Java - What is the output: Autoboxing/Unboxing and Method Overloading

Question

Suppose you have two methods in a class.

public void test(Integer iObject) {
        System.out.println("Integer=" + iObject);
}

public void test(int iValue) {
        System.out.println("int=" + iValue);
}

Suppose you make two calls to the test() method.

test(101);
test(new Integer(101));

Click to view the answer

int=101
Integer=101

Note

For primitive parameter value, for example, test(10);

Try to find a method with the primitive type argument. If there is no exact match, try widening the primitive type to find a match.
If the previous step fails, box the primitive type and try to find a match.

For reference parameter value, test(new Integer(101));

Try to find a method with the reference type argument. If there is match, call that method. A match does not have to be exact. It should follow the subtype and super type assignment rule.
If the previous step fails, unbox the reference type to the corresponding primitive type and try to find an exact match, or widen the primitive type and find a match.

Demo

public class Main {
  public static void main(String[] args) {
    test(101);/*  w  w  w .j  a  va 2 s  .  com*/
    test(new Integer(101));

  }

  public static void test(Integer iObject) {
    System.out.println("Integer=" + iObject);
  }

  public static void test(int iValue) {
    System.out.println("int=" + iValue);
  }
}

Java - What is the output: Autoboxing/Unboxing and Method Overloading

Question

Note

Demo

Result

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