Number for your parameter type. Integer does inherit Number and you will not have the same problem.

answered Jun 27 '11 at 18:31

Marcelo

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@Tiago I included a workaround. Use `Number` instead of `Comparable` for your parameter type. – Marcelo Jun 27 '11 at 18:32
It isn't really a problem, you are correctly getting an error based on how the Integer class is defined. – Hunter McMillen Jun 27 '11 at 18:34
Thainks. I can cast `Number` to `Comparable` within the constructor then? – Tiago Veloso Jun 27 '11 at 18:37
@Tiago: No: Number doesn't implement Comparable (see http://stackoverflow.com/q/480632/771837 for a fairly good rationale). Nitin's answer to your question should work for you. – trutheality Jun 27 '11 at 18:47

score 0 · Answer 3 · 2011-06-27T18:35:40.643

0

That is because Integer doesn't implement Comparable<Number> it is an implements Comparable<Integer>.

All Implemented Interfaces: Serializable, Comparable<Integer>

Relevant Integer JavaDoc.

edited Jun 27 '11 at 18:35

answered Jun 27 '11 at 18:30

Problems in casting variables

3 Answers3