Question

Java does not have concept of operator overloading.

Still + operator behaves as addition operator with numbers and concatenate operator with strings. This is similar to the operator overloading behavior.

So, does Java have operator overloading?

Answer 1

Does the Java language overload some operators?

YES! As you’ve found out, the operator + can mean two different things, string concatenation or numeric addition. This is, by definition, an operator overload.

Here’s the list of all Java operators:

JLS 3.12 Operators

The following 37 tokens are the operators, formed from ASCII characters:

  =     >     <     !     ~     ?      :
  ==    <=    >=    !=    &&    ||     ++     --
  +     -     *     /     &     |      ^      %     <<     >>    >>>
  +=    -=    *=    /=    &=    |=     ^=     %=    <<=    >>=   >>>=

Some of those operators are overloaded. Here are some examples:

System.out.println(   3 + 4 + "X"     ); // prints "7X"
System.out.println(   3 + (4 + "X")   ); // prints "34X"
System.out.println(   "X" + 3 + 4     ); // prints "X34"
System.out.println(   "X" + (3 + 4)   ); // prints "X7"

System.out.println(0 == 0);                           // prints "true"
System.out.println(new Integer(0) == new Integer(0)); // prints "false"

System.out.println(true & (1 & 2) == 12); // prints "false"

---

Can we overload the operators defined in the Java language?

ABSOLUTELY NOT! All Java operators mean exactly as specified by the language specification. There is no "extra-linguistic" semantics: a Java operator can NEVER do something that isn’t specified by the language.

That is, unless the language changes, the following are guaranteed truths:

• someString + whatever is ALWAYS string concatenation
• referenceType == anotherReferenceType is ALWAYS reference equality
• No funky things like 3 * "a lady" or "my heart" / 2 or even 10**3 ~= 999

As the above snippet shows, however, even the current state of operator overloading can still be quite confusing, especially for beginners. By not allowing extra-linguistic overloads, at least this confusion is limited: once a programmer learns about what all the operators in the Java language do in various overloaded scenarios, their exact semantics in all Java code becomes clear and precise.

Operator overloading can be quite confusing. Some think that it’s "bad" enough as it is. To allow users to overload the Java operators to do something outside the language specification can only lead to even more confusion.

Here’s an excerpt from Java Puzzlers, Puzzle 30: Son of Looper:

The lesson for language designers is the same as [two other puzzles]. Operator overloading can be confusing. Perhaps the + operator should not have been overloaded for string concatenation. It may well be worth providing a string concatenation operator, but it doesn’t have to be +.

---

Do you need C++ to support operator overloading in Java?

NOPE! This has nothing to do with it at all. All that the Java compiler needs to do is parse the program source code according to the grammatical rules of the language, and determine, for each operator, what the types of the operands are. This information is enough to deduce what the meaning of the operator is, and to then act accordingly as specified by the language.

---

Appendix

JLS References

• 15.18 Additive Operators
  • 15.18.1 String Concatenation Operator +
  • 15.18.2 Additive Operators (+ and -) for Numeric Types
• 15.21 Equality Operators
  • 15.21.1 Numerical Equality Operators == and !=
  • 15.21.2 Boolean Equality Operators == and !=
  • 15.21.3 Reference Equality Operators == and !=
• 15.22 Bitwise and Logical Operators
  • 15.22.1 Integer Bitwise Operators &, ^, and |
  • 15.22.2 Boolean Logical Operators &, ^, and |

Revealing questions

• why does a char + another char = a weird number
• Is it guaranteed that new Integer(i) == i in Java?
• When comparing two Integers in Java does auto-unboxing occur?
• Java String.equals versus ==