[SOLVED] Java operator overloading

[tim]

Hello everyone.

I have an interesting question to ask. I have seen on a few forums that operator overloading is not possible in Java. So then, why can we use the following?

Code:

    Integer a = new Integer(1);
    Integer b = new Integer(2);
    Integer c = a + b;

I've tested this and it seems to work. Could this be due to Java's auto-boxing? In this case a, b and c are all instances of Integer. And the + means the addition of two Integer objects, which then returns an Integer object. Could it be that the objects, a and b, are converted to their primitive type equivalents, added and then converted back to an Integer object?

Just a strange thing I picked up. Thank you.

[Zosden]

I believe but could be wrong, but operator overloading refers like in C++ where you change what an operator does. In C++ I could make the + really mean multiply or concatenate two strings. Java has no such function other that what is already defined in java.

[tim]

Hello Zosden

Can you explain why my code snippet compiles and runs successfully?

Thank you.

[Zosden]

Because I do believe correct me if I'm wrong but Integer can be made into int form in java. They are almost the same to a programmer in many aspects.

[CaptainMorgan]

Based on the JavaDoc's first line:

Quote:

The Integer class wraps a value of the primitive type int in an object. An object of type Integer contains a single field whose type is int.

I would then deduce that a resolves to 1, b resolves to 2 and c resolves to the value of 1 + 2, or 3.

Code:

    Integer a = new Integer(1);
    Integer b = new Integer(2);
    Integer c = a + b;
  
    System.out.println(c);

Output:
3

[tim]

Thank you CaptainMorgan and Zosden