Operator overloading in Java

java operator-overloading
Vipul picture Vipul · Nov 6, 2009 · Viewed 210.7k times · Source

Please can you tell me if it is possible to overload operators in Java? If it is used anywhere in Java could you please tell me about it.

Answer

Jon Skeet picture Jon Skeet · Nov 6, 2009

No, Java doesn't support user-defined operator overloading. The only aspect of Java which comes close to "custom" operator overloading is the handling of + for strings, which either results in compile-time concatenation of constants or execution-time concatenation using StringBuilder/StringBuffer. You can't define your own operators which act in the same way though.

For a Java-like (and JVM-based) language which does support operator overloading, you could look at Kotlin or Groovy. Alternatively, you might find luck with a Java compiler plugin solution.

Related questions

Why doesn't Java offer operator overloading?

Coming from C++ to Java, the obvious unanswered question is why didn't Java include operator overloading? Isn't Complex a, b, c; a = b + c; much simpler than Complex a, b, c; a = b.add(c);? Is there a known reason …

Read More
Compare two objects with "<" or ">" operators in Java

How to make two objects in Java comparable using "<" or ">" e.g. MyObject<String> obj1= new MyObject<String>(“blablabla”, 25); MyObject<String> obj2= new MyObject<String>(“nannaanana”, 17); if (obj1 > obj2) …

Read More
How do I convert a String to an int in Java?

How can I convert a String to an int in Java? My String contains only numbers, and I want to return the number it represents. For example, given the string "1234" the result should be the number 1234.

Read More