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Home/ Questions/Q 6233619
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T10:19:59+00:00 2026-05-24T10:19:59+00:00

Suppose I have a method called foo taking 2 Object as parameter. Both objects

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Suppose I have a method called foo taking 2 Object as parameter. Both objects are of the same type and both implements comparable interface.

void foo(Object first, Object second){

    if (!first.getClass().isInstance(second))   //first and second of the same type
        return;

    Comparable firstComparable = (Comparable)first;  //WARNING
    Comparable secondComparable = (Comparable)second;  //WARNING

    int diff = firstComparable.compareTo(secondComparable);  //WARNING
}

The first 2 warning are:

Comparable is a raw type. References to generic type Comparable
should be parameterized

The last warning:

Type safety: The method compareTo(Object) belongs to the raw type
Comparable. References to generic type Comparable should be
parameterized

How could I refactor my code in order to remove these warnings?

EDIT:
Can I do that without changing foo method’s signature?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-24T10:20:00+00:00Added an answer on May 24, 2026 at 10:20 am

    You have to tell the compiler that they are the same type and comparable. If you can’t change the signature you can add a method for backward compatibility.

    @SuppressWarnings("unchecked")
    static void foo(Object first, Object second) {
        foo((Comparable) first, (Comparable) second);
    }
    
    static <T extends Comparable<T>> void foo(T first, T second){
        int diff = first.compareTo(second); // no warning.
    }
    
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