How does + operator behaves differently with numbers and strings in Java?

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

It's basically operator overloading - just built into the language.

"Java does not have concept of operator overloading" is only true inasmuch developers cannot overload operators.

The language spec can, and strictly speaking, all the arithmetic operators are overloaded to handle calculations that involve more than one numerical type. And even there, it sometimes creates confusion (such as having to cast one operand to double if you want a divsion of int values to yield fractional results).

Answer 2

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"

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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 +.

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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.

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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 ==

Answer 3

The + operator IS overloaded. Java just prevents YOU from overloading it yourself.

Answer 4

Concat operator is a special support provided in Java. A quote from Javadoc below.

The Java language provides special support for the string concatenation operator ( + ), and for conversion of other objects to strings. String concatenation is implemented through the StringBuffer class and its append method. String conversions are implemented through the method toString, defined by Object and inherited by all classes in Java. For additional information on string concatenation and conversion, see Gosling, Joy, and Steele, The Java Language Specification.

For information, see this

Answer 5

The polymorphic treatment of operators in Java is a form of operator overloading, as that term is traditionally used. For example, read the wikipedia page on operator overloading, and you will see Java style of overloading mentioned in the first paragraph.

"In computer programming, operator overloading (less commonly known as operator ad-hoc polymorphism) is a specific case of polymorphism in which some or all of operators like +, =, or == have different implementations depending on the types of their arguments. Sometimes the overloadings are defined by the language; sometimes the programmer can implement support for new types."

Answer 6

The functionality you are talking about is just built-in in Java. You still don't have control on operator overloading.

Answer 7

Java does not allow programmers to overload operators, but has built-in support for operators on some of its built-in types (I think String is the only object with support for any operator, other than auto-boxing stuff).

Answer 8

There is no operator overloading in Java, in the sense that the programmer can't overload any operator himself. However, the Java language has some kind of integrated operator overloading, which assigns different behaviors to the + operator, depending on the context.

The operator gets compiled into different bytecode instructions, depending on the operands:

For String "addition", a StringBuilder is created and used behind the scenes

For int addition, the JVM command iadd is used.

For double addition, the JVM command dadd is used.

etc...

Answer 9

I feel this behaves more like implicit conversion of one type to another than like operator overloading.